int b=0,a=1;b= ++a + ++a; what is the value of b? what is the calculation for it? [duplicate]

Question

This question already has an answer here:

• Why are these constructs using pre and post-increment undefined behavior? 14 answers

int main()
{
    int b=0,a=1;//initialize a and b
    b=++a + ++a;// calculate assign the value of b
    print f("%d",b);
    return 0;
}

1. what is the value of b?
2. and what is the calculation for it?

Answer 1

This is Undefined Behaviour.Lack of sequence point.

For more info have a look here and output-of-multiple-post-and-pre-increments-in-one-statement.

Answer 2

It would seem that having a sequence point is immaterial in the expression b=++a + ++a;

That is, whether the first ++a is evaluated first or the second ++a is evaluated first in either case a is incremented twice and then the + operator takes effect, so the eventual equation is either b = 2 + 3; or b = 3 + 2 thus b = 5.

When I get home I will try this on with my C compiler.

Blastfurnace's comment about both being evaluated before the + operator takes effect is correct, and now that I think about it, obvious.

That is, + is lower in precedence than ++a. It could be argued that this statement is NOT ambiguous in that switching the order of evaluation (R to L or L to R, both with respect to precedence) results in the same answer.

No one will claim this is well written code, interesting on several points for a discussion, but not something that should be endorsed.