void指针作为参数 [英] void pointer as argument

问题描述

下面的C代码片段：

[...] 
void f1(void* a){
  printf("f(a) address = %p \n",a);
  a = (void*)(int*)malloc(sizeof(int));

  printf("a address = %p \n",a);
  *(int*)a = 3;

  printf("data = %d\n",*(int*)a);
}

void f(void){
  void* a1=NULL;
  printf("a1 address = %p \n",a1);

  f1(a1);

  printf("a1 address = %p \n",a1);
  printf("Data.a1 = %d\n",*(int*)a1);
}
[...]

结果

a1 address = (nil) 
f(a) address = (nil) 
a address = 0xb3f010 
data = 3
a1 address = (nil) 
Segmentation fault (core dumped)

为什么不 A1 保持已分配给它的函数地址？

Why doesn't a1 keep the address that has been assigned to it in the function?

推荐答案

由于这是C，你不能通过引用传递指针而不指针传递到指针（如无效** ，而不是无效* 指向指针）。您需要返回新指针。发生了什么：

As this is C, you cannot pass the pointer by reference without passing in a pointer to the pointer (e.g., void ** rather than void * to point to the pointer). You need to return the new pointer. What is happening:

f(a1);

推动指针（ NULL ）的 A 值作为栈参数值。 A 拿起这个值，然后重新分配本身就是一个新的值（的malloc ED地址）。因为它是按值传递，没有为改变A1 。

Pushes the value of the pointer (NULL) as the stack parameter value for a. a picks up this value, and then reassigns itself a new value (the malloced address). As it was passed by value, nothing changes for a1.

如果这是C ++，你可以达到你做引用传递指针想要什么：

If this were C++, you could achieve what you wanted by doing passing the pointer by reference:

void f(void *&a);

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