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What will be the output of the program? #include<stdio.h> int main() { int fun(int); int i = fun(10); printf("%d\n", --i); return 0; } int fun(int i) { return (i++); } ?->(Show Answer!)
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1. What will be the output of the program? #include<stdio.h> int main() { int fun(int); int i = fun(10); printf("%d\n", --i); return 0; } int fun(int i) { return (i++); }





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  • By: guest on 01 Jun 2017 06.01 pm
    Step 1: int fun(int); Here we declare the prototype of the function fun(). Step 2: int i = fun(10); The variable i is declared as an integer type and the result of the fun(10) will be stored in the variable i. Step 3: int fun(int i){ return (i++); } Inside the fun() we are returning a value return(i++). It returns 10. because i++ is the post-increement operator. Step 4: Then the control back to the main function and the value 10 is assigned to variable i. Step 5: printf("%d\n", --i); Here --i denoted pre-increement. Hence it prints the value 9.
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