CODEWITHSUNDEEP
csizeof
LocalHost
LocalHost

Reputation: 910

Address as a operand in sizeof() operator

I was just trying an example and i tried to check the output when an address is passed as an argument in the sizeof operator and i got output of 4. Now my question is when you pass a pointer in sizeof operator why is it showing 4 bytes of memory when actually there is no pointer variable, it's just only an address?

#include<stdio.h>
int main()
{
  int a=1;
  int c;
  c=sizeof(&a);
  printf("%d\n",c);
  return 0;
}

Upvotes: 1

Views: 233

Answers (2)

paxdiablo
paxdiablo

Reputation: 882296

It's because sizeof returns the size of a type, as per C11 6.5.3.4 The sizeof and _Alignof operators /2 (my emphasis):

The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand.

The type of &a where a is an int is covered by 6.5.3.2 Address and indirection operators /3 in the same standard:

The unary & operator yields the address of its operand. If the operand has type "type", the result has type "pointer to type".

In other words, int a; sizeof(&a) is functionally equivalent to sizeof(int *).

Upvotes: 7

Sourav Ghosh
Sourav Ghosh

Reputation: 134376

sizeof operator works on the type of the operand.

Quoting C11, chapter §6.5.3.4 (emphasis mine)

The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. [....]

&a is of type int *, so your statement is the same as sizeof(int *). The result, is the size of a pointer (to integer), in your platform.

That said, sizeof produces a type of size_t as result,

  • use a variable of type size_t
  • use %zu to print the result.

Upvotes: 4

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