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Functions - Programming

Q1:

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;

A 3, 4, 4, 3

B 4, 3, 4, 3

C 3, 3, 4, 4

D 3, 4, 3, 4

ANS:B - 4, 3, 4, 3

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".