Why is it an error to initialize a variable with multiple values using comma operator?

Question

If I write the statement

int i=1,2,3;

why does here the comma acts as a separator rather than operator since we have comma operator having associativity from left to right so according to me first value is initialized using i=1 but it doesn't work like this , what's the reason behind this ?

see [Why does this use of comma work in a expression but fail in a declaration?](http://stackoverflow.com/q/26505722/1708801) — Shafik Yaghmour, Jul 29 '15 at 12:07
similar expressions like `int i = 3, y;` would be ambiguous: is it declaring y as an int or part of the initialization? — ryanpattison, Jul 29 '15 at 13:00

M.M · Answer 1 · 2015-07-29T12:27:53.877

Because the C language grammar says that an initializer must be an assignment-expression. The latter includes all expressions except those formed from two other expressions and the comma operator:

expression :
    assignment-expression
    expression , assignment-expression

So 0,1,2 is not a valid initializer for i. Since 1 is not a valid declarator either, this code does not match any syntax rules, making it a syntax error.

I'd guess that the grammar was designed this way on purpose, to avoid the possibility of any situations where a comma is ambiguous between a separator and an operator.

Why is it an error to initialize a variable with multiple values using comma operator?

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