c - 为什么将取消引用指针设置为等于原语是非法的？

Question

为什么设置取消引用的指针的值会引发段错误 11？为了让我的意思更清楚，请看下面的代码:

#include <stdio.h>

int *ptr;
*ptr = 2;

int main(){
  printf("%d\n", *ptr);
  return 0;
}

我以为 *ptr=2 会将指针 ptr 指向的右值设置为 2。事实不是这样吗？如果对于那些 C 专家程序员来说，这真的很简单/显而易见，我深表歉意。

是否只允许将取消引用的指针(即 *ptr)设置为某个值(如果该值具有内存地址)？即喜欢做:

int k = 7;
int *ptr = k;

然后:

*ptr = 2;

Answer 1

这里的问题是 ptr 没有指向分配的空间。请参阅以下内容:

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

int main(void){

  // Create a pointer to an integer.
  // This pointer will point to some random (likely unallocated) memory address.
  // Trying set the value at this memory address will almost certainly cause a segfault.
  int *ptr;

  // Create a new integer on the heap, and assign its address to ptr.
  // Don't forget to call free() on it later!
  ptr = malloc(sizeof(*ptr)); 

  // Alternatively, we could create a new integer on the stack, and have
  // ptr point to this.
  int value;
  ptr = &value;

  // Set the value of our new integer to 2.
  *ptr = 2;

  // Print out the value at our now properly set integer.
  printf("%d\n", *ptr);

  return 0;
}