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What will be the output of the program? #include<stdio.h> int reverse(int); int main() { int no=5; reverse(no); return 0; } int reverse(int no) { if(no == 0) return 0; else printf("%d,", no); reverse (no--); } ?->(Show Answer!)
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1. What will be the output of the program? #include<stdio.h> int reverse(int); int main() { int no=5; reverse(no); return 0; } int reverse(int no) { if(no == 0) return 0; else printf("%d,", no); reverse (no--); }





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  • By: guest on 01 Jun 2017 06.01 pm
    Step 1: int no=5; The variable no is declared as integer type and initialized to 5. Step 2: reverse(no); becomes reverse(5); It calls the function reverse() with '5' as parameter. The function reverse accept an integer number 5 and it returns '0'(zero) if(5 == 0) if the given number is '0'(zero) or else printf("%d,", no); it prints that number 5 and calls the function reverse(5);. The function runs infinetely because the there is a post-decrement operator is used. It will not decrease the value of 'n' before calling the reverse() function. So, it calls reverse(5) infinitely. Note: If we use pre-decrement operator like reverse(--n), then the output will be 5, 4, 3, 2, 1. Because before calling the function, it decrements the value of 'n'.
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