C++ does sizeof calculate every call if object is static

Question

Lets say I have a static array of an object which size cannot change.

struct vector2 { //8 Bytes
    float x, y;
};

Does the computer calculate the size every time sizeof is called or is this stored somewhere? If not, does it have the same performance as it was stored somewhere?

std::cout <<  sizeof(vector2) << std::endl;
std::cout <<  sizeof(vector2) << std::endl;

Is this as fast as

Byte sizeOfVector2 = sizeof(vector2);
std::cout <<  sizeOfVector2 << std::endl;
std::cout <<  sizeOfVector2 << std::endl;

The size of a type in C++ is known at compile time. It's as if replacing the call to sizeof with an integer literal. — DeiDei, Mar 04 '17 at 21:35
Don't worry about it. *"Premature optimization is the root of all evil."* — Baum mit Augen, Mar 04 '17 at 21:37
Related: https://stackoverflow.com/questions/2615203/is-sizeof-in-c-evaluated-at-compilation-time-or-run-time — Baum mit Augen, Mar 04 '17 at 21:43
@BaummitAugen Okay, true, but I didn't even think about that it could be calculated at compiletime. — Kiryu144, Mar 04 '17 at 21:47

score 6 · Accepted Answer · edited Mar 04 '17 at 21:40

6

sizeof is calculated at compile time.

edited Mar 04 '17 at 21:40

Baum mit Augen

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answered Mar 04 '17 at 21:36

doron

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Shahar Bental · Answer 2 · 2017-03-04T21:43:45.267

2

The sizeof operator is computed at compile time.

edited Mar 04 '17 at 21:43

answered Mar 04 '17 at 21:37

Shahar Bental

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@hvd correct, didn't notice that, removed it from the answer – Shahar Bental Mar 04 '17 at 21:44

C++ does sizeof calculate every call if object is static

2 Answers2