花了两天做了一下Data Lab，本来以为被书上的课后习题覆盖了，没想到还是有两题比较难的，isAsciiDigit和howManyBits两题借鉴了网上的思路。
实验材料在http://csapp.cs.cmu.edu/3e/labs.html可以找到。
总结
bitXor
送分题，有数字逻辑基础轻松解决。
/*
* bitXor - x^y using only ~ and &
* Example: bitXor(4, 5) = 1
* Legal ops: ~ &
* Max ops: 14
* Rating: 1
*/
int bitXor(int x, int y)
{
return ~(x & y) & ~(~x & ~y);
}
tmin
送分题。
/*
* tmin - return minimum two's complement integer
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 4
* Rating: 1
*/
int tmin(void)
{
return 1 << 31;
}
isTmax
简单，有边界情况，运行结果提示后轻松解决。
/*
* isTmax - returns 1 if x is the maximum, two's complement number,
* and 0 otherwise
* Legal ops: ! ~ & ^ | +
* Max ops: 10
* Rating: 1
*/
int isTmax(int x)
{
return !~((x + 1) ^ x) & !!((x + x + 1) ^ x);
}
allOddBits
由于我是先做书上习题的，做这道题被带到奇偶性那个思维惯性中去了，用的相同的办法解决，实际上直接用掩码更加简单，操作数也更少。
/*
* allOddBits - return 1 if all odd-numbered bits in word set to 1
* where bits are numbered from 0 (least significant) to 31 (most significant)
* Examples allOddBits(0xFFFFFFFD) = 0, allOddBits(0xAAAAAAAA) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 12
* Rating: 2
*/
int allOddBits(int x)
{
x &= x >> 16;
x &= x >> 8;
x &= x >> 4;
x &= x >> 2;
return !!(x & 2);
}
negate
送分题。
/*
* negate - return -x
* Example: negate(1) = -1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 5
* Rating: 2
*/
int negate(int x)
{
return ~x + 1;
}
isAsciiDigit
没想到思路，看了网上别人的答案思路，通过检测加上某个值是否溢出来判断区间，学到了！
/*
* isAsciiDigit - return 1 if 0x30 <= x <= 0x39 (ASCII codes for characters '0' to '9')
* Example: isAsciiDigit(0x35) = 1.
* isAsciiDigit(0x3a) = 0.
* isAsciiDigit(0x05) = 0.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 15
* Rating: 3
*/
int isAsciiDigit(int x)
{
int sign = x >> 31 & 1;
int sign1 = ((x + (~0x30) + 1) >> 31) & 1;
int sign2 = ((x + (~0x39)) >> 31) & 1;
return !sign1 & sign2 & !sign;
}
conditional
简单。
/*
* conditional - same as x ? y : z
* Example: conditional(2,4,5) = 4
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 16
* Rating: 3
*/
int conditional(int x, int y, int z)
{
x = ~(!x) + 1;
return (~x & y) | (x & z);
}
isLessOrEqual
书上习题有类似的思路，了解套路就不难了。
/*
* isLessOrEqual - if x <= y then return 1, else return 0
* Example: isLessOrEqual(4,5) = 1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 24
* Rating: 3
*/
int isLessOrEqual(int x, int y)
{
int diff = y + (~x) + 1;
int signx = (x >> 31) & 1;
int signy = (y >> 31) & 1;
int signdiff = (diff >> 31) & 1;
return (!signy & signx) | (!(signy & !signx) & !signdiff);
}
logicalNeg
我还是用的折半位操作的做法，稍微复杂了一点。
/*
* logicalNeg - implement the ! operator, using all of
* the legal operators except !
