What is the output of the following code?#include <stdio.h>int main() { int i = 0; for (; i < 5; i++); printf("%d ", i); return 0;}
Question
What is the output of the following code?#include <stdio.h>int main() { int i = 0; for (; i < 5; i++); printf("%d ", i); return 0;}
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Solution
The output of the code will be "5".
Here's the step by step explanation:
- The integer variable
iis initialized to 0. - The
forloop will incrementiby 1 each time it loops untiliis no longer less than 5. Note that the loop body is empty, so it's just incrementingiand not doing anything else. - Once
iis equal to 5, theforloop conditioni < 5is no longer true, so the loop ends. - The
printffunction then prints the current value ofi, which is 5. - The program ends with a return statement of 0, indicating successful execution.
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