Why can't variables defined in a conditional be constructed with arguments?

Question

The question is simple. Why does this compile:

bool b(true);
if (b) { /* */ }

And this compile:

if (bool b = true) { /* */         


        

Answer 1

It's a bit technical. There's no reason why what you want couldn't be allowed, it just isn't. It's the grammar.

An if statement is a selection statement, and it takes the grammatical form:

if (condition) statement

Here, condition can be either:

• expression or
• type-specifier-seq declarator = assignment-expression

And there you have it. Allowing a declaration in a condition is a special case, and it must follow that form or your program is ill-formed. They could have probably allow direct-initialization instead of copy-initialization, but there isn't really any motivation to do so now. As Johannes Schaub points out, this change would break existing code, so it's pretty much never going to happen.

Let_Me_Be notes that C++11 added a third form (I'm ignoring attributes here):

decl-specifier-seq declarator braced-init-list

So if (bool b{true}) is fine. (This can't possibly break any valid existing code.)

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Note your question seems to do with efficiency: don't worry. The compiler will elide the temporary value and just construct the left-hand side directly. This, however, requires your type be copyable (or movable in C++11).