A 'C' Test: The 0x10 Best Questions for Would-be Embedded Programmers（想成为嵌入式程序员应知道的0x10个基本问题）

http://www.embedded.com/2000/0005/0005feat2.htm

 

Nigel Jones

 

Pencils up, everyone. Here's a test to identify potential embedded programmers or embedded programmers with potential

 

A n obligatory and significant part of the recruitment process for embedded systems programmers seems to be the "C test." Over the years, I have had to both take and prepare such tests and, in doing so, have realized that these tests can be informative for both the interviewer and interviewee. Furthermore, when given outside the pressure of an interview situation, these tests can also be quite entertaining.

From the interviewee's perspective, you can learn a lot about the person who has written or administered the test. Is the test designed to show off the writer's knowledge of the minutiae of the ANSI standard rather than to test practical know-how? Does it test ludicrous knowledge, such as the ASCII values of certain characters? Are the questions heavily slanted towards your knowledge of system calls and memory allocation strategies, indicating that the writer may spend his time programming computers instead of embedded systems? If any of these are true, then I know I would seriously doubt whether I want the job in question.

From the interviewer's perspective, a test can reveal several things about the candidate. Primarily, you can determine the level of the candidate's knowledge of C. However, it's also interesting to see how the person responds to questions to which they don't know the answers. Do they make intelligent choices backed up with good intuition, or do they just guess? Are they defensive when they are stumped, or do they exhibit a real curiosity about the problem and see it as an opportunity to learn something? I find this information as useful as their raw performance on the test.

With these ideas in mind, I have attempted to construct a test that is heavily slanted towards the requirements of embedded systems. This is a lousy test to give to someone seeking a job writing compilers! The questions are almost all drawn from situations I have encountered over the years. Some of them are tough; however, they should all be informative.

This test may be given to a wide range of candidates. Most entry-level applicants will do poorly on this test, while seasoned veterans should do very well. Points are not assigned to each question, as this tends to arbitrarily weight certain questions. However, if you choose to adapt this test for your own uses, feel free to assign scores.

Preprocessor 

1. Using the #define statement, how would you declare a manifest constant that returns the number of seconds in a year? Disregard leap years in your answer.

#define SECONDS_PER_YEAR (60 * 60 * 24 * 365)UL

I'm looking for several things here:

• Basic knowledge of the #define syntax (for example, no semi-colon at the end, the need to parenthesize, and so on)
• An understanding that the pre-processor will evaluate constant expressions for you. Thus, it is clearer, and penalty-free, to spell out how you are calculating the number of seconds in a year, rather than actually doing the calculation yourself
• A realization that the expression will overflow an integer argument on a 16-bit machine-hence the need for the L, telling the compiler to treat the variable as a Long
• As a bonus, if you modified the expression with a UL (indicating unsigned long), then you are off to a great start. And remember, first impressions count!

2. Write the "standard" MIN macro-that is, a macro that takes two arguments and returns the smaller of the two arguments.

#define MIN(A,B) ((A) <= (B) ? (A) : (B))

The purpose of this question is to test the following:

• Basic knowledge of the #define directive as used in macros. This is important because until the inline operator becomes part of standard C, macros are the only portable way of generating inline code. Inline code is often necessary in embedded systems in order to achieve the required performance level
• Knowledge of the ternary conditional operator. This operator exists in C because it allows the compiler to produce more optimal code than an if-then-else sequence. Given that performance is normally an issue in embedded systems, knowledge and use of this construct is important
• Understanding of the need to very carefully parenthesize arguments to macros
• I also use this question to start a discussion on the side effects of macros, for example, what happens when you write code such as:

 

least = MIN(*p++, b);

3. What is the purpose of the preprocessor directive #error?

Either you know the answer to this, or you don't. If you don't, see Reference 1. This question is useful for differentiating between normal folks and the nerds. Only the nerds actually read the appendices of C textbooks to find out about such things. Of course, if you aren't looking for a nerd, the candidate better hope she doesn't know the answer.

Infinite loops 
4. Infinite loops often arise in embedded systems. How does you code an infinite loop in C?

There are several solutions to this question. My preferred solution is:

while(1) { } 

Many programmers seem to prefer:

for(;;) { }

This construct puzzles me because the syntax doesn't exactly spell out what's going on. Thus, if a candidate gives this as a solution, I'll use it as an opportunity to explore their rationale for doing so. If their answer is basically, "I was taught to do it this way and I haven't thought about it since," it tells me something (bad) about them.

A third solution is to use a goto :

Loop: ... goto Loop; 

Candidates who propose this are either assembly language programmers (which is probably good), or else they are closet BASIC/FORTRAN programmers looking to get into a new field.