* Examples: logicalNeg(3) = 0, logicalNeg(0) = 1
* Legal ops: ~ & ^ | + << >>
* Max ops: 12
* Rating: 4
*/
int logicalNeg(int x)
{
x |= x >> 16;
x |= x >> 8;
x |= x >> 4;
x |= x >> 2;
x |= x >> 1;
return ~x & 1;
}
howManyBits
思路没想到，看了别人的答案后懂了，类似于二分法，明明自己会二分，只能感叹智商太低很多地方不能学以致用。
/* howManyBits - return the minimum number of bits required to represent x in
* two's complement
* Examples: howManyBits(12) = 5
* howManyBits(298) = 10
* howManyBits(-5) = 4
* howManyBits(0) = 1
* howManyBits(-1) = 1
* howManyBits(0x80000000) = 32
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 90
* Rating: 4
*/
int howManyBits(int x)
{
int sign = (x & (1 << 31)) >> 31, i16, i8, i4, i2, i1, iz;
x = (sign & ~x) | (~sign & x);
iz = ~!!x + 1;
i16 = (~!!(x >> 16) + 1);
x = (i16 & (x >> 16)) | (~i16 & x);
i8 = ~!!(x >> 8) + 1;
x = (i8 & (x >> 8)) | (~i8 & x);
i4 = ~!!(x >> 4) + 1;
x = (i4 & (x >> 4)) | (~i4 & x);
i2 = ~!!(x >> 2) + 1;
x = (i2 & (x >> 2)) | (~i2 & x);
i1 = ~!!(x >> 1) + 1;
return (i16 & 16) + (i8 & 8) + (i4 & 4) + (i2 & 2) + (i1 & 1) + 1 + (iz & 1);
}
floatScale2
见第二章习题答案，一模一样，分类讨论即可。
/*
* floatScale2 - Return bit-level equivalent of expression 2*f for
* floating point argument f.
* Both the argument and result are passed as unsigned int's, but
* they are to be interpreted as the bit-level representation of
* single-precision floating point values.
* When argument is NaN, return argument
* Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
* Max ops: 30
* Rating: 4
*/
unsigned floatScale2(unsigned uf)
{
unsigned sign = uf >> 31, exp = uf >> 23 & 0xff, frac = uf & 0x7fffff;
if (exp == 0xff)
return uf;
if (exp == 0) // Denormalized
{
frac <<= 1;
}
else if (exp == 0xff - 1) // Normalized max
{
exp = 0xff;
frac = 0;
}
else
{
exp++;
}
return sign << 31 | exp << 23 | frac;
}
floatFloat2Int
见第二章习题答案，一模一样，分类讨论即可。
/*
* floatFloat2Int - Return bit-level equivalent of expression (int) f
* for floating point argument f.
* Argument is passed as unsigned int, but
* it is to be interpreted as the bit-level representation of a
* single-precision floating point value.
* Anything out of range (including NaN and infinity) should return
* 0x80000000u.
* Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
* Max ops: 30
* Rating: 4
*/
int floatFloat2Int(unsigned uf)
{
unsigned sign = uf >> 31, exp = uf >> 23 & 0xff, frac = uf & 0x7fffff;
if (exp == 0x7ffff) //oo or NaN
return 0x80000000u;
if (exp < 127) // floats less then 1.0 or denormalized
return 0;
exp -= 127;
if (exp > 30) // too big
return 0x80000000u;
frac |= 1 << 23;
frac = frac << 7 >> 30 - exp;
return sign ? (~frac + 1) : frac;
}
floatPower2
见第二章习题答案，一模一样，分类讨论即可。
/*
* floatPower2 - Return bit-level equivalent of the expression 2.0^x
* (2.0 raised to the power x) for any 32-bit integer x.
*
* The unsigned value that is returned should have the identical bit
* representation as the single-precision floating-point number 2.0^x.
* If the result is too small to be represented as a denorm, return
* 0. If too large, return +INF.
*
* Legal ops: Any integer/unsigned operations incl. ||, &&. Also if, while
* Max ops: 30
* Rating: 4
*/
unsigned floatPower2(int x)
{
unsigned exp, frac;
unsigned u;
if (x < -149) // Too small. Return 0.0.
