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1. What will be the output of the program? #include<stdio.h> int check(int); int main() { int i=45, c; c = check(i); printf("%d\n", c); return 0; } int check(int ch) { if(ch >= 45) return 100; else return 10; }





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  • By: guest on 01 Jun 2017 06.01 pm
    Step 1: int check(int); This prototype tells the compiler that the function check() accepts one integer parameter and returns an integer value. Step 2: int l=45, c; The variable i and c are declared as an integer type and i is initialized to 45. The function check(i) return 100 if the given value of variable i is >=(greater than or equal to) 45, else it will return 10. Step 3: c = check(i); becomes c = check(45); The function check() return 100 and it get stored in the variable c.(c = 100) Step 4: printf("%d\n", c); It prints the value of variable c. Hence the output of the program is '100'.
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