我需要将字符串“ on”替换为“ in”，strstr（）函数返回一个指向字符串的指针，所以我认为将新值分配给该指针可以正常工作，但是没有实现
#include <stdio.h>
#include <string.h>
int main(void) {
char *m = "cat on couch";
*strstr(m, "on") = "in";
printf("%s\n", m);
}
答案 0 :(得分：4)
如果两个子串的长度相同，则用另一个子串替换很容易：
strstr memcpy用新的子字符串覆盖它。*strstr(m, "on") = "in";分配指针是不正确的，应该会生成编译器警告。您可以使用gcc -Wall -Werror避免此类错误。char的初始化数组才能对其进行修改。这是更正的版本：
#include <stdio.h>
#include <string.h>
int main(void) {
char m[] = "cat on couch";
char *p = strstr(m, "on");
if (p != NULL) {
memcpy(p, "in", 2);
}
printf("%s\n", m);
return 0;
}
如果替换的时间较短，则代码会稍微复杂一些：
#include <stdio.h>
#include <string.h>
int main(void) {
char m[] = "cat is out roaming";
char *p = strstr(m, "out");
if (p != NULL) {
memcpy(p, "in", 2);
memmove(p + 2, p + 3, strlen(p + 3) + 1);
}
printf("%s\n", m);
return 0;
}
在一般情况下，它甚至更加复杂，并且数组必须足够大以适应长度差：
#include <stdio.h>
#include <string.h>
int main(void) {
char m[30] = "cat is inside the barn";
char *p = strstr(m, "inside");
if (p != NULL) {
memmove(p + 7, p + 6, strlen(p + 6) + 1);
memcpy(p, "outside", 7);
}
printf("%s\n", m);
return 0;
}
这是处理所有情况的通用函数：
#include <stdio.h>
#include <string.h>
char *strreplace(char *s, const char *s1, const char *s2) {
char *p = strstr(s, s1);
if (p != NULL) {
size_t len1 = strlen(s1);
size_t len2 = strlen(s2);
if (len1 != len2)
memmove(p + len2, p + len1, strlen(p + len1) + 1);
memcpy(p, s2, len2);
}
return s;
}
int main(void) {
char m[30] = "cat is inside the barn";
printf("%s\n", m);
printf("%s\n", strreplace(m, "inside", "in"));
printf("%s\n", strreplace(m, "in", "on"));
printf("%s\n", strreplace(m, "on", "outside"));
return 0;
}
答案 1 :(得分：2)
此方法存在一些问题。首先，关闭swapouts指向只读内存，因此尝试覆盖那里的内存是未定义的行为。
第二行：$ top -l 1 | head
Processes: 797 total, 4 running, 1 stuck, 792 sleeping, 1603 threads
2019/05/08 09:48:40
Load Avg: 54.32, 41.08, 34.69
CPU usage: 62.2% user, 36.89% sys, 1.8% idle
SharedLibs: 258M resident, 65M data, 86M linkedit.
MemRegions: 78888 total, 6239M resident, 226M private, 2045M shared.
PhysMem: 15G used (2220M wired), 785M unused.
VM: 3392G vsize, 1299M framework vsize, 0(0) swapins, 0(0) swapouts.
Networks: packets: 24484543/16G in, 24962180/7514M out.
Disks: 21281572/769G read, 20527776/242G written.