{
exp = 0;
frac = 0;
}
else if (x < -126) // Denormalized result
{
exp = 0;
frac = 1 << x + 149;
}
else if (x < 128) // Normalized result
{
exp = x + 127;
frac = 0;
}
else // Too big. Return +oo
{
exp = 255;
frac = 0;
}
u = exp << 23 | frac; // Pack exp and frac into 32 bits
return u;
}
总体来说，Data Lab中的题目都是比较经典的，有难有简单，收获比较大。
结果
操作数
得分
代码
全部代码：
/*
* CS:APP Data Lab
*
*
*
* bits.c - Source file with your solutions to the Lab.
* This is the file you will hand in to your instructor.
*
* WARNING: Do not include the header; it confuses the dlc
* compiler. You can still use printf for debugging without including
* , although you might get a compiler warning. In general,
* it's not good practice to ignore compiler warnings, but in this
* case it's OK.
*/
#if 0
/*
* Instructions to Students:
*
* STEP 1: Read the following instructions carefully.
*/
You will provide your solution to the Data Lab by
editing the collection of functions in this source file.
INTEGER CODING RULES:
Replace the "return" statement in each function with one
or more lines of C code that implements the function. Your code
must conform to the following style:
int Funct(arg1, arg2, ...) {
/* brief description of how your implementation works */
int var1 = Expr1;
...
int varM = ExprM;
varJ = ExprJ;
...
varN = ExprN;
return ExprR;
}
Each "Expr" is an expression using ONLY the following:
1. Integer constants 0 through 255 (0xFF), inclusive. You are
not allowed to use big constants such as 0xffffffff.
2. Function arguments and local variables (no global variables).
3. Unary integer operations ! ~
4. Binary integer operations & ^ | + << >>
Some of the problems restrict the set of allowed operators even further.
Each "Expr" may consist of multiple operators. You are not restricted to
one operator per line.
You are expressly forbidden to:
1. Use any control constructs such as if, do, while, for, switch, etc.
2. Define or use any macros.
3. Define any additional functions in this file.
4. Call any functions.
5. Use any other operations, such as &&, ||, -, or ?:
6. Use any form of casting.
7. Use any data type other than int. This implies that you
cannot use arrays, structs, or unions.
You may assume that your machine:
1. Uses 2s complement, 32-bit representations of integers.
2. Performs right shifts arithmetically.
3. Has unpredictable behavior when shifting if the shift amount
is less than 0 or greater than 31.
EXAMPLES OF ACCEPTABLE CODING STYLE:
/*
* pow2plus1 - returns 2^x + 1, where 0 <= x <= 31
*/
int pow2plus1(int x) {
/* exploit ability of shifts to compute powers of 2 */
return (1 << x) + 1;
}
/*
* pow2plus4 - returns 2^x + 4, where 0 <= x <= 31
*/
int pow2plus4(int x) {
/* exploit ability of shifts to compute powers of 2 */
int result = (1 << x);
result += 4;
return result;
}
FLOATING POINT CODING RULES
For the problems that require you to implement floating-point operations,
the coding rules are less strict. You are allowed to use looping and
conditional control. You are allowed to use both ints and unsigneds.
You can use arbitrary integer and unsigned constants. You can use any arithmetic,
logical, or comparison operations on int or unsigned data.
You are expressly forbidden to:
1. Define or use any macros.
2. Define any additional functions in this file.
3. Call any functions.
4. Use any form of casting.
5. Use any data type other than int or unsigned. This means that you
cannot use arrays, structs, or unions.
6. Use any floating point data types, operations, or constants.
NOTES:
1. Use the dlc (data lab checker) compiler (described in the handout) to
check the legality of your solutions.
2. Each function has a maximum number of operations (integer, logical,
or comparison) that you are allowed to use for your implementation
of the function. The max operator count is checked by dlc.