Data declarations 

5. Using the variable a, give definitions for the following: 
a) An integer 
b) A pointer to an integer 
c) A pointer to a pointer to an integer 
d) An array of 10 integers 
e) An array of 10 pointers to integers 
f) A pointer to an array of 10 integers 
g) A pointer to a function that takes an integer as an argument and returns an integer 
h) An array of ten pointers to functions that take an integer argument and return an integer

The answers are: 

a) int a; // An integer 
b) int *a; // A pointer to an integer 
c) int **a; // A pointer to a pointer to an integer 
d) int a[10]; // An array of 10 integers 
e) int *a[10]; // An array of 10 pointers to integers 
f) int (*a)[10]; // A pointer to an array of 10 integers 
g) int (*a)(int); // A pointer to a function a that takes an integer argument and returns an integer 
h) int (*a[10])(int); // An array of 10 pointers to functions that take an integer argument and return an integer

People often claim that a couple of these are the sorts of thing that one looks up in textbooks-and I agree. While writing this article, I consulted textbooks to ensure the syntax was correct. However, I expect to be asked this question (or something close to it) when I'm being interviewed. Consequently, I make sure I know the answers, at least for the few hours of the interview. Candidates who don't know all the answers (or at least most of them) are simply unprepared for the interview. If they can't be prepared for the interview, what will they be prepared for?

Static 

6. What are the uses of the keyword static?

This simple question is rarely answered completely. Static has three distinct uses in C:

• A variable declared static within the body of a function maintains its value between function invocations
• A variable declared static within a module, (but outside the body of a function) is accessible by all functions within that module. It is not accessible by functions within any other module. That is, it is a localized global
• Functions declared static within a module may only be called by other functions within that module. That is, the scope of the function is localized to the module within which it is declared

Most candidates get the first part correct. A reasonable number get the second part correct, while a pitiful number understand the third answer. This is a serious weakness in a candidate, since he obviously doesn't understand the importance and benefits of localizing the scope of both data and code.

Const 
7. What does the keyword const mean?

As soon as the interviewee says "const means constant," I know I'm dealing with an amateur. Dan Saks has exhaustively covered const in the last year, such that every reader of ESP should be extremely familiar with what const can and cannot do for you. If you haven't been reading that column, suffice it to say that constmeans "read-only." Although this answer doesn't really do the subject justice, I'd accept it as a correct answer. (If you want the detailed answer, read Saks' columns-carefully!)

If the candidate gets the answer correct, I'll ask him these supplemental questions:

What do the following declarations mean?

const int a; int const a; const int *a; int * const a; int const * a const;

The first two mean the same thing, namely a is a const (read-only) integer. The third means a is a pointer to a const integer (that is, the integer isn't modifiable, but the pointer is). The fourth declares a to be a const pointer to an integer (that is, the integer pointed to by a is modifiable, but the pointer is not). The final declaration declares a to be a const pointer to a const integer (that is, neither the integer pointed to by a, nor the pointer itself may be modified). If the candidate correctly answers these questions, I'll be impressed. Incidentally, you might wonder why I put so much emphasis on const, since it is easy to write a correctly functioning program without ever using it. I have several reasons:

• The use of const conveys some very useful information to someone reading your code. In effect, declaring a parameter const tells the user about its intended usage. If you spend a lot of time cleaning up the mess left by other people, you'll quickly learn to appreciate this extra piece of information. (Of course, programmers who useconst , rarely leave a mess for others to clean up.)
• const has the potential for generating tighter code by giving the optimizer some additional information
• Code that uses const liberally is inherently protected by the compiler against inadvertent coding constructs that result in parameters being changed that should not be. In short, they tend to have fewer bugs

Volatile 
8. What does the keyword volatile mean? Give three different examples of its use.

A volatile variable is one that can change unexpectedly. Consequently, the compiler can make no assumptions about the value of the variable. In particular, the optimizer must be careful to reload the variable every time it is used instead of holding a copy in a register. Examples of volatile variables are:

• Hardware registers in peripherals (for example, status registers)
• Non-automatic variables referenced within an interrupt service routine
• Variables shared by multiple tasks in a multi-threaded application

Candidates who don't know the answer to this question aren't hired. I consider this the most fundamental question that distinguishes between a C programmer and an embedded systems programmer. Embedded folks deal with hardware, interrupts, RTOSes, and the like. All of these require volatile variables. Failure to understand the concept of volatile will lead to disaster.

