Question
#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;
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;
Answer: Option B
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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 mainfunction.
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 mainfunction.
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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