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

问题描述

以下 C 代码段:

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

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

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

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

  f1(a1);

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

结果

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

a1为什么不在函数中保留已经分配给它的地址?

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

推荐答案

因为这是 C 语言，如果不传入指向指针的指针，就不能通过引用传递指针(例如，void **而不是 void * 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);

将指针的值 (NULL) 作为 a 的堆栈参数值推送.a 获取该值，然后为自己重新分配一个新值(malloc ed 地址).因为它是按值传递的，所以 a1 没有任何变化.

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++，你可以通过引用传递指针来实现你想要的:

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

void f(void *&a);

这篇关于void 指针作为参数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！