尝试将指针指向函数中的指针时出现段错误 [英] Segfault when trying to index pointer to pointers in function

问题描述

我正在尝试使用数组(malloc-ed)，即自定义结构的arr.该数组通过引用传递给函数.每当我在运行时尝试在函数中索引除arr[0]以外的任何内容时，都会遇到段错误(例如(*arr[1])->i = 3;).为什么会这样?

I'm trying to do something with an array (malloc-ed), namely arr of a custom struct. The array is passed by reference to a function. I get a segfault whenever I tried to index anything other than arr[0] in the function at runtime (e.g (*arr[1])->i = 3;). Why is this happening?

完整的源代码是:

#include <stdio.h>
#include <stdlib.h>
#define SIZE 100

typedef struct{
    int i;
    float f;
}foo;

void doSomething(foo ***arr);

int main()
{
  foo **arr = (foo**) malloc (SIZE * sizeof(foo*));
  int i;
  for(i = 0; i < SIZE; i++)
    arr[i] = (foo*)malloc(sizeof(foo));

  arr[1]->i = 1;
  printf("Before %d\n",arr[1]->i );
  doSomething(&arr);
  printf("After %d\n",arr[1]->i );
  return 0;
}

void doSomething(foo ***arr)
{
  (*arr[1])->i = 3;
}

推荐答案

您的问题是那行

(*arr[1])->i = 3;

因为下标运算符的评估位于之前，因此，它等于取消引用的评估以下:

Because the subscripting operator's evaluation precedes the dereferencing's evaluation it is equivalent to the following:

(*(arr[1]))->i = 3;

这显然是错误的.您需要

This is obviously wrong. You need

(*arr)[1]->i = 3;

因此.

注意:

• 不转换malloc 的结果
• 添加#include <stdlib.h>来解决警告
• 无需添加额外的间接级别(foo***指向foo**)；只需按值复制

• (除了大写字母外)，对于您而言，一个好的旧一维数组实际上应该足够了

• do not cast the result of malloc
• add #include <stdlib.h> to resolve the warning
• adding an extra level of indirection (foo*** pointing to foo**) is unnecessary; just copy by value

• (in addition to the upper note) a good old 1D array should actually be sufficient in your case

在malloc

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