1. What will be the output of the program? #include<stdio.h> int i; int fun1(int); int fun2(int); int main() { extern int j; int i=3; fun1(i); printf("%d,", i); fun2(i); printf("%d", i); return 0; } int fun1(int j) { printf("%d,", ++j); return 0; } int fun2(int i) { printf("%d,", ++i); return 0; } int j=1;





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  • By: guest on 01 Jun 2017 12.01 pm
    Step 1: int i; The variable i is declared as an global and integer type. Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value. Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value. Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file. Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3. Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomes fun1(3) hence it prints '4' then the control is given back to the main function. Step 7: printf("%d,", i); It prints the value of local variable i. So, it prints '3'. Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints '4' then the control is given back to the main function. Step 9: printf("%d,", i); It prints the value of local variable i. So, it prints '3'. Hence the output is "4 3 4 3".
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