Why is sizeof considered an operator?

Question

Why is sizeof considered an operator and not a function?

What property is necessary to qualify as an operator?

Blench

Answer 1

Because the C standard says so, and it gets the only vote.

As consequences:

• The operand of sizeof can be a parenthesised type, sizeof (int), instead of an object expression.
• The parentheses are unnecessary: int a; printf("%d\n", sizeof a); is perfectly fine. They're often seen, firstly because they're needed as part of a type cast expression, and secondly because sizeof has very high precedence, so sizeof a + b isn't the same as sizeof (a+b). But they aren't part of the invocation of sizeof, they're part of the operand.
• You can't take the address of sizeof.
• The expression which is the operand of sizeof is not evaluated at runtime (sizeof a++ does not modify a).
• The expression which is the operand of sizeof can have any type except void, or function types. Indeed, that's kind of the point of sizeof.

A function would differ on all those points. There are probably other differences between a function and a unary operator, but I think that's enough to show why sizeof could not be a function even if there was a reason to want it to be.

Marchelle

Wow, just what I was thinking! – Invar

I believe things are more complex nowadays due to variable-length arrays (VLA). IIRC, the standard would even allow sizeof to have side effects if there is a VLA in the expression. – Manufacture

@glglgl No, that doesn't make any sense. In that context, (int) is nothing fancy - just a name of a type inside parentheses. Parentheses here are a part of the syntax of sizeof - they are required when taking the size of a type, but not required when taking the size of an expression. See e.g. here – Automation

@Automation My comment was long ago, I suppose I was trying to be sarcastic then. – Sheeb

The standard uses two notations for sizeof: sizeof unary-expression and sizeof ( type-name ) — so in the C11 standard it is not deemed to be a 'cast' but a parenthesized type name. The net result is much the same. (For comparison, a cast expression is ( type-name ) cast-expression.) And I hate the way that comment Markdown works differently from Q&A Markdown! – Thanet

printf("%d\n", sizeof a) is not perfectly fine. %d is the wrong specifier. It has undefined behavior. – Thoughtful

Answer 2

It can be used as a compile-time constant, which is only possible if it's an operator rather than a function. For instance:

union foo {
    int i;
    char c[sizeof(int)];
};

Syntactically if it weren't an operator then it would have to be a preprocessor macro since functions can't take types as arguments. That would be a difficult macro to implement since sizeof can take both types and variables as an argument.

Pasadena

+1 but note that it is not a compile-time constant when the argument is a VLA - variable length array. – Thanet

Answer 3

Because the C standard says so, and it gets the only vote.

And the standard is probably correct because sizeof takes a type and

In general, if either the domain or codomain (or both) of a function contains elements significantly more complex than real numbers, that function is referred to as an operator. Conversely, if neither the domain nor the codomain of a function contain elements more complicated than real numbers, that function is likely to be referred to simply as a function. Trigonometric functions such as cosine are examples of the latter case.

Additionally, when functions are used so often that they have evolved faster or easier notations than the generic F(x,y,z,...) form, the resulting special forms are also called operators. Examples include infix operators such as addition "+" and division "/", and postfix operators such as factorial "!". This usage is unrelated to the complexity of the entities involved.

(Wikipedia)

Hadron

This probably explains the motivation of the C standard (and other programming languages) in using the terms "operator" and "function" as they do. – Marchelle

Answer 4

Because it's not a function. You can use it like that:

int a;
printf("%d\n", sizeof a);

Function does have entry point, code, etc. Function is to be run at runtime (or inlined), sizeof has to be determined at compile-time.

Cowling

Answer 5

Because:

• when you pass a value to a function, the size of the object is not passed to the function, so a sizeof "function" would have no way of determining the size
• in C, functions can only accept one type of argument; sizeof() needs to accept all sorts of differnet things (variables as well as types! You can't pass a type to a function in C)
• calling a function involves making a copy of the arguments and other unnecessary overhead

Pharsalus

Answer 6

Because it is a compile-time operator that, in order to calculate the size of an object, requires type information that is only available at compile-time. This doesn't hold for C++.

Publus

Answer 7

There is small diference from function - value of sizeof is resolved on compile time, but not at runtime!

Commonwealth

Except for VLA - variable length array - arguments. – Thanet

Answer 8

sizeof operator is compile time entity not runtime and don't need parenthesis like a function. When code is compiled then it replace the value with the size of that variable at compile time but in function after function gets execute then we will know the returning value.

Rogers

Answer 9

sizeof() operator is a compile time would be occurrence. It can be used to determine the parameters or arguments.

Gragg

Answer 10

Sizeof(), I think obviously it's both a function and an operator. Why? Because a function holds parentheses for entry at the stage of entry. But mainly also an operator cause operators are action character, therefore sizeof is an action statement that acts on the operand in the parentheses.

Affer

The standard already states clearly that sizeof is an operator – Hissing