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/* | |
* Exercise 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 <stdio.h> | |
#include <stdlib.h> | |
void squeeze(char s1[], const char s2[]); | |
int indexof(const char s[], char c); |
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/* | |
* Exercise 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 <stdio.h> | |
int any(const char s1[], const char s2[]); |
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/* | |
* Exercise 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. | |
*/ | |
#include <stdio.h> | |
unsigned int setbits(unsigned int x, int p, int n, unsigned int y); | |
unsigned int setbits(unsigned int x, int p, int n, unsigned int y) { |
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/* | |
* Exercise 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. | |
*/ | |
#include <stdio.h> | |
unsigned int invert(unsigned int x, int p, int n); | |
unsigned int invert(unsigned int x, int p, int n) { |
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/* | |
* Exercise 2-9. In a two's complement number system, x &= (x-1) deletes the | |
* right most 1-bit in x. Explain why. Use this observation to write a faster | |
* version of bitcount. | |
*/ | |
#include <stdio.h> | |
int bitcount(unsigned int x); | |
int bitcount(unsigned int x) { |
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/* | |
* Exercise 2-8. Write a function rightrot(x,n) that returns the value of the | |
* integer x rotated to the right by n positions. | |
*/ | |
#include <stdio.h> | |
unsigned int rightrot(unsigned int x, int n); | |
unsigned int rightrot(unsigned int x, int n) { | |
while (n > 0) { |
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/* | |
* Exercise 3-2. Write a function escape(s,t) that converts characters like | |
* newline and tab into visible escape sequences like \n and \t as it copies | |
* the string t to s. Use a switch. Write a function for the other direction | |
* as well, converting escape sequences into the real characters. | |
*/ | |
#include <stdio.h> | |
#define MAXLEN 1000 | |
int escape(char s[], const char t[]); |
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#include <ctype.h> | |
/* | |
* 将十进制字符串转换为数值。可处理前导的 + 和 - | |
* | |
* 如果有空白符,则跳过 | |
* 如果有符号,则读取符号 | |
* 取整数部分,并执行转换 | |
* 当遇到第一个不能转换为数字的字符时,整个处理过程终止 | |
*/ |
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/* | |
* Exercise 3-3. Write a function expand(s1,s2) that expands shorthand | |
* notations like a-z in the string s1 into the equivalent complete list | |
* abc...xyz in s2. Allow for letters of either case and digits, and be | |
* prepared to handle cases like a-b-c and a-z0-9 and -a-z. Arrange that | |
* a leading or trailing - is taken literally. | |
*/ | |
#include <stdio.h> |
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/* | |
* Exercise 3-4. In a two's complement number representation, our version of | |
* itoa does not handle the largest negative number, that is, the value of n | |
* equal to -(2^wordsize - 1). Explain why not. Modify it to print that value | |
* correctly, regardless of the machine on which it runs. | |
*/ | |
#include <stdio.h> | |
#include <limits.h> | |
#include <string.h> |