On the (dubious) assumption that the interviewee gets this question correct, I like to probe a little deeper to see if they really understand the full significance of volatile . In particular, I'll ask them the following additional questions:

• Can a parameter be both const and volatile ? Explain.
• Can a pointer be volatile ? Explain.
• What's wrong with the following function?:

int square(volatile int *ptr) { return *ptr * *ptr; }

The answers are as follows:

• Yes. An example is a read-only status register. It is volatile because it can change unexpectedly. It is const because the program should not attempt to modify it
• Yes, although this is not very common. An example is when an interrupt service routine modifies a pointer to a buffer
• This one is wicked. The intent of the code is to return the square of the value pointed to by *ptr . However, since *ptr points to a volatile parameter, the compiler will generate code that looks something like this:

 

int square(volatile int *ptr) { int a,b; a = *ptr; b = *ptr; return a * b; }

Because it's possible for the value of *ptr to change unexpectedly, it is possible for a and b to be different. Consequently, this code could return a number that is not a square! The correct way to code this is:

long square(volatile int *ptr) { int a; a = *ptr; return a * a; } 

Bit manipulation 
9. Embedded systems always require the user to manipulate bits in registers or variables. Given an integer variable a, write two code fragments. The first should set bit 3 of a. The second should clear bit 3 of a. In both cases, the remaining bits should be unmodified.

These are the three basic responses to this question:

• No idea. The interviewee cannot have done any embedded systems work
• Use bit fields. Bit fields are right up there with trigraphs as the most brain-dead portion of C. Bit fields are inherently non-portable across compilers, and as such guarantee that your code is not reusable. I recently had the misfortune to look at a driver written by Infineon for one of their more complex communications chips. It used bit fields and was completely useless because my compiler implemented the bit fields the other way around. The moral: never let a non-embedded person anywhere near a real piece of hardware!
• Use #defines and bit masks. This is a highly portable method and is the one that should be used. My optimal solution to this problem would be:

#define BIT3 (0x1 << 3)  static int a;

void set_bit3(void) { a |= BIT3; } void clear_bit3(void) { a &= ~BIT3; } 

Some people prefer to define a mask together with manifest constants for the set and clear values. This is also acceptable. The element that I'm looking for is the use of manifest constants, together with the |= and &= ~ constructs

Accessing fixed memory locations 
10. Embedded systems are often characterized by requiring the programmer to access a specific memory location. On a certain project it is required to set an integer variable at the absolute address 0x67a9 to the value 0xaa55. The compiler is a pure ANSI compiler. Write code to accomplish this task.

This problem tests whether you know that it is legal to typecast an integer to a pointer in order to access an absolute location. The exact syntax varies depending upon one's style. However, I would typically be looking for something like this:

int *ptr; ptr = (int *)0x67a9; *ptr = 0xaa55;

A more obscure approach is:

*(int * const)(0x67a9) = 0xaa55;

Even if your taste runs more to the second solution, I suggest the first solution when you are in an interview situation.

Interrupts 
11. Interrupts are an important part of embedded systems. Consequently, many compiler vendors offer an extension to standard C to support interrupts. Typically, this new keyword is __interrupt. The following code uses __interrupt to define an interrupt service routine (ISR). Comment on the code.

__interrupt double compute_area(double radius) { double area = PI * radius * radius; printf("/nArea = %f", area); return area; }

This function has so much wrong with it, it's hard to know where to start:

• ISRs cannot return a value. If you don't understand this, you aren't hired
• ISRs cannot be passed parameters. See the first item for your employment prospects if you missed this
• On many processors/compilers, floating-point operations are not necessarily re-entrant. In some cases one needs to stack additional registers. In other cases, one simply cannot do floating point in an ISR. Furthermore, given that a general rule of thumb is that ISRs should be short and sweet, one wonders about the wisdom of doing floating-point math here
• In a vein similar to the third point, printf() often has problems with reentrancy and performance. If you missed points three and four, I wouldn't be too hard on you. Needless to say, if you got these last two points, your employment prospects are looking better and better

Code examples 
12. What does the following code output and why?

void foo(void)

{

    unsigned int a = 6;

    int b = -20;

   (a+b > 6) ? puts("> 6") : puts("<= 6");

}

This question tests whether you understand the integer promotion rules in C-an area that I find is very poorly understood by many developers. Anyway, the answer is that this outputs "> 6." The reason for this is that expressions involving signed and unsigned types have all operands promoted to unsigned types. Thus -20 becomes a very large positive integer and the expression evaluates to greater than 6. This is a very important point in embedded systems where unsigned data types should be used frequently (see Reference 2). If you get this one wrong, you are perilously close to not getting the job.

13. Comment on the following code fragment.

unsigned int zero = 0; unsigned int compzero = 0xFFFF; /*1's complement of zero */

On machines where an int is not 16 bits, this will be incorrect. It should be coded:

unsigned int compzero = ~0;

This question really gets to whether the candidate understands the importance of word length on a computer. In my experience, good embedded programmers are critically aware of the underlying hardware and its limitations, whereas computer programmers tend to dismiss the hardware as a necessary annoyance.

By this stage, candidates are either completely demoralized-or they're on a roll and having a good time. If it's obvious that the candidate isn't very good, then the test is terminated at this point. However, if the candidate is doing well, then I throw in these supplemental questions. These questions are hard, and I expect that only the very best candidates will do well on them. In posing these questions, I'm looking more at the way the candidate tackles the problems, rather than the answers. Anyway, have fun...

