Why in C does the function sizeof() output the size of right most operand when more than one operands are passed separated by comma?

Question

I have the following code in C:

#include <stdio.h> 

void main() {
    printf("%d %d\n",sizeof(5),sizeof(5,5));   
    printf("%d %d\n",sizeof(5),sizeof(5.0,5)); 
    printf("%d %d\n",sizeof(5),sizeof(5,5.0)); 
}

And I get the output:

4 4

4 4

4 8

I understand that sizeof(5) would return me the size of integer and sizeof(5.0) would return the size of a double, but why does it give the size of the rightmost operand in case more than one arguments are passed separated by comma? Why not the first argument or the collective size of all the arguments?

I am compiling online using OnlineGDB.com compiler for C.

Thanks for your help.

btw sizeof is not a function, it's an operator https://en.cppreference.com/w/c/language/sizeof — Mat, Apr 05 '20 at 19:40
... also because it is an operator as @Mat points out, and it takes _only one_ argument, the comma here is the _comma operator_ in a _single expression_ not a function parameter delimiter. There are not two arguments, there is a single argument that is an expression involving the comma operator. — Clifford, Apr 05 '20 at 19:47
@Clifford thanks for elaborating. That is a very useful clarification for me. — frodobuddy, Apr 05 '20 at 20:06
@Clifford: Because `sizeof` is an operator, it does not have any arguments or parameters, just one operand. — Eric Postpischil, Apr 05 '20 at 20:26
I'm reopening this question because the only answer up-to-date does not touch on the core of the misconception in the question: The nature of the `sizeof` operator. I've also added an answer that directly addresses this misconception, as it is a very common one. — cmaster - reinstate monica, Apr 05 '20 at 21:57
It is always relevant, and here it indicates that you don't care much about *details*. — wildplasser, Apr 06 '20 at 14:50

cmaster - reinstate monica · Accepted Answer · 2020-04-05T21:58:15.287

The simple reason is: Because sizeof is not a function! It is an operator that takes some expression on its right. Syntactically, it behaves the same as the return operator. The parentheses are only added by the programmers for clarity, and are not needed in most cases:

sizeof(foo);       //no surprise, take the size of the variable/object
sizeof foo;        //same as above, the parentheses are not needed

sizeof(void*);     //you cannot pass a type to a function, but you can pass it to the sizeof operator
sizeof void*;      //same as above

typedef char arrayType[20]
arrayType* bar;    //pointer to an array
sizeof(*bar);      //you cannot pass an array to a function, but you can pass it to the sizeof operator
sizeof*bar;        //same as above

//compare to the behavior of `return`:
return foo;     //no surprise
return(foo);    //same as above, any expression may be enclosed in parentheses

So, what happens when you say sizeof(5, 5.0)? Well, since sizeof is an operator, the parentheses are not a function call, but rather interpreted like the parentheses in 1*(2 + 3) == 5. In both cases, the ( follows an operator, and is thus not interpreted as a function call. As such, the comma does not separate function call arguments (because there is no function call), rather it's interpreted as the comma operator. And the comma operator is defined to evaluate both its operands, and then return the value of the last operand. The operator nature of the sizeof dictates how the expression on its right is parsed.

Ardent Coder · Answer 2 · 2020-04-05T19:50:27.103

2

Because the associativity of the comma operator is left to right.

Only the rightmost expression is used and the rest are discarded (although it's side effects have to do with sequencing).

Therefore,

sizeof(5.0,5) is equivalent to sizeof(5)

and

sizeof(5,5.0) is equivalent to sizeof(5.0)

edited Apr 05 '20 at 19:50

answered Apr 05 '20 at 19:39

Ardent Coder

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Why in C does the function sizeof() output the size of right most operand when more than one operands are passed separated by comma?

2 Answers2