C 程序设计语言——第二章练习题
C 程序设计语言——第二章——Types, Operators and Expressions——练习题
- 2.1 at the end of this chapter summarizes precedence and associativity for all operators.
- 2.2 Write a loop equivalent to the for loop above without using && or ||.
- 2.3 Write a function htoi(s), which converts a string of hexadecimal digits (including an optional 0x or 0X) into its equivalent integer value. The allowable digits are 0 through 9, a through f, and A through F.
- 2.4 Write an alternative version of squeeze(s1,s2) that deletes each character in s1 that matches any character in the string s2.
- 2.5 Write the function any(s1,s2), which returns the first location in a string s1 where any character from the string s2 occurs, or -1 if s1 contains no characters from s2. (The standard library function strpbrk does the same job but returns a pointer to the location.)
- 2.6 Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged.
- 2.7 Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the others unchanged.
- 2.8 Write a function rightrot(x,n) that returns the value of the integer x rotated to the right by n positions.
- 2.9 In a two's complement number system, x &= (x-1) deletes the rightmost 1-bitin x. Explain why. Use this observation to write a faster version of bitcount.
- 2.10 Rewrite the function lower, which converts upper case letters to lower case, with a conditional expression instead of if-else.
2.1 at the end of this chapter summarizes precedence and associativity for all operators.
参考：C语言运算符优先级
优先级总结:
括号，指向成员类优先级 > 单目运算符 > 算数运算符 （乘除求余 高于 加减）> 移位运算符 > 关系运算符 > 按位运算符 > 逻辑运算符 > 条件运算符 > 赋值运算符 > 逗号运算符
结合性：
只有单目运算符，条件运算符（？ ：）和赋值运算符是从右向左，其余从左往右。
2.2 Write a loop equivalent to the for loop above without using && or ||.
分析：
题中的for循环如下：
for(i = 0; i < lim - 1 && (c = getchar())!='\n' && c != EOF;++i)
s[i] = c;
要求实现相同的功能但不用逻辑运算符：
int i = 0;
char c;
while(i < lim - 1)
{
c = getchar();
if(c == EOF)
break;
else if (c == '\n')
break;
else
s[i++] = c;
}
2.3 Write a function htoi(s), which converts a string of hexadecimal digits (including an optional 0x or 0X) into its equivalent integer value. The allowable digits are 0 through 9, a through f, and A through F.
分析：
- 提示输入十六进制字符数，注意大小，整型范围最大为 unsigned long，有8字节，算上十六进制前缀和字符串结束符，输入字符串长度最大19，需进行判断，长度超过重新输入。
- 判断字符串是不是合法：以 0x 或 0X 前缀开始，字符位于 0 ～9，a ～f， A ～ F。
- 字符转化为转换，首先将十六进制字符转换为对应的数需要分情况，这里大小写可以分开分情况探讨，一共三种情况，或者用 tolower（）函数转化为小写字符。
- 十六进制转化为十进制：
十六进制数 0xa1a2a3a4，对应的十进制数计算为：
a4 + a3 * 16 + a2 * 16 * 16 + a1 * 16 * 16 * 16 =
a4 + 16* (a3 + 16 * (a2 + 16 * a1))
#include
#include
#include
#define LEN 19 //unsigned long有8位，要16个十六进制字符，加上两个字符表示前缀，一个字符串结束符
int s_gets(char *s, int n);
void eatline(void);
int check(char *s);
unsigned long htoi(char *s);
int main(void)
{
char s[LEN];
int ok, good;
unsigned long val;
puts("Please enter a string of hexadecimal digis (optional 0x or 0X):");
while(((good = s_gets(s, LEN)) == 0) || (ok = check(s)) == 0)
{
if(good == 0)
{
printf("Please enter no more than %d hexadecimal characters, "
"enter again:\n",LEN-1);
continue;
}
else if(ok == 0)
{
puts("Not a string of hexadecimal digis, enter again:");
continue;
}
}
val = htoi(s);
printf("%s corresponds to %lu\n",s, val);
return 0;
}
void eatline(void)
{
while(getchar() != '\n')
continue;
}
int s_gets(char *s, int n)
{
char *find;
int ok = 0; //检查输入是否超过，超过则对应整数范围超过，需重新输入，不能直接将后面截断
