what's the mechanism of sizeof() in C/C++?

It seems sizeof is not a real function?

for example, if you write like this:

int i=0;
printf("%d\n", sizeof(++i));
printf("%d\n", i);

You may get output like:

4
0

And when you dig into the assemble code, you'll find sth like this:

movl     $4, %esi
leaq     LC0(%rip), %rdi
xorl %eax, %eax
call     _printf

So, the compiler put directly the constant "4" as parameters of printf add call it. Then what does sizeof do?

You know, there's a reason why there are standard documents (3.8MB PDF); C99, section 6.5.3.4, §2:

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 result is an integer. If the type of the operand is a variable length array type, the operand is evaluated; otherwise, the operand is not evaluated and the result is an integer constant.

In response to ibread's comment, here's an example for the C99 variable length array case:

#include <stdio.h>

size_t sizeof_int_vla(size_t count)
{
    int foo[count];
    return sizeof foo;
}

int main(void)
{
    printf("%u", (unsigned)sizeof_int_vla(3));
}

The size of foo is no longer known at compile-time and has to be determined at run-time. The generated assembly looks quite weird, so don't ask me about implementation details...

sizeof is an operator, not a function.

It's usually evaluated as compile time - the exception being when it's used on C99-style variable length arrays.

Your example is evaluating sizeof(int), which is of course known at compile time, so the code is replaced with a constant and therefore the ++ doesn't exist at run-time to be executed.

int i=0;
cout << sizeof(++i) << endl;
cout << i << endl;

It's also worth noting that since it's an operator, it can be used without the brackets on values:

int myVal;
cout << sizeof myVal << endl;
cout << sizeof(myVal) << endl;

Are equivalent.

Sizeof analyzes the passed expression to find its type. It then returns the size of the type.

Because the size of a type is always known at compile time, it is put into the machine code as a constant.

It is replaced with the constant (4 in your case) at compile time. Because it takes 4 bytes to hold an int on your platform.

And your code will not compile, instead of giving you any output ;-) Because of sizoef ;-)

The size of returned type is calculated at compile time, there is no runtime overhead

P Shved

In C++ sizeof() calculates size of the type of the expression within it and replaces the whole "sizeof() function call" with a constant during compilation.

The expression within sizeof() is never evaluated during the program execution. And it may not even be a type name. Check these examples out:

struct X { int i; double j;};
int call_to_undefined_function();

sizeof(10/0);
sizeof( ((struct X*)NULL)->j );
sizeof( call_to_undefined_function() + 100 );
sizeof( call_to_undefined_function() + 100.0 );
sizeof( double() / int() );

Exactly what it's meant to do: puts directly the constant "the size of the variable/constant/type/etc" into the code

sizeof() returns the size in bytes of whatever you pass as an argument to it. In a 32-bit architecture sizeof(int) would return 4, while sizeof(char) would return 1.

You said it yourself: 'sizeof' is not a function. It is a built-in operator with special syntax and semantics (see the previous responses). To remember better that it is not a function, you might want to get rid of the habit to use superfluous braces and prefer to use the "braceless" 'sizeof' with expressions as in the following example

printf("%d\n", sizeof ++i);

This is exactly equivalent to your original version.

来源：https://stackoverflow.com/questions/1581839/whats-the-mechanism-of-sizeof-in-c-c

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