Why asterisk is used with data type in type conversion wit a pointer?

Question

Why the asterisk is used with the "int" keyword in a type conversion with a pointer but not with a variable?

In the following program why the asterisk is used two times with the "int" keyword while type conversion with a pointer but not with a float variable.

#include <stdio.h>

int main()
{
   float a = 2.5;
   int b;

   b = (int) a; // Why asterisk is not used with "int"?

   void *ab_ptr = &b;
    
   printf("Value of 'a' is %d\n", *(int*)ab_ptr); // Why two times asterisk is used here?

   return 0;
}

Answer 1

Maybe the following example will clarify by breaking down your code into two separate statements -

#include <stdio.h>

int main()
{
    float a = 2.5;
    int b;

    b = (int)a; // Why asterisk is not used with "int"?

    void* ab_ptr = &b;

    printf("Value of 'a' is %d\n", *(int*)ab_ptr); // Why two times asterisk is used here?

    // Use multiple statements for illustration

    int *iptr = (int*)ab_ptr;  // cast void pointer to pointer to an int

    int x = *iptr; // dereference int pointer

    printf("Value of 'a' is %d\n", x);
}

Perhaps using additional parenthesis would also clarify the meaning of the printf statemet -

    printf("Value of 'a' is %d\n", *((int*)ab_ptr)); // Why two times asterisk is used here?

   b = (int) a; // Why asterisk is not used with "int"?

No asterisk is needed because casting a variable from one type to another is not a dereferencing operation. There are no pointers involved here.

Answer 2

You need to recognize that not all asterisks are dereference operators. There a multiple cases where C uses the same character to have multiple meanings depending on context.

A pointer has two "properties" of interest. One is the value of the pointer. This the address of the object pointed to. The other is the value of the object the pointer points to.

You use the dereference operator when and only when you want to work with the value pointed to rather than the value of the pointer itself.

Your first assignment statement does not use any pointers. You access the value of directly, convert it to an int value, and store the converted value directly into b. No dereference operator needed.

The asterisk in the definition of ab_ptr is NOT a dereference operator. It used to inform the compiler that ab_ptr is a pointer rather than an object of the type pointed to.

In your call to printf, the parenthetical expression before ab_ptr is called a cast operator. It causes the compiler to generate the code necessary to convert the value of ab_ptr (the address it contains) from one type to another. The asterisk in this expression is also NOT a dereference operator. It simply specifies that the result of the conversion should be have type pointer to int and not int. Thus the result of the conversion is a pointer. You do not care what this value actually is. You want to access the value pointed to. That is why there is a dereference operator to the left of this expression.

The asterisks used in your code serve multiple purposes. The ones that are part of type specification simply identify pointers. The (one) dereference operator applied to a pointer value causes the compiler to generate code to access the value pointed to rather than the pointer value itself.

Answer 3

in the line:

   printf("Value of 'a' is %d\n", *(int*)ab_ptr);// cast ab_ptr as (int*) then deference

could have been written clearer:

printf("Value of 'a' is %d\n", (int)*ab_ptr); // cast the deference

note: the cast is required, so the compiler knows how many bytes to copy to the stack, as this is a pass by value.

Answer 4

I think no one had got my question. I want to know why asterisk is used with the datatype keyword when explicit conversion of pointer. Why this expression:

 *(int*)ab_ptr); 

can't be written as:

 **(int)ab_ptr); 

Answer 5

Hi,

**(int)ab_ptr);

(int)ab_ptr

You explicitly cast the pointer to an int. But,

*(int)ab_ptr

operand of '*' must be a pointer.

It is syntactically incompatible and your code will not compile.

Best regards,

Minxin Yu

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