Note that assignment ('=') is not counted; you may use as many of
these as you want without penalty.
3. Use the btest test harness to check your functions for correctness.
4. Use the BDD checker to formally verify your functions
5. The maximum number of ops for each function is given in the
header comment for each function. If there are any inconsistencies
between the maximum ops in the writeup and in this file, consider
this file the authoritative source.
/*
* STEP 2: Modify the following functions according the coding rules.
*
* IMPORTANT. TO AVOID GRADING SURPRISES:
* 1. Use the dlc compiler to check that your solutions conform
* to the coding rules.
* 2. Use the BDD checker to formally verify that your solutions produce
* the correct answers.
*/
#endif
//1
/*
* bitXor - x^y using only ~ and &
* Example: bitXor(4, 5) = 1
* Legal ops: ~ &
* Max ops: 14
* Rating: 1
*/
int bitXor(int x, int y)
{
return ~(x & y) & ~(~x & ~y);
}
/*
* tmin - return minimum two's complement integer
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 4
* Rating: 1
*/
int tmin(void)
{
return 1 << 31;
}
//2
/*
* isTmax - returns 1 if x is the maximum, two's complement number,
* and 0 otherwise
* Legal ops: ! ~ & ^ | +
* Max ops: 10
* Rating: 1
*/
int isTmax(int x)
{
return !~((x + 1) ^ x) & !!((x + x + 1) ^ x);
}
/*
* allOddBits - return 1 if all odd-numbered bits in word set to 1
* where bits are numbered from 0 (least significant) to 31 (most significant)
* Examples allOddBits(0xFFFFFFFD) = 0, allOddBits(0xAAAAAAAA) = 1
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 12
* Rating: 2
*/
int allOddBits(int x)
{
x &= x >> 16;
x &= x >> 8;
x &= x >> 4;
x &= x >> 2;
return !!(x & 2);
}
/*
* negate - return -x
* Example: negate(1) = -1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 5
* Rating: 2
*/
int negate(int x)
{
return ~x + 1;
}
//3
/*
* isAsciiDigit - return 1 if 0x30 <= x <= 0x39 (ASCII codes for characters '0' to '9')
* Example: isAsciiDigit(0x35) = 1.
* isAsciiDigit(0x3a) = 0.
* isAsciiDigit(0x05) = 0.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 15
* Rating: 3
*/
int isAsciiDigit(int x)
{
int sign = x >> 31 & 1;
int sign1 = ((x + (~0x30) + 1) >> 31) & 1;
int sign2 = ((x + (~0x39)) >> 31) & 1;
return !sign1 & sign2 & !sign;
}
/*
* conditional - same as x ? y : z
* Example: conditional(2,4,5) = 4
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 16
* Rating: 3
*/
int conditional(int x, int y, int z)
{
x = ~(!x) + 1;
return (~x & y) | (x & z);
}
/*
* isLessOrEqual - if x <= y then return 1, else return 0
* Example: isLessOrEqual(4,5) = 1.
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 24
* Rating: 3
*/
int isLessOrEqual(int x, int y)
{
int diff = y + (~x) + 1;
int signx = (x >> 31) & 1;
int signy = (y >> 31) & 1;
int signdiff = (diff >> 31) & 1;
return (!signy & signx) | (!(signy & !signx) & !signdiff);
}
//4
/*
* logicalNeg - implement the ! operator, using all of
* the legal operators except !
