查找并替换C中出现的所有子字符串
查找并替换C中出现的所有子字符串
提问于 2016-09-08 22:23:51
我试图在C中的字符串数组中找到并替换所有出现的子字符串，我想我已经有了大部分的逻辑，但是我不知道我在哪里搞砸了剩下的部分。
下面是相关的代码--我要替换的字符串在searchStr中，我试图用replaceStr替换它。字符串数组称为buff。之后，我不需要将修改后的字符串保存回数组中，只需将修改后的字符串打印到控制台。
for (size_t i = 0; i < numLines; i++) {
char *tmp = buff[i];
char finalStr[MAX_STR_LEN * 2];
char temporaryString[MAX_STR_LEN];
int match = 0;
while ((tmp = strstr(tmp, searchStr))) {
match = 1;
char temporaryString[MAX_STR_LEN];
char tmp2[MAX_STR_LEN];
printf("Buff[i]: %s", buff[i]);
sprintf(temporaryString, "%s", strstr(tmp, searchStr) + strlen(searchStr)); // Grab everything after the match
printf("Behind: %s", temporaryString);
strncpy(tmp2, buff[i], tmp - buff[i]); // Grab everything before the match
strcat(finalStr, tmp2);
printf("In Front: %s\n", finalStr);
strcat(finalStr, replaceStr); // Concat everything before with the replacing string
tmp = tmp + strlen(searchStr);
buff[i] = tmp; // Move buff pointer up so that it looks for another match in the remaining part of the string
}
if (match) {
strcat(finalStr, temporaryString); // Add on any remaining bytes
printf("Final: %s\n", finalStr);
}
}如果有大量的printf在那里，那么我就可以看到所有需要调试的地方。
示例案例：
如果我在字符串what4is4this上运行searchStr = 4和replaceStr = !!!，这是控制台中的输出.我也在使用//添加注释
Buff[i]: what4is4this // Just printing out the current string before we attempt to replace anything
Behind: is4this // Looking good here
In Front: hat // Why is it cutting off the 'w'?
Buff[i]: is4this // Good - this is the remaining string we need to look through
Behind: this // Again, looking good
In Front: hat!!!isat // It should be 'is'
Final: hat!!!isat! // final should be 'what!is!!!this'伙计们有什么想法吗？我把头发扯掉，试图解决这个问题
谢谢!
回答 1
Stack Overflow用户
发布于 2016-09-08 23:39:51
这是一个不健康的组合指针杂耍和未定义的行为，但评论者已经告诉你。如果你把它简化一点，好好地、节俭地使用指针，你就可以做以下事情：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
// ALL CHECKS OMMITTED!
#define MAX_STR_LEN 1024
int main(int argc, char **argv)
{
char *buff, *cpbuff;
char *searchStr;
char *replaceStr;
// pointers too the two parts with the search string in between
char *tmp, *after;
// the final output (a fixed length is not good,
// should be dynamically allocated)
char finalStr[MAX_STR_LEN * 2] = { '\0' };
if (argc < 4) {
fprintf(stderr, "Usage: %s string searchstr replacestr\n", argv[0]);
exit(EXIT_FAILURE);
}
buff = malloc(strlen(argv[1]) + 1);
strcpy(buff, argv[1]);
searchStr = malloc(strlen(argv[2]) + 1);
strcpy(searchStr, argv[2]);
replaceStr = malloc(strlen(argv[3]) + 1);
strcpy(replaceStr, argv[3]);
// Keep a finger on the start of buff
cpbuff = buff;
while (1) {
printf("Buff: %s\n", buff);
// Grab everything after the match
after = strstr(buff, searchStr);
// No further matches? Than we're done
if (after == NULL) {
strcat(finalStr, buff);
break;
}
// assuming strlen(searchStr) >= 1
tmp = buff;
// mark the end of the first part
tmp[after - buff] = '\0';
// set the after pointer to the start of the second part
after = after + strlen(searchStr);
printf("Behind: %s\n", after);
printf("Before: %s\n\n", tmp);
// Put the first part to it's final place
strcat(finalStr, tmp);
// concat the replacement string
strcat(finalStr, replaceStr);
// Set buff to the start of the second part
buff = after + strlen(searchStr) - 1;
}
printf("Final: %s\n", finalStr);
// set the buff pointer back to it's start
buff = cpbuff;
free(buff);
free(searchStr);
free(replaceStr);
exit(EXIT_SUCCESS);
}可以称为滥用指针算法的单个点是标记第一部分结尾的行。通过测量和使用所涉及的字符串的长度并与它们一起进行算术，可以避免这一问题。它是比较慢的，承认，所以它取决于您的个别用例。
它仍然比我喜欢的复杂，但这是一个开始。
我希望您现在可以把它扩展到几个输入行。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/39401070
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