fgets(s, n, stdin);
if(s)
{
find = strchr(s, '\n');
if(find)
{ *find = '\0';
ok = 1;
}
else
{
if(getchar() == '\n')
ok = 1;
else
eatline();
}
}
return ok;
}
int check(char *s)
{
int i = 0;
int flag = 0;
if(s[i++] == '0')
{
if(s[i] == 'x' || s[i] == 'X')
flag = 1;
}
while(s[++i] != '\0' && flag == 1)
{
if(s[i] < '0' || (s[i] > '9' && s[i] < 'A') || (s[i] > 'F' && s[i] < 'a') || s[i] > 'f')
flag = 0;
}
return flag;
}
unsigned long htoi(char *s)
{
unsigned long n = 0;
int h_d;
for(int i = 2; s[i] != '\0'; i++)
{
if(s[i] >= '0' && s[i] <= '9')
h_d = s[i] - '0';
else if(s[i] >= 'A' && s[i] <= 'F')
h_d = s[i] - 'A' + 10;
else if(s[i] >= 'a' && s[i] <= 'f')
h_d = s[i] - 'a' + 10;
n = 16 * n + h_d;
}
return n;
}
/*
// 进制转换另一种方法
// 加上头文件 ctype.h
unsigned long pow_i(int x, int y)
{
unsigned long sum = 1;
for(int i = 0; i < y; i++)
sum *= x;
return sum;
}
unsigned long htoi(char *s)
{
unsigned long n = 0;
int i, j;
int h_d;
int len = strlen(s);
for(i = len - 1, j = 0; i > 1; i--, j++)
{
s[i] = tolower(s[i]);
if(s[i] >= '0' && s[i] <= '9')
h_d = s[i] - '0';
else
h_d = s[i] - 'a' + 10;
n += s[i] * pow_i(16, j);
}
return n;
}
*/
2.4 Write an alternative version of squeeze(s1,s2) that deletes each character in s1 that matches any character in the string s2.
#include
#include
#define LEN 41
char *s_gets(char *s, int n);
void squeeze(char *s1, char *s2);
int main(void)
{
char s1[LEN], s2[LEN];
printf("Please enter two strings (no more than %d characters:\n",LEN-1);
s_gets(s1, LEN);
s_gets(s2, LEN);
printf("Delete each charater in \"%s\" that matches any character in \"%s\".\n",s1,s2);
printf("\"%s\" becomes ",s1);
squeeze(s1, s2);
printf("\"%s\"\n",s1);
return 0;
}
char *s_gets(char *s, int n)
{
char *ret_val, *find;
ret_val = fgets(s, n, stdin);
if(ret_val)
{
find = strchr(s, '\n');
if(find)
*find = '\0';
else
while(getchar() != '\n')
continue;
}
return ret_val;
}
void squeeze(char *s1, char *s2)
{
int i, j, k;
for(i = 0; s2[i] != '\0'; i++)
{
for(j = 0, k = 0; s1[j] != '\0'; j++)
if(s1[j] != s2[i])
s1[k++] = s1[j];
s1[k] = '\0';
}
}
2.5 Write the function any(s1,s2), which returns the first location in a string s1 where any character from the string s2 occurs, or -1 if s1 contains no characters from s2. (The standard library function strpbrk does the same job but returns a pointer to the location.)
#include
#include
#define LEN 41
char *s_gets(char *s, int n);
int mystrpbrk(char *s1, char *s2);
int main(void)
{
char s1[LEN], s2[LEN];
int pos;
printf("Please enter two strings (no more than %d characters):\n",LEN-1);
s_gets(s1, LEN);
s_gets(s2, LEN);
pos = mystrpbrk(s1, s2);
if(pos != -1)
printf("The first character in \"%s\" that also belongs to \"%s\" is %c.\n",s1, s2, s1[pos]);
else
printf("\"%s\" does not have the same characters as \"%s\".\n",s1, s2);
return 0;
}
char *s_gets(char *s, int n)
{
char *ret_val, *find;
ret_val = fgets(s, n, stdin);
if(ret_val)
{
find = strchr(s, '\n');
if(find)
*find = '\0';
else
while(getchar() != '\n')
continue;
}
return ret_val;
}
int mystrpbrk(char *s1, char *s2)
{
int i, j;
for(i = 0; s1[i] != '\0'; i++)
for(j = 0; s2[j] != '\0'; j++)
if(s1[i] == s2[j])
return i;
return -1;
}
2.6 Write a function setbits(x,p,n,y) that returns x with the n bits that begin at position p set to the rightmost n bits of y, leaving the other bits unchanged.
