What is the scope and evaluation of this if

Asked Sep 16 '21 at 11:24

Active Sep 16 '21 at 11:24

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When cleaning up some code that I have found online, I came across with this weird c++ line:

if (int i = 1) std::cout << i;

With LLVM it compiled fine and the console output is 1, but how does the scope is handled in here, shouldn't the i variable be only accessible inside the conditional (inside the parenthesis)? And how is that possible to be evaluated to true, isn't an assignment a void operation and with no value, so 0/false? What is going on with this line?

asked Sep 16 '21 at 11:24

Rui Pedro

Also: assignment isn't void, it returns the value – L. Scott Johnson Sep 16 '21 at 11:33
More precisely, assignment operator returns a (reference to) left hand side, so the return value here is `i`. – Yksisarvinen Sep 16 '21 at 11:34
@L.ScottJohnson assignment of fundamental types and sensibly defined classes returns the value. You can have `void Foo::operator=(...)` if you want – Caleth Sep 16 '21 at 12:01

What is the scope and evaluation of this if

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