This is a beginners question.
Following is the C++ code that I was working with
int main() {
int x=5, y=5;
cout<<x--;
cout<<",";
cout<<--x;
cout<<",";
cout<<y--<<","<<--y;
return 0;
}
When run in Turbo C++ 3.0 the following is the output displayed by the code:
5,3,4,4
When compiled with Code::Blocks on Windows (uses MinGW with GCC/G++) the following is the output displayed:
5,3,4,3
Previously, I have heard that sometimes different compilers behave differently to certain problems, but I don't understand why this result is being displayed. Since logically, as I think, the output should be:
5,3,5,3
Can you please tell me the reason for such output logically.
Thank you!
There is no right or wrong output. Your code has undefined behavior and anything is possible.
The reason lies in paragraph 1.9.15 (n3337) of C++ standard (emphasis mine):
Except where noted, evaluations of operands of individual operators and of subexpressions of individual
expressions are unsequenced. [Note: In an expression that is evaluated more than once during the execution
of a program, unsequenced and indeterminately sequenced evaluations of its subexpressions need not be
performed consistently in different evaluations. —end note ] The value computations of the operands of an
operator are sequenced before the value computation of the result of the operator. If a side effect on a scalar
object is unsequenced relative to either another side effect on the same scalar object or a value computation
using the value of the same scalar object, the behavior is undefined.
In this line
cout << y-- << "," << --y;
you've introduced two side effects (increments by postfix and prefix, respectively) and two value computations (results of y-- and --y) on a scalar object (y) where they are unsequenced. Thus, your program is ill-formed and just about anything is a possible output.
Read more about this here.
cout<<y--<<","<<--y; is an expression with two unsequenced side effects on y, and therefore has undefined behavior.