* Examples: logicalNeg(3) = 0, logicalNeg(0) = 1
* Legal ops: ~ & ^ | + << >>
* Max ops: 12
* Rating: 4
*/
int logicalNeg(int x)
{
x |= x >> 16;
x |= x >> 8;
x |= x >> 4;
x |= x >> 2;
x |= x >> 1;
return ~x & 1;
}
/* howManyBits - return the minimum number of bits required to represent x in
* two's complement
* Examples: howManyBits(12) = 5
* howManyBits(298) = 10
* howManyBits(-5) = 4
* howManyBits(0) = 1
* howManyBits(-1) = 1
* howManyBits(0x80000000) = 32
* Legal ops: ! ~ & ^ | + << >>
* Max ops: 90
* Rating: 4
*/
int howManyBits(int x)
{
int sign = (x & (1 << 31)) >> 31, i16, i8, i4, i2, i1, iz;
x = (sign & ~x) | (~sign & x);
iz = ~!!x + 1;
i16 = (~!!(x >> 16) + 1);
x = (i16 & (x >> 16)) | (~i16 & x);
i8 = ~!!(x >> 8) + 1;
x = (i8 & (x >> 8)) | (~i8 & x);
i4 = ~!!(x >> 4) + 1;
x = (i4 & (x >> 4)) | (~i4 & x);
i2 = ~!!(x >> 2) + 1;
x = (i2 & (x >> 2)) | (~i2 & x);
i1 = ~!!(x >> 1) + 1;
return (i16 & 16) + (i8 & 8) + (i4 & 4) + (i2 & 2) + (i1 & 1) + 1 + (iz & 1);
}
//float
/*
* floatScale2 - Return bit-level equivalent of expression 2*f for
* floating point argument f.
* Both the argument and result are passed as unsigned int's, but
* they are to be interpreted as the bit-level representation of
* single-precision floating point values.
* When argument is NaN, return argument
* Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
* Max ops: 30
* Rating: 4
*/
unsigned floatScale2(unsigned uf)
{
unsigned sign = uf >> 31, exp = uf >> 23 & 0xff, frac = uf & 0x7fffff;
if (exp == 0xff)
return uf;
if (exp == 0) // Denormalized
{
frac <<= 1;
}
else if (exp == 0xff - 1) // Normalized max
{
exp = 0xff;
frac = 0;
}
else
{
exp++;
}
return sign << 31 | exp << 23 | frac;
}
/*
* floatFloat2Int - Return bit-level equivalent of expression (int) f
* for floating point argument f.
* Argument is passed as unsigned int, but
* it is to be interpreted as the bit-level representation of a
* single-precision floating point value.
* Anything out of range (including NaN and infinity) should return
* 0x80000000u.
* Legal ops: Any integer/unsigned operations incl. ||, &&. also if, while
* Max ops: 30
* Rating: 4
*/
int floatFloat2Int(unsigned uf)
{
unsigned sign = uf >> 31, exp = uf >> 23 & 0xff, frac = uf & 0x7fffff;
if (exp == 0x7ffff) //oo or NaN
return 0x80000000u;
if (exp < 127) // floats less then 1.0 or denormalized
return 0;
exp -= 127;
if (exp > 30) // too big
return 0x80000000u;
frac |= 1 << 23;
frac = frac << 7 >> 30 - exp;
return sign ? (~frac + 1) : frac;
}
/*
* floatPower2 - Return bit-level equivalent of the expression 2.0^x
* (2.0 raised to the power x) for any 32-bit integer x.
*
* The unsigned value that is returned should have the identical bit
* representation as the single-precision floating-point number 2.0^x.
* If the result is too small to be represented as a denorm, return
* 0. If too large, return +INF.
*
* Legal ops: Any integer/unsigned operations incl. ||, &&. Also if, while
* Max ops: 30
* Rating: 4
*/
unsigned floatPower2(int x)
{
unsigned exp, frac;
unsigned u;
if (x < -149) // Too small. Return 0.0.
{
exp = 0;
frac = 0;
}
else if (x < -126) // Denormalized result
{
exp = 0;
frac = 1 << x + 149;
}
else if (x < 128) // Normalized result
{
exp = x + 127;
frac = 0;
}
else // Too big. Return +oo
{
exp = 255;
frac = 0;
}
u = exp << 23 | frac; // Pack exp and frac into 32 bits
return u;
}