将指针作为参数传递给C中的函数

Question

Something I stumbled upon and made me wonder. Why does this work?

void foo (int* a)
{
    int x = 3;
    *a = x;
}

int main()
{
    int a;
    foo(&a);
    return 0;
}

But this causes a segmentation fault (both on Visual Studio 2008 and gcc)?

void foo (int* a)
{
    int x = 3;
    *a = x;
}

int main()
{
    int* a;
    foo(a);
    return 0;
}

Is it something defined in the language or just an implementation issue?

Answer 1

When you declare

int* a;

You are declaring a pointer variable a but you are not making it point to anything. Then in the function, you do

*a = x;

Which dereferences the pointer and tries to assign what it points to the value of x . But since it doesn't point to anything, you get undefined behaviour, manifested in a segmentation fault.

You should do this:

int i; // the actual integer variable
int* a = &i; // a points to i

The difference between that and the first one is that int a; declares a real integer variable, then you take its address with &a and passes it to the function. The pointer a inside the function foo points to the variable a in main , and so dereferencing it and assigning to it is perfectly fine.

Answer 2

int a;

Assigns memory as soon as you declare it but this not the case with int *a;

int *a; 

is pointer declaration (MEMORY not yet allocated for that).

int *a = (int*)malloc(sizeof(int)); // allocate memory