Why sizeof is an operator rather than macro or function

[manjuks]

Hi All,

Why sizeof is called as an operator, why it can not be function or a macro?
My assumption is type information is available at compile time, and to find out the size we need the type of object. Is it correct? or tell me why it must be an operator?

Thanks & Regards,
Manjunath

[Banfa]

It operates on types, neither functions or macros operate on types they operate on variables or constants.

Only the compiler can determine the result of sizeof from parsing the code.

[donbock]

sizeof is evaluated at compile-time. Functions are evaluated at run-time. Therefore sizeof can't be a function.

In C, macros perform simple textual substitution. This means that any macro can be replaced by its expansion without altering the program. If sizeof were a macro, then what does it expand to? Sizeof can't be a macro.

[weaknessforcats]

And the sizeof operator only determines the size of the type or variable as it appears on the stack. That is, if you allocate memory on the heap using
malloc():

MyStuff* var = malloc(1000);

then

printf("%d", sizeof(var));

gives you the size of the pointer and not the size of what it points at.