What will be the output of the following C code?#include <stdio.h>int main() { int i; for (i = 0; i < 5; i++); printf("%d\n", i); return 0;}
Question
What will be the output of the following C code?
#include <stdio.h>
int main() {
int i;
for (i = 0; i < 5; i++);
printf("%d\n", i);
return 0;
}
Solution
The output of the given C code will be "5".
Here's the step by step explanation:
-
The program starts with the inclusion of the standard input/output library using
#include <stdio.h>. -
The
main()function is declared, which is the entry point of the program. -
Inside the
main()function, an integeriis declared. -
A
forloop is initiated whereiis set to 0 and the condition isi < 5. The loop incrementsiby 1 (i++) each time it runs. However, there is no code to execute within the loop (indicated by the semicolon;right after the loop) so it simply incrementsifrom 0 to 5. -
Once
iequals 5, the conditioni < 5is no longer true, so the loop ends. -
The
printffunction is then called to print the value ofi, which is now 5, followed by a newline character\n. -
The program then returns 0, indicating successful execution, and ends.
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