Dynamic memory allocation 

14. Although not as common as in non-embedded computers, embedded systems do still dynamically allocate memory from the heap. What are the problems with dynamic memory allocation in embedded systems?

Here, I expect the user to mention memory fragmentation, problems with garbage collection, variable execution time, and so on. This topic has been covered extensively in ESP, mainly by P.J. Plauger. His explanations are far more insightful than anything I could offer here, so go and read those back issues! Having lulled the candidate into a sense of false security, I then offer up this tidbit:

What does the following code fragment output and why?

char *ptr; if ((ptr = (char *)malloc(0)) == NULL) puts("Got a null pointer"); else puts("Got a valid pointer");

This is a fun question. I stumbled across this only recently when a colleague of mine inadvertently passed a value of 0 to malloc and got back a valid pointer! That is, the above code will output "Got a valid pointer." I use this to start a discussion on whether the interviewee thinks this is the correct thing for the library routine to do. Getting the right answer here is not nearly as important as the way you approach the problem and the rationale for your decision.

Typedef 
15. Typedef is frequently used in C to declare synonyms for pre-existing data types. It is also possible to use the preprocessor to do something similar. For instance, consider the following code fragment:

#define dPS struct s * typedef struct s * tPS;
  

The intent in both cases is to define dPS and tPS to be pointers to structure s. Which method, if any, is preferred and why?

This is a very subtle question, and anyone who gets it right (for the right reason) is to be congratulated or condemned ("get a life" springs to mind). The answer is the typedef is preferred. Consider the declarations:

dPS p1,p2; tPS p3,p4; 

The first expands to:

struct s * p1, p2;
 

which defines p1 to be a pointer to the structure and p2to be an actual structure, which is probably not what you wanted. The second example correctly defines p3 and p4to be pointers.

Obscure syntax 
16. C allows some appalling constructs. Is this construct legal, and if so what does this code do?

int a = 5, b = 7, c; c = a+++b;

This question is intended to be a lighthearted end to the quiz, as, believe it or not, this is perfectly legal syntax. The question is how does the compiler treat it? Those poor compiler writers actually debated this issue, and came up with the "maximum munch" rule, which stipulates that the compiler should bite off as big (and legal) a chunk as it can. Hence, this code is treated as:

c = a++ + b;

Thus, after this code is executed, a = 6, b = 7, and c = 12.

If you knew the answer, or guessed correctly, well done. If you didn't know the answer then I wouldn't consider this to be a problem. I find the greatest benefit of this question is that it is good for stimulating questions on coding styles, the value of code reviews, and the benefits of using lint.

Well folks, there you have it. That was my version of the C test. I hope you had as much fun taking it as I had writing it. If you think the test is a good test, then by all means use it in your recruitment. Who knows, I may get lucky in a year or two and end up being on the receiving end of my own work.

Nigel Jones is a consultant living in Maryland. When not underwater, he can be found slaving away on a diverse range of embedded projects. He enjoys hearing from readers and can be reached at [email protected].

References

• Jones, Nigel, "In Praise of the #error directive," Embedded Systems Programming, September 1999, p. 114.
• Jones, Nigel, " Efficient C Code for Eight-bit MCUs ," Embedded Systems Programming, November 1998, p. 66.
• 
  

 

参考译文：http://topic.csdn.net/u/20080726/22/a20815b5-eb2c-4943-a10d-e7695fa30b87.html

C语言测试是招聘嵌入式系统程序员过程中必须而且有效的方法。这些年，我既参加也组织了许多这种测试，在这过程中我意识到这些测试能为面试者和被面试者提供许多有用信息，此外，撇开面试的压力不谈，这种测试也是相当有趣的。 

　　从被面试者的角度来讲，你能了解许多关于出题者或监考者的情况。这个测试只是出题者为显示其对ANSI标准细节的知识而不是技术技巧而设计吗？这是个愚蠢的问题吗？如要你答出某个字符的ASCII值。这些问题着重考察你的系统调用和内存分配策略方面的能力吗？这标志着出题者也许花时间在微机上而不是在嵌入式系统上。如果上述任何问题的答案是"是"的话，那么我知道我得认真考虑我是否应该去做这份工作。 

　　从面试者的角度来讲，一个测试也许能从多方面揭示应试者的素质：最基本的，你能了解应试者C语言的水平。不管怎么样，看一下这人如何回答他不会的问题也是满有趣。应试者是以好的直觉做出明智的选择，还是只是瞎蒙呢？当应试者在某个问题上卡住时是找借口呢，还是表现出对问题的真正的好奇心，把这看成学习的机会呢？我发现这些信息与他们的测试成绩一样有用。 

　　有了这些想法，我决定出一些真正针对嵌入式系统的考题，希望这些令人头痛的考题能给正在找工作的人一点帮助。这些问题都是我这些年实际碰到的。其中有些题很难，但它们应该都能给你一点启迪。 