不会更改指向的字符串，而是重新分配指针。
解决方案：
$ df -h
Filesystem Size Used Avail Use% Mounted on
/dev/disk1s1 466G 444G 19G 97% /
/dev/disk1s4 466G 3.1G 19G 14% /private/var/vm
/dev/disk2s1 932G 546G 387G 59% /Volumes/usbhd
com.apple.TimeMachine.2019-05-06-225547@/dev/disk1s1 466G 441G 19G 96% /Volumes/com.apple.TimeMachine.localsnapshots/Backups.backupdb/py???s MacBook Air/2019-05-06-225547/Macintosh HD
com.apple.TimeMachine.2019-05-02-082105@/dev/disk1s1 466G 440G 19G 96% /Volumes/com.apple.TimeMachine.localsnapshots/Backups.backupdb/py???s MacBook Air/2019-05-02-082105/Macintosh HD
请注意，如果您只是使用普通m，它将在strstr(m, "on") = "in"之后终止，因此#include <stdio.h>
#include <string.h>
int main(void)
{
char m[] = "cat on couch";
memcpy(strstr(m, "on"), "in", 2);
printf("%s\n", m);
}
是必需的。 strcpy也可以使用，但是在使用this discussion之前，您应该先阅读它。
还应该知道，如果要处理程序中不是硬编码常量的字符串，则应始终检查"cat in"，memcpy的返回值以及与之相关的函数strncpy。
答案 2 :(得分：1)
此函数使用替换字符串对子字符串的所有实例执行通用模式替换。它为结果分配正确大小的缓冲区。对于与javascript replace（）语义相对应的空子字符串的情况，行为已得到很好的定义。尽可能使用memcpy代替strcpy。
/*
* strsub : substring and replace substring in strings.
*
* Function to replace a substring with a replacement string. Returns a
* buffer of the correct size containing the input string with all instances
* of the substring replaced by the replacement string.
*
* If the substring is empty the replace string is written before each character
* and at the end of the string.
*
* Returns NULL on error after setting the error number.
*
*/
char * strsub (char *input, char *substring, char *replace)
{
int number_of_matches = 0;
size_t substring_size = strlen(substring), replace_size = strlen(replace), buffer_size;
char *buffer, *bp, *ip;
/*
* Count the number of non overlapping substring occurences in the input string. This
* information is used to calculate the correct buffer size.
*/
if (substring_size)
{
ip = strstr(input, substring);
while (ip != NULL)
{
number_of_matches++;
ip = strstr(ip+substring_size, substring);
}
}
else
number_of_matches = strlen (input) + 1;
/*
* Allocate a buffer of the correct size for the output.
*/
buffer_size = strlen(input) + number_of_matches*(replace_size - substring_size) + 1;
if ((buffer = ((char *) malloc(buffer_size))) == NULL)
{
errno=ENOMEM;
return NULL;
}
/*
* Rescan the string replacing each occurence of a match with the replacement string.
* Take care to copy buffer content between matches or in the case of an empty find
* string one character.
*/
bp = buffer;
ip = strstr(input, substring);
while ((ip != NULL) && (*input != '\0'))
{
if (ip == input)
{
memcpy (bp, replace, replace_size+1);
bp += replace_size;
if (substring_size)
input += substring_size;
else
*(bp++) = *(input++);
ip = strstr(input, substring);
}
else
while (input != ip)
*(bp++) = *(input++);
}
/*
* Write any remaining suffix to the buffer, or in the case of an empty find string
* append the replacement pattern.
*/
if (substring_size)
strcpy (bp, input);
else
memcpy (bp, replace, replace_size+1);
return buffer;
}
出于测试目的，我包括一个使用替换功能的主程序。
#define BUFSIZE 1024
char * read_string (const char * prompt)
{
char *buf, *bp;
if ((buf=(char *)malloc(BUFSIZE))==NULL)
{
error (0, ENOMEM, "Memory allocation failure in read_string");
return NULL;
}
else
bp=buf;
printf ("%s\n> ", prompt);
while ((*bp=getchar()) != '\n')bp++;
*bp = '\0';
return buf;
}
int main ()
{
char * input_string = read_string ("Please enter the input string");
char * pattern_string = read_string ("Please enter the test string");
char * replace_string = read_string ("Please enter the replacement string");
char * output_string = strsub (input_string, pattern_string, replace_string);
printf ("Result :\n> %s\n", output_string);
free (input_string);
free (pattern_string);
free (replace_string);
free (output_string);
exit(0);
}