Operator precedence in j = j++ * ++j

Answer 1

The expression

j = j++ * ++j

...is undefined. There is no way to know what the order of these operations are. It could result in anything, or make your CPU burst into flames. It will differ from compiler to compiler, and even within a particular compiler depending on optimisation levels or versions.

Answer 2

The operator precedence should be post-increment then pre-increment then multiply although I bet this comes down to one of those compiler-dependent / undefined things.

This precedence is based on the table at Wikipedia which is pretty common across the web which states that post-increment has precedence 2, pre-increment has precedence 3, and multiply has precedence 5.

My intuitive look is:

j = j++ * ++j (j is 2)
j = 2 * ++j (j is now 3)
j = 2 * 4 (j is now 4)
j = 8

Which is obviously wrong.

The correct answer is obviously that this is undefined.

Answer 3

int j=2; j= j++ * ++j; printf("%d", j);

j=j++*++j invokes Undefined Behavior as you are trying to change the value of j more than once between two sequence points.

Precedence only affects which tokens are considered to be the operands of each operator, but it does not affect in any way the order of evaluation.

Answer 4

I can guess that passed through the following way of computation:

• pre-increment j by one
• multiply j by itself
• post-increment j by one

The exact order of these computations is undefined by the standard and may be compiler-specific.