这以什么顺序发生

Question

Lets say I have an instance called box1 , box2 and run the code below.

if(box1->getSize() > box2->copyBox(box1)->getSize())

getSize() returns size of box, copyBox(box) copies the data of box1 to box2 not the address.

In what order does the code happen? I thought

1. box1->getSize() : The size of box1 is returned
2. box2->copyBox(box1) : box2 now shares the same address as box1 as in they're the same instance
3. box2->getSize() : The size of box2 is returned
4. > operator : size of box1 and box2 is compared

I can't find what the orders are with VS2017 debugger. Can anyone tell me a way to find the order with the debugger or at least what the orders are in this example? Thanks.

Answer 1

No.

copyBox cannot change the address of box2 .

getSize(10) is not in the above expression you are breaking down.

There are no guarantees about the order of evaluation of the lhs and the rhs of > .

Given exprA > exprB , the compiler could evaluate exprB first or exprA . Prior to C++17 it could even evakuate part of exprB , pause, do par of exprA , the continue in exprB ; this may have changed in C++17 (it did in some similar contexts, and I am not certain here).

It must evaluate both exprA and exprB before > .

This unspecified execution order exists to permit different compilers solving the problem differently. It gives freedom to optimize, both in a given expression, and in how the compiler handles low level details like calling conventions.