What does i = (i, ++i, 1) + 1; do?

Question

After reading this answer about undefined behavior and sequence points, I wrote a small program:

#include <stdio.h>

int main(void) {
  int i = 5;
  i = (i, ++i, 1) + 1;
  printf("%d\n", i);
  return 0;
}

The output is 2. Oh God, I didn't see the decrement coming! What is happening here?

Also, while compiling the above code, I got a warning saying:

px.c:5:8: warning: left-hand operand of comma expression has no effect

  [-Wunused-value]   i = (i, ++i, 1) + 1;
                        ^

Why? But probably it will be automatically answered by the answer of my first question.

Answer 1

In the expression (i, ++i, 1), the comma used is the comma operator

the comma operator (represented by the token ,) is a binary operator that evaluates its first operand and discards the result, and then evaluates the second operand and returns this value (and type).

Because it discards its first operand, it is generally only useful where the first operand has desirable side effects. If the side effect to the first operand does not takes place, then the compiler may generate warning about the expression with no effect.

So, in the above expression, the leftmost i will be evaluated and its value will be discarded. Then ++i will be evaluated and will increment i by 1 and again the value of the expression ++i will be discarded, but the side effect to i is permanent. Then 1 will be evaluated and the value of the expression will be 1.

It is equivalent to

i;          // Evaluate i and discard its value. This has no effect.
++i;        // Evaluate i and increment it by 1 and discard the value of expression ++i
i = 1 + 1;  

Note that the above expression is perfectly valid and does not invoke undefined behavior because there is a sequence point between the evaluation of the left and right operands of the comma operator.