setbits(x,p,n,y) 函数返回对x执行下列操作后的结果值: 将x中从第p位开始的n个(二进制)位设置为y中最右边n位的值，x的其余各位保持不变。
分析：
- 将 x 中需要改变的第 p 位到第 p-n+1 位置0，其余不变。
mask1 = ~(~(~0 << n) << (p-n+1));
x = x & mask1;
- 将 y 的最低 n 位移到 p 到 p-n+1 位处，其余位全为0
y = y & (~(~0 << n)) << (p-n+1);
- 将 y 的 p 到 p-n+1 位的值替换 x 对应的位，其余不变
x = x | y; // 0 | x = x;
unsigned setbits(unsigned x, int p, int n, unsigned y)
{
unsigned mask1;
mask1 = ~(~(~0 << n) << (p-n+1));
x = x & mask1;
y = y & (~(~0 << n)) << (p-n+1);
x = x | y; // 0 | x = x;
return x;
}
2.7 Write a function invert(x,p,n) that returns x with the n bits that begin at position p inverted (i.e., 1 changed into 0 and vice versa), leaving the others unchanged.
invert(x,p,n) 函数返回对x执行下列操作后的结果值: 将x中从第p位开始的n个(二进制)位取反，x的其余各位保持不变。
分析：
取反考虑异或 ^，有如下关系：
1 ^ 1 == 0; // 1 与 x 异或，使 x 取反
1 ^ 0 == 1;
0 ^ 0 == 0; // 0 与 x 异或， x 不变
0 ^ 1 == 1;
因此，得到：
mask = ~(~0 << n) << (p-n+1); // 将 p 到 p-n+1 置1，其余0;
x = x ^ mask;
函数：
unsigned invert(unsigned x, int p, int n)
{
unsigned mask1;
mask1 = ~(~(~0 << n) << (p-n+1));
x = x ^ mask;
return x;
}
2.8 Write a function rightrot(x,n) that returns the value of the integer x rotated to the right by n positions.
rightrot(x,n) 函数返回将x循环右移(即从最右端移出的位将从最左端移入)n(二进制)
位后所得到的值。
分析：
- 存储右移 n 位后最低端的 n-1 位到 0 位，用于以后将其填充到最高位。
unsigned y = ~(~0 << n) & x;
- 将第一步存储的数值移到最高位端，因此需要知道总共比特数
int len(void)
{
unsigned i
int len = 0;
for(i = 1; i != 0; i << 1)
++len;
return len;
}
// len 就是unsigned 的位数
y = y << len - n;
- 将 y 存储的高位值替换 x 右移后空出的位
x = y | (x >> n);
函数：
int len(void)
{
unsigned i
int len = 0;
for(i = 1; i != 0; i << 1)
++len;
return len;
}
unsigned y = ~(~0 << n) & x;
y = y << len - n;
x = y | (x >> n);
return x;
2.9 In a two’s complement number system, x &= (x-1) deletes the rightmost 1-bitin x. Explain why. Use this observation to write a faster version of bitcount.
分析：
- x - 1会将原来 x 的最后一个1变为0，因此 x & (x - 1)会将 x 中最右边一个1 去掉。
- bitcount() 函数统计数值的二进制表示中值为 1 的个数。原来书例子中的实现方式为：
int bitcount(unsigned x)
{
int b;
for(b = 0; x != 0; x >>= 1)
if (x & 01)
b++;
return b;
}
改进：
int bitcount(unsigned x)
{
int b = 0;
while(x != 0)
{
b++;
x &= (x - 1);
}
return b;
}
2.10 Rewrite the function lower, which converts upper case letters to lower case, with a conditional expression instead of if-else.
将一个字符串中的大写字符转化为小写字符，ASCLL 码中大些字母范围为 65 ～ 90，小写字母范围为 97 ～ 122，对于大些字母 c，将其转化为对应的小写字母：c + ‘a’ - 'A‘
void * lower(char *s)
{
for(int i = 0; s[i] != '\0'; i++)
s[i] = (s[i] >= 'A' && s[i] <= 'Z') ? (s[i] + 'a' - 'A') : s[i]);
}