查找子字符串并在C中替换它

Question

#include <stdio.h>
#include <assert.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char *argv[]) {     

char * str = "Testing replace text...\n\n";


        char* buffer = malloc(sizeof(char));
        char* insertPoint = &buffer[0];
        char* copy = str; 
        char* p = strstr(str, "epl");
        char* g = "gard";
        int size = 0; 

        size = p-copy; //p = 9, which is the number of elemts till the first element of the substring
                       //want to allocate this space, and then increment insertPoint, by that amt(it'll be pointing
                       // to nothing)
        buffer = realloc(buffer, size);
        printf("Size: %d\n", size);
        memcpy(insertPoint, copy, size);
        printf("COPY: %s\n", buffer);

        copy += size;
        buffer = realloc(buffer, size+strlen(g));
        insertPoint += size;
        printf("%c", *insertPoint);
        memcpy(insertPoint, g, strlen(g)); //insert after the 9 letters, the string the size of g
        size += strlen(g); //size if the size of the buffer
        printf("Size2: %d\n", size);
        printf("COPY2: %s\n", buffer);

  return EXIT_SUCCESS;
}

只是一些快速的实验代码;我只是试图用“gard”替换str中的子串epl，但当我打印出来时，我正在打印的字符串缓冲区没有变化，这意味着第一个字符串im打印工作，它将所有字母放入缓冲区之前发生子字符串，但是当我尝试用子字符串替换时，它不起作用。我正在测试各个指针，它们看起来都是正确的......不确定发生了什么，有什么见解？谢谢......完全可运行的程序。

Answer 1

我认为代码中的问题是因为strlen不包括终止零。我试图修复你的代码，但最后我发现重新编写它更容易（并使用更合理的变量名称）。

以下简单的四个步骤可行。 strlen的连续使用可能会被变量所取代，但为了清楚起见，我将它们留下了。 （另外，一个好的编译器可以通过保留调用来很好地优化这段代码。）

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int main(int argc, char *argv[])
{     
    char *str = "Testing replace text...\n\n";
    char* buffer;
    char* find_str = "epl";
    char* repl_str = "gard";

    char *find_str_pos = strstr (str, find_str);

    /* 1. create new buffer of the correct size */
    buffer = malloc (strlen(str) - strlen(find_str) + strlen(repl_str) + 1);
    /* 2. copy first part */
    memcpy (buffer, str, find_str_pos - str);
    /* 3. add new text */
    memcpy (buffer + (find_str_pos - str), repl_str, strlen(repl_str));
    /* 4. append original text */
    memcpy (buffer + (find_str_pos - str) + strlen(repl_str), find_str_pos + strlen(find_str), strlen(find_str_pos) - strlen(repl_str) + 1);

    printf ("-> [%s]\n", buffer);

    return EXIT_SUCCESS;

}

Answer 2

替换“epl”后，您不会附加剩余文本“ace text ...”。

你能试试下面的代码吗？

#include<stdio.h>
#include<string.h>
#include<stdlib.h>

int main() {

    char *source_string = "Testing replace text...";
    char *search_string = "epl";
    char *replace_string = "gard";

    int search_length = strlen(search_string);
    int replace_length = strlen(replace_string);

    // Find start position of search string
    char *start = strstr(source_string, search_string); // start pointing to "eplace text..."
    int intial_length = start - source_string; // intial_length = 9

    // Get remaining text which should append after replace
    char *remaining_string = (start + search_length); // remaining_string pointing to "ace text..."
    int remaining_length = strlen(remaining_string); // remaining_length = 11

    // Find total length of string after replacing text
    int total_string_length = intial_length + replace_length + remaining_length; // 24

    char *buffer = (char *)malloc(total_string_length + 1); // +1 for null pointer
    char *current_index = buffer;

    // Add initial text
    memcpy(current_index, source_string, intial_length);
    current_index += intial_length;

    // Add replace text
    memcpy(current_index, replace_string, replace_length);
    current_index += replace_length;

    // Add remaining text
    memcpy(current_index, remaining_string, remaining_length);
    current_index += remaining_length;

    memcpy(current_index, "\0", 1); // add null pointer at last
    printf("Final Output: %s", buffer); // Final Output: Testing rgardace text...

    return 0;
}