How it works.. printf("number %d",b=(a++,a++)); printf("b=%d",b);

Question

how the expression in 1st printf evaluated and how this all work ..with output

Answer 1

The tricky part of the snippet that you have provided is b=(a++, a++), Let me explain it by expanding upon the snippet itself.

Consider the following version

1. int a = 1;
2. int b = 2;
3.
4. // b = (Expr1, Expr2)
5. b = (a++, a++);
6. printf("b is ℅d\n", b);
7. printf("a is %d", a);

In line 4, what we are going to expect to being assigned to variable b is the Expr1 and the rest is discarded, that is, b = a++. Both Expr1 and Expr2 are evaluated before assignment, so, at the end of the day, the result of the two print statements would be as

b is 2
a is 3

Answer 2

Would you mind posting more of the code?