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

Question

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);


        

Answer 1

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

Answer 2

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.

Answer 3

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.