在调用函数中取消引用指针时出现分段错误

Question

When I run the following code, I get the segmentation error fault and I cannot rectify my mistake.

    void test(int *c)
{
    c++;
    *c = 10;
    cout<<*c<<endl;
}

int main()
{
    int a =2;
    int *b = &a;
    test(b);
    cout<<*b;
    return 0;
}

I think that b should point to value '2'. But instead it gives an error.

Answer 1

With c++ you are incrementing the value of the pointer and not the value of the integer. It will point to a new location that was not allocated by you, and it will raise the segmentation fault error. The new location may be used by an other program or by the system.....

Answer 2

For what you move the pointer in the test ? You move it behind the memory of the variable a and reach undefined behavior.

  c
  |
+---+---+
| 2 |   |
+---+---+
      |
     ++c

Answer 3

void test(int *c)
{
   c++;
   *c = 10;

*c = 10 is undefined behavior since you are writing into memory that does not belong to you.

Answer 4

In function test , 'c' has address of 'a' say 1000. Now you increment this address say it gets to 1004 ,now this memory address is not allocated by you .

When you try to access memory which was not allocated by you , Segmentation fault occurs .

Now when you try to print *c ,it might print 10 but it is undefined behavior which can even crash the PC at times.

To get a better idea ,just think about free() when you free() the memory and then access a variable you might still get the value but it is undefined behavior ,anything can happen .