C指针的初始化和取消引用，这是怎么回事? [英] C pointer initialization and dereferencing, what's wrong here?

问题描述

这应该超级简单，但是我不确定为什么编译器会在这里抱怨.

This should be super simple, but I'm not sure why the compiler is complaining here.

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

int main(int argc, char *argv[])
{
  int *n = 5;

  printf ("n: %d", *n);

  exit(0);
}

收到以下投诉:

foo.c:在"main"函数中:
foo.c:6: 警告:初始化会使指针 从没有强制转换的整数

foo.c: In function ‘main’:
foo.c:6: warning: initialization makes pointer from integer without a cast

我只想打印指针n引用的值.我在printf()语句中取消引用它，但遇到了分段错误.用gcc -o foo foo.c编译.

I just want to print the value that the pointer n references. I'm dereferencing it in the printf() statement and I get a segmentation fault. Compiling this with gcc -o foo foo.c.

推荐答案

您将指针设置为内存地址5，使其指向地址5处的任何内容.您可能想使其指向存储值5的地址.例如:

You set the pointer to memory address 5, so that it points to whatever at address 5 might be. You probably wanted to make it point to an address where the value 5 is stored. For example:

int v = 5;    // Store the value 5 in a normal variable
int *n = &v;  // Make n contain the address of v, so that it points to the 
              // contents of v

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