　　这个测试适于不同水平的应试者，大多数初级水平的应试者的成绩会很差，经验丰富的程序员应该有很好的成绩。为了让你能自己决定某些问题的偏好，每个问题没有分配分数，如果选择这些考题为你所用，请自行按你的意思分配分数。 

预处理器（Preprocessor） 
1 . 用预处理指令#define 声明一个常数，用以表明1年中有多少秒（忽略闰年问题） 
　　#define SECONDS_PER_YEAR (60 * 60 * 24 * 365)UL 
我在这想看到几件事情： 
1) #define 语法的基本知识（例如：不能以分号结束，括号的使用，等等） 
2)懂得预处理器将为你计算常数表达式的值，因此，直接写出你是如何计算一年中有多少秒而不是计算出实际的值，是更清晰而没有代价的。 
3) 意识到这个表达式将使一个16位机的整型数溢出-因此要用到长整型符号L,告诉编译器这个常数是的长整型数。 
4) 如果你在你的表达式中用到UL（表示无符号长整型），那么你有了一个好的起点。记住，第一印象很重要。 

2 . 写一个"标准"宏MIN ，这个宏输入两个参数并返回较小的一个。 
　　#define MIN(A,B) （（A） <= (B) ? (A) : (B)) 这个测试是为下面的目的而设的： 
1) 标识#define在宏中应用的基本知识。这是很重要的。因为在  嵌入(inline)操作符 变为标准C的一部分之前，宏是方便产生嵌入代码的唯一方法，对于嵌入式系统来说，为了能达到要求的性能，嵌入代码经常是必须的方法。 
　　2)三重条件操作符的知识。这个操作符存在C语言中的原因是它使得编译器能产生比if-then-else更优化的代码，了解这个用法是很重要的。 
3) 懂得在宏中小心地把参数用括号括起来 
4) 我也用这个问题开始讨论宏的副作用，例如：当你写下面的代码时会发生什么事？ 
　　least = MIN(*p++, b); 

3. 预处理器标识#error的目的是什么？ 
　　如果你不知道答案，请看参考文献1。这问题对区分一个正常的伙计和一个书呆子是很有用的。只有书呆子才会读C语言课本的附录去找出象这种问题的答案。当然如果你不是在找一个书呆子，那么应试者最好希望自己不要知道答案。 

死循环（Infinite loops） 
　　4. 嵌入式系统中经常要用到无限循环，你怎么样用C编写死循环呢？ 这个问题用几个解决方案。  
我首选的方案是： 
　　while(1) 
　　{ 

　　} 
　　一些程序员更喜欢如下方案： 
　　for(;;) 
　　{ 

　　} 
　　这个实现方式让我为难，因为这个语法没有确切表达到底怎么回事。如果一个应试者给出这个作为方案，我将用这个作为一个机会去探究他们这样做的基本原理。如果他们的基本答案是："我被教着这样做，但从没有想到过为什么。"这会给我留下一个坏印象。 
　　第三个方案是用 goto 
　　Loop: 
　　... 
　　goto Loop; 
应试者如给出上面的方案，这说明或者他是一个汇编语言程序员（这也许是好事）或者他是一个想进入新领域的BASIC/FORTRAN程序员。 

数据声明（Data declarations） 
5. 用变量a给出下面的定义 
　　a) 一个整型数（An integer） 
　　b)一个指向整型数的指针（ A pointer to an integer） 
　　c)一个指向指针的的指针，它指向的指针是指向一个整型数（ A pointer to a pointer to an intege）r 
　　d)一个有10个整型数的数组（ An array of 10 integers） 
　　e) 一个有10个指针的数组，该指针是指向一个整型数的。（An array of 10 pointers to 
　　integers） 
　　f) 一个指向有10个整型数数组的指针（ A pointer to an array of 10 integers） 
　　g) 一个指向函数的指针，该函数有一个整型参数并返回一个整型数（A pointer to a function 
　　that takes an integer as an argument and returns an  integer） 
　　h) 一个有10个指针的数组，该指针指向一个函数，该函数有一个整型参数并返回一个整型数（ An array of  ten pointers to functions that take an integer argument  and return an integer ） 
　　答案是： 
　　a) int a; // An integer 
　　b) int *a; // A pointer to an integer 
　　c) int **a; // A pointer to a pointer to an integer 
　　d) int a[10]; // An array of 10 integers 
　　e) int *a[10]; // An array of 10 pointers to integers 
　　f) int (*a)[10]; // A pointer to an array of 10 integers 
　　g) int (*a)(int); // A pointer to a function a that 
　　takes an integer argument and returns an integer 
　　h) int (*a[10])(int); // An array of 10 pointers to 
　　functions that take an integer argument and return an 
　　integer 
人们经常声称这里有几个问题是那种要翻一下书才能回答的问题，我同意这种说法。当我写这篇文章时，为了确定语法的正确性，我的确查了一下书。但是当我被面试的时候，我期望被问到这个问题（或者相近的问题）。因为在被面试的这段时间里，我确定我知道这个问题的答案。应试者如果不知道所有的答案（或至少大部分答案），那么也就没有为这次面试做准备，如果该面试者没有为这次面试做准备，那么他又能为什么出准备呢？ 
Static 
　　6. 关键字static的作用是什么？ 
　　这个简单的问题很少有人能回答完全。在C语言中，关键字static有三个明显的作用： 
1)在函数体，一个被声明为静态的变量在这一函数被调用过程中维持其值不变。 
　　2)  在模块内（但在函数体外），一个被声明为静态的变量可以被模块内所用函数访问，但不能被模块外其它函数访问。它是一个本地的全局变量。 
3)  在模块内，一个被声明为静态的函数只可被这一模块内的其它函数调用。那就是，这个函数被限制在声明它的模块的本地范围内使用。 
大多数应试者能正确回答第一部分，一部分能正确回答第二部分，同是很少的人能懂得第三部分。这是一个应试者的严重的缺点，因为他显然不懂得本地化数据和代码范围的好处和重要性。 

