How to interpret assignment to pointer in C-CodePudding

In C, the statement int *var_name; by convention means that *var_name is of type int, which implies that var_name is an integer pointer.

But now consider this sequence of statements.

int a = 5;
int *var_name = &a;

But here how are we assigning address of a to *var_name which is of type int?

CodePudding user response：

You are not assigning the address to *var_name, but to var_name. And this variable is of type "pointer to int", as you found.

Oh, and it is an initialization, not an assignment.

CodePudding user response：

That's just the way the syntax in C works.

int *p = &a;

is completely equivalent to:

int *p;
p = &a;

It's not so difficult to understand why it would not make sense if it worked the way you thought. Consider this code:

int *p;
p = &a;
*p = 42;

Now imagine flipping the last two statements, like this:

int *p;
*p = 42; // To which address do we write 42? p is not initialized.
p = &a;

So although the meaning of int *p = &a;, which is int *p; p = &a; might be unintuitive, it's actually the only sane way to define it.

CodePudding user response：

In this declaration

int *var_name = &a;

there is declared the variable var_name of the pointer type int * and this variable (pointer) is initialized by the address of the variable a.

It seems you mixing the symbol * in declarations and in expressions.

In declarations it denotes a pointer. In expressions it denotes the dereference operator.

Consider the following demonstration program

#include <stdio.h>

int main( void )
{
    int a = 5;
    int *var_name = &a;

    printf( "The address stored in var_name is %p\n", ( void * )var_name );
    printf( "The address of the variable a  is %p\n", ( void * )&a );

    printf( "The value of the object pointed to by var_name is %d\n", *var_name );
    printf( "The value stored in the variable a is %d\n", a );
}

Its output might look like

The address stored in var_name is 00EFFAC0
The address of the variable a  is 00EFFAC0
The value of the object pointed to by var_name is 5
The value stored in the variable a is 5

That is in this line

    int *var_name = &a;

*var_name means a declarator of a pointer type.

In this call

printf( "The value of the object pointed to by var_name is %d\n", *var_name );

*var_name means an expression with teh dereference operator *.