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Why this code gives infinite loop ?
int i=0; while(i<=5){ printf("*"); i=i++; } https://code.sololearn.com/c812sBYy2WvG/?ref=app Ps: I am using gcc compiler. https://code.sololearn.com/c812sBYy2WvG/?ref=app
3 Respuestas
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As Schindlabua and Chirag Kumar said, because the C++ standard doesn't specify the exact order of evaluation, the order is unspecified and is compiler implementation dependant.
My guess is that `i` is pre-incremented first, and the old value of `i` returned by `i++` is assigned to `i`.
Example:
int i = 0;
i = i++;
Step 1: i = [i++], i++ => int temp; temp = i; i = i + 1; return temp; // returns 0
Step 2: i is now 1, but i is assigned to the old value of i, i = 0
Step 3: Rinse and repeat.
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You don't need to assign here. This is enough:
printf("*");
i++;
because `i++` is shorthand for `i = i + 1`!
When you do `i=i++` it is actually undefined behaviour and anything can happen. Maybe it will run forever, maybe not.
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It's unexplainable C compiler himself can't explain why is this code giving infinite loop.
So you can just use
i++; instead of i=i++;