Const 
7．关键字const有什么含意？ 
我只要一听到被面试者说："const意味着常数"，我就知道我正在和一个业余者打交道。去年Dan 
　　Saks已经在他的文章里完全概括了const的所有用法，因此ESP(译者：Embedded Systems 
　　Programming)的每一位读者应该非常熟悉const能做什么和不能做什么.如果你从没有读到那篇文章，只要能说出const意味着"只读"就可以了。尽管这个答案不是完全的答案，但我接受它作为一个正确的答案。（如果你想知道更详细的答案，仔细读一下Saks的文章吧。） 
　　如果应试者能正确回答这个问题，我将问他一个附加的问题：下面的声明都是什么意思？ 
　　const int a; 
　　int const a; 
　　const int *a; 
　　int * const a; 
　　int const * a const; 
　　/******/ 
　　前两个的作用是一样，a是一个常整型数。第三个意味着a是一个指向常整型数的指针（也就是，整型数是不可修改的，但指针可以）。第四个意思a是一个指向整型数的常指针（也就是说，指针指向的整型数是可以修改的，但指针是不可修改的）。最后一个意味着a是一个指向常整型数的常指针（也就是说，指针指向的整型数是不可修改的，同时指针也是不可修改的）。如果应试者能正确回答这些问题，那么他就给我留下了一个好印象。顺带提一句，也许你可能会问，即使不用关键字 ，也还是能很容易写出功能正确的程序，那么我为什么还要如此看重关键字const呢？我也如下的几下理由： 
　　1) 关键字const的作用是为给读你代码的人传达非常有用的信息，实际上，声明一个参数为常量是为了告诉了用户这个参数的应用目的。如果你曾花很多时间清理其它人留下的垃圾，你就会很快学会感谢这点多余的信息。（当然，懂得用const的程序员很少会留下的垃圾让别人来清理的。） 
2) 通过给优化器一些附加的信息，使用关键字const也许能产生更紧凑的代码。 
　　3) 合理地使用关键字const可以使编译器很自然地保护那些不希望被改变的参数，防止其被无意的代码修改。简而言之，这样可以减少bug的出现。 

Volatile 
8. 关键字volatile有什么含意?并给出三个不同的例子。 
一个定义为volatile的变量是说这变量可能会被意想不到地改变，这样，编译器就不会去假设这个变量的值了。精确地说就是，优化器在用到这个变量时必须每次都小心地重新读取这个变量的值，而不是使用保存在寄存器里的备份。下面是volatile变量的几个例子： 
1) 并行设备的硬件寄存器（如：状态寄存器） 
2) 一个中断服务子程序中会访问到的非自动变量(Non-automatic variables) 
3) 多线程应用中被几个任务共享的变量 
　　回答不出这个问题的人是不会被雇佣的。我认为这是区分C程序员和嵌入式系统程序员的最基本的问题。搞嵌入式的家伙们经常同硬件、中断、RTOS等等打交道，所有这些都要求用到volatile变量。不懂得volatile的内容将会带来灾难。 假设被面试者正确地回答了这是问题（嗯，怀疑是否会是这样），我将稍微深究一下，看一下这家伙是不是直正懂得volatile完全的重要性。 
1)一个参数既可以是const还可以是volatile吗？解释为什么。 
2); 一个指针可以是volatile 吗？解释为什么。 
3); 下面的函数有什么错误： 
int square(volatile int *ptr) 
　　{ 
　　return *ptr * *ptr; 
　　} 
下面是答案： 
　　1)是的。一个例子是只读的状态寄存器。它是volatile因为它可能被意想不到地改变。它是const因为程序不应该试图去修改它。 
2); 是的。尽管这并不很常见。一个例子是当一个中服务子程序修该一个指向一个buffer的指针时。 
　　3) 这段代码有点变态。这段代码的目的是用来返指针*ptr指向值的平方，但是，由于*ptr指向一个volatile型参数，编译器将产生类似下面的代码： 
int square(volatile int *ptr) 
　　{ 
　　int a,b; 
　　a = *ptr; 
　　b = *ptr; 
　　return a * b; 
　　} 
由于*ptr的值可能被意想不到地该变，因此a和b可能是不同的。结果，这段代码可能返不是你所期望的平方值！正确的代码如下： 
　　long square(volatile int *ptr) 
　　{ 
　　int a; 
　　a = *ptr; 
　　return a * a; 
　　} 

