Is the comma in a variable list a sequence point?

问题

In the following type of code is there a sequence point between each variable construction, or is the result undefined?

int a = 0;
int b = a++, c = a++;

I wasn't able to find in the standard a specific reference to a sequence point here. Does that mean it is undefined, or just that I failed in my search? The completion of an expression is a sequence point, but does the above initialization also count?

回答1:

I believe behavior is well-defined because of 8[dcl.decl]/3

Each init-declarator in a declaration is analyzed separately as if it was in a declaration by itself.

Which is even additionally explained in a footnote as

A declaration with several declarators is usually equivalent to the corresponding sequence of declarations each with a single declarator. That is

T D1, D2, ... Dn;

is usually equvalent to

T D1; T D2; ... T Dn;

回答2:

As you suspect there is a sequence point after each initializer expression, because they're full expressions (1.9/16, 1.9/12).

来源：https://stackoverflow.com/questions/6414030/is-the-comma-in-a-variable-list-a-sequence-point