在的malloc函数 - 分段错误 [英] malloc in function — segmentation error

问题描述

本程序完美的作品：

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

#define MAX_NUM 2

int tempfunction (char **comments)
{
    char str1[]="First string\n";
    char str2[]="This is the second string\n";

    *(comments+0)=(char *) malloc(strlen(str1)+1);
    *(comments+1)=(char *) malloc(strlen(str2)+1);

    strcpy(*(comments+0), str1);
    strcpy(*(comments+1), str2);
    return 0;
}

int main(void)
{
    char **comments;

    /* This is the section I am talking about */
    comments=(char **) malloc(MAX_NUM*sizeof(char *));
    if (comments==NULL)
    {
        printf("\n### ERROR: malloc failed.\n");
        exit(EXIT_FAILURE);
    }
    /* Upto here............................. */

    tempfunction(comments);
    printf("%s%s", comments[0], comments[1]);
    return 0;
}

但为了将来方便，我愿把的malloc在 tempfunction 在部分。当我这样做，我得到一个分段错误。

But for future convenience I would like to put the malloc section inside the tempfunction. When I do that, I get a segmentation fault error.

我想这可能是由于初始化这样的，而不是的char **意见; 我写的：

I thought it might be due to initialization, so instead of char **comments; I write:

char a = 'a';
char *P = &a;
char **comments = &P;

但是，它仍然无法正常工作。我会很感激，如果你能帮助我理解为什么会这样，以及如何解决它。

But, it still doesn't work. I would be very grateful if you could help me understand why this happens and how to fix it.

推荐答案

尝试：

int tempfunction (char ***comments)
{
  char str1[]="First string\n";
  char str2[]="This is the second string\n";

  *comments = malloc(MAX_NUM * sizeof(**comments)); /* you must check the return of malloc */

  (*comments)[0] = strdup(str1);
  (*comments)[1] = strdup(str2);

  return 0;
}

和你调用它像：

tempfunction(&comments);

当然，你必须在最后，以避免内存泄漏释放

Of course you'll have to free at the end in order to avoid memory leaks

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