Why do I have to parenthesize an initializing expression that's a comma expression?

Question

Boiling a problem I'm having down to its essence, I can initialize a variable as an int by first executing a do-nothing lambda in a comma expression like this:

int main(){
  auto x = ( []{}(), 10 );          // same effect as auto x = 10;
}

But if I don't parenthesize the initializing expression,

int main(){
  auto y = []{}(), 10;              // won't compile
}

all of gcc, clang, and MSVC complain about trying to initialize y with a void expression.

Why do I have to parenthesize the comma expression to use it as an initializer?

Answer 1

In a declaration, the , symbol separates declarators. A simpler example:

int i = 2, j = 3;     // OK: declares `i` and `j`
int i = 2, 3;         // Error: `3` is not a declarator

In the second case, it looks ambiguous. Is the , separating declarators, or is the , part of an expression 2, 3?

To resolve this ambiguity we can consult the language grammar (C++14 [dcl.decl]):

simple-declaration:
    decl-specifier-seq_opt init-declarator-list_opt ;
    attribute-specifier-seq decl-specifier-seq_opt init-declarator-list ;

init-declarator-list:
    init-declarator
    init-declarator-list , init-declarator

init-declarator:
    declarator initializer_opt

The way grammars work, this means that when parsing a declaration the longest possible sequence that matches init-declarator , is considered. (This is sometimes called "maximum munch principle"). So int i = 2, matches init-declarator , . Then 3 fails to match init-declarator, so parsing fails.