位操作（Bit manipulation） 
9. 嵌入式系统总是要用户对变量或寄存器进行位操作。给定一个整型变量a，写两段代码，第一个设置a的bit 3，第二个清除a 的bit 3。在以上两个操作中，要保持其它位不变。 对这个问题有三种基本的反应 
1)不知道如何下手。该被面者从没做过任何嵌入式系统的工作。 
　　2) 用bit fields。Bit  fields是被扔到C语言死角的东西，它保证你的代码在不同编译器之间是不可移植的，同时也保证了的你的代码是不可重用的。我最近不幸看到 Infineon为其较复杂的通信芯片写的驱动程序，它用到了bit fields因此完全对我无用，因为我的编译器用其它的方式来实现bit  fields的。从道德讲：永远不要让一个非嵌入式的家伙粘实际硬件的边。 
3) 用 #defines 和 bit masks  操作。这是一个有极高可移植性的方法，是应该被用到的方法。最佳的解决方案如下： 
　　#define BIT3 (0x1 < < 3) 
　　static int a; 
void set_bit3(void) 
　　{ 
　　a |= BIT3; 
　　} 
　　void clear_bit3(void) 
　　{ 
　　a &= ~BIT3; 
　　} 

　　一些人喜欢为设置和清除值而定义一个掩码同时定义一些说明常数，这也是可以接受的。我希望看到几个要点：说明常数、|=和&=~操作。 

访问固定的内存位置（Accessing fixed memory locations） 
10. 嵌入式系统经常具有要求程序员去访问某特定的内存位置的特点。 
在某工程中，要求设置一绝对地址为0x67a9的整型变量的值为0xaa66。编译器是一个纯粹的ANSI编译器。写代码去完成这一任务。这一问题测试你是否知道为了访问一绝对地址把一个整型数强制转换（typecast）为一指针是合法的。这一问题的实现方式随着个人风格不同而不同。典型的类似代码如下： 
　　int *ptr; 
　　ptr = (int *)0x67a9; 
　　*ptr = 0xaa55; 
　　A more obscure approach is: (  一个较晦涩的方法是)： 
　　*(int * const)(0x67a9) = 0xaa55; 
　　即使你的品味更接近第二种方案，但我建议你在面试时使用第一种方案。 
中断（Interrupts） 
　　11. 
中断是嵌入式系统中重要的组成部分，这导致了很多编译开发商提供一种扩展—让标准C支持中断。具代表事实是，产生了一个新的关键字 __interrupt。下面的代码就使用了__interrupt关键字去定义了一个中断服务子程序(ISR)，请评论一下这段代码的。 
__interrupt double compute_area (double radius) 
　　{ 
　　double area = PI * radius * radius; 
　　printf("/nArea = %f", area); 
　　return area; 
　　} 
这个函数有太多的错误了，以至让人不知从何说起了： 
1)ISR 不能返回一个值。如果你不懂这个，那么你不会被雇用的。 
2) ISR 不能传递参数。如果你没有看到这一点，你被雇用的机会等同第一项。 
3) 在许多的处理器/编译器中，浮点一般都是不可重入的。有些处理器/编译器需要让额处的寄存器入栈，有些处理器/编译器就是不允许在ISR中做浮点运算。此外，ISR应该是短而有效率的，在ISR中做浮点运算是不明智的。 
4) 与第三点一脉相承，printf()经常有重入和性能上的问题。如果你丢掉了第三和第四点，我不会太为难你的。不用说，如果你能得到后两点，那么你的被雇用前景越来越光明了。 

代码例子（Code examples） 
12 . 下面的代码输出是什么，为什么？ 
　　void foo(void) 
　　{ 
　　unsigned int a = 6; 
　　int b = -20; 
　　(a+b > 6) ? puts("> 6") : puts(" <= 6"); 
　　} 
　　这个问题测试你是否懂得C语言中的整数自动转换原则，我发现有些开发者懂得极少这些东西。不管如何，这无符号整型问题的答案是输出是 ">6"。原因是当表达式中存在有符号类型和无符号类型时所有的操作数都自动转换为无符号类型。因此-20变成了一个非常大的正整数，所以该表达式计算出的结果大于6。这一点对于应当频繁用到无符号数据类型的嵌入式系统来说是丰常重要的。如果你答错了这个问题，你也就到了得不到这份工作的边缘。 

