C char 数组、指针、malloc、free

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我正在尝试理解指针，我有这个简单的例子

void third(char ****msg) {
    ***msg = malloc(5 * sizeof (char));
    printf("\nthe msg in third is :%s ", ***msg);
    strcpy(***msg, "third");
    printf("\nthe msg in third after is: %s", ***msg);
    // free(***msg);
}

void change(char***msg) {
    **msg = malloc(5 * sizeof (char));
    printf("\nthe msg in change is :%s ", **msg);
    strcpy(**msg, "change");
    printf("\nthe msg in change after is: %s", **msg);
    third(&msg);
    //   free(**msg);
}

void test(char ** msg) {
    *msg = malloc(5 * sizeof (char));
    printf("\n the msg in test is: %s", *msg);

    strcpy(*msg, "test");
    printf("\nthe msg in test after is: %s\n", *msg);

    change(&msg);
    free(*msg);
}

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

    char * msg;
    test(&msg);
    printf("\nthe msg back in main is: %s", msg);
}

我可以说它工作正常，但是你能告诉我何时以及如何需要释放分配的内存吗？因为如果我从函数更改和第三个中删除//并运行它，我会遇到错误。有没有办法在每个函数的第一个打印语句中获取消息的内容 - 请参阅 otuput:

the msg in test is: 
the msg in test after is: test
the msg in change is :0�� 
the msg in change after is: change
the msg in third is :P�� 
the msg in third after is: third
the msg back in main is: 

有没有办法让消息发生变化:测试然后第三个消息是:更改

Answer 1

忘记那个程序吧，它有太多问题了:

• 您不需要多层间接。两级就够了。如果该函数需要更改地址，只需将指针传递给下一个函数即可。
• 您尝试在字符串初始化之前打印该字符串。
• 您尝试在释放字符串后使用该字符串。
• 通过重复调用 malloc 而不清理旧内容，会造成内存泄漏。请改用 realloc。
• 您分配的内存量不正确，因此数组不够大，无法容纳您 strcpy 到其中的字符串。另请注意，您需要为空终止符分配足够的空间。

等等。这是一个固定版本:

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

void third(char** msg) {
    const char str[] = "third";

    *msg = realloc(*msg, sizeof(str));
    printf("the msg in third is :%s\n", *msg);

    strcpy(*msg, str);
    printf("the msg in third after is: %s\n", *msg);
}

void change(char** msg) {
    const char str[] = "change";

    *msg = realloc(*msg, sizeof(str));
    printf("the msg in change is :%s\n", *msg);

    strcpy(*msg, str);
    printf("the msg in change after is: %s\n", *msg);

    third(msg);
}

void test(char** msg) {
    const char str[] = "test";

    *msg = malloc(sizeof(str));
    printf("the msg in test is just garabage at this point, no need to print it.\n");

    strcpy(*msg, str);
    printf("the msg in test after is: %s\n", *msg);

    change(msg);
}

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

    char* msg;
    test(&msg);
    printf("the msg back in main is: %s\n", msg);

    free(msg);
}