指针初始化导致分段错误 [英] Pointer initialisation gives segmentation fault

问题描述

我写了一个C程序如下:

I wrote a C program as follows:

int *a; /* pointer variable declaration */

int b; /* actual variable declaration */

*a=11;

a=&b;/* store address of b in pointer variable*/

运行程序时出现分段错误.

It gives a segmentation fault when running the program.

我把代码改成如下:

int *a; /* pointer variable declaration */

int b; /* actual variable declaration */

a=&b;/* store address of b in pointer variable*/

*a=11;

现在一切正常.

如果有人知道，请解释为什么它在 CASE 1 中给出了分段错误.

If anyone knows please explain why it is giving a segmentation fault in CASE 1.

推荐答案

CASE .1

int *a; /* pointer variable declaration */

int b; /* actual variable declaration */

*a=11;//Not valid means you are not owner of the address where a now pointing it is unknown and accessing this will segfault/

a=&b;/* store address of b in pointer variable*/

这将是分段错误，因为您使用的地址不是有效地址，并且您存储的 11 是非法的.

This is going to be segmentation fault because the address you are using is not a valid address and there you are storing 11 which is illegal.

                               b
      +-------+            +--------+
      |       +            |   11   |
      |Unknown|            |        |
      +---+---+            +---+----+
          |                    |
          |                    |
          +                    +
          a                    a
CASE .2

int *a; /* pointer variable declaration */

int b; /* actual variable declaration */

a=&b;/* store address of b in pointer variable*/

*a=11;

现在它可以正常工作了，因为 b 的地址是有效的，而你存储的 11 是合法的.

Now its working fine because the address of b is valid an there you are storing 11 which is legal.

上述情况也不是正确的指针声明方式

Also above cases are not correct way of pointer declaration

  int *a  = NUll;
  a = malloc(sizeof(int));
  *a=5;
  free(a);//must

或

 int *a  = NUll;
 int b;
 a = &b;
 *a=5;

这将消除多次难以找到的分段错误.

This will remove segmentation fault many times which is hard to find .

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