13. 评价下面的代码片断： 
　　unsigned int zero = 0; 
　　unsigned int compzero = 0xFFFF; 
　　/*1's complement of zero */ 
　　对于一个int型不是16位的处理器为说，上面的代码是不正确的。应编写如下： 
unsigned int compzero = ~0; 
　　这一问题真正能揭露出应试者是否懂得处理器字长的重要性。在我的经验里，好的嵌入式程序员非常准确地明白硬件的细节和它的局限，然而PC机程序往往把硬件作为一个无法避免的烦恼。 
到了这个阶段，应试者或者完全垂头丧气了或者信心满满志在必得。如果显然应试者不是很好，那么这个测试就在这里结束了。但如果显然应试者做得不错，那么我就扔出下面的追加问题，这些问题是比较难的，我想仅仅非常优秀的应试者能做得不错。提出这些问题，我希望更多看到应试者应付问题的方法，而不是答案。不管如何，你就当是这个娱乐吧... 

动态内存分配（Dynamic memory allocation） 
14. 
尽管不像非嵌入式计算机那么常见，嵌入式系统还是有从堆（heap）中动态分配内存的过程的。那么嵌入式系统中，动态分配内存可能发生的问题是什么？这里，我期望应试者能提到内存碎片，碎片收集的问题，变量的持行时间等等。这个主题已经在ESP杂志中被广泛地讨论过了（主要是 P.J. Plauger, 他的解释远远超过我这里能提到的任何解释），所有回过头看一下这些杂志吧！让应试者进入一种虚假的安全感觉后，我拿出这么一个小节目：下面的代码片段的输出是什么，为什么？ 
　　char *ptr; 
　　if ((ptr = (char *)malloc(0)) == NULL) 
　　puts("Got a null pointer"); 
　　else 
　　puts("Got a valid pointer"); 
　　这是一个有趣的问题。最近在我的一个同事不经意把0值传给了函数malloc，得到了一个合法的指针之后，我才想到这个问题。这就是上面的代码，该代码的输出是"Got  a valid pointer"。我用这个来开始讨论这样的一问题，看看被面试者是否想到库例程这样做是正确。得到正确的答案固然重要，但解决问题的方法和你做决定的基本原理更重要些。 
Typedef 
　　15 Typedef 
在C语言中频繁用以声明一个已经存在的数据类型的同义字。也可以用预处理器做类似的事。例如，思考一下下面的例子： 
　　#define dPS struct s * 
　　typedef struct s * tPS; 
　　以上两种情况的意图都是要定义dPS 和 tPS 作为一个指向结构s指针。哪种方法更好呢？（如果有的话）为什么？ 
　　这是一个非常微妙的问题，任何人答对这个问题（正当的原因）是应当被恭喜的。答案是：typedef更好。思考下面的例子： 
　　dPS p1,p2; 
　　tPS p3,p4; 
第一个扩展为 
　　struct s * p1, p2; 
上面的代码定义p1为一个指向结构的指，p2为一个实际的结构，这也许不是你想要的。第二个例子正确地定义了p3  和p4 两个指针。 

晦涩的语法 
　　16 . C语言同意一些令人震惊的结构,下面的结构是合法的吗，如果是它做些什么？ 
　　int a = 5, b = 7, c; 
　　c = a+++b; 
　　这个问题将做为这个测验的一个愉快的结尾。不管你相不相信，上面的例子是完全合乎语法的。问题是编译器如何处理它？水平不高的编译作者实际上会争论这个问题，根据最处理原则，编译器应当能处理尽可能所有合法的用法。因此，上面的代码被处理成：c = a++ + b; 
因此, 这段代码持行后a = 6, b = 7, c = 12。 
　　如果你知道答案，或猜出正确答案，做得好。如果你不知道答案，我也不把这个当作问题。我发现这个问题的最大好处是这是一个关于代码编写风格，代码的可读性，代码的可修改性的好的话题。 
好了，伙计们，你现在已经做完所有的测试了。这就是我出的C语言测试题，我怀着愉快的心情写完它，希望你以同样的心情读完它。如果是认为这是一个好的测试，那么尽量都用到你的找工作的过程中去吧。天知道也许过个一两年，我就不做现在的工作，也需要找一个。 

作者介绍: 
　　Nigel Jones      是一个顾问，现在住在Maryland，当他不在水下时，你能在多个范围的嵌入项目中找到他。他很高兴能收到读者的来信，他的email地址是: [email protected] 
参考文献 
1) Jones, Nigel, "In Praise of the #error directive," 
　　Embedded Systems Programming, September 1999, p. 114. 
2) Jones, Nigel, " Efficient C Code for Eight-bit MCUs 
　　," Embedded Systems Programming, November 1998, p. 66.