Question

#include<stdio.h> int functionl (int x, int y); int main() int ij=2,k; for(i=1;i<=5; i++) k = function1(ij); printf("k=%d\n",k); return 0; int function] (int x, int y) int z; z=x*2+y; return z;

Answer 1

CODE:

#include<stdio.h>
int function1(int x,int y); //function declaration
int main()
{
int i,j=2,k;   
for(i=1;i<=5;i++) //iterates the loop for 5 times   
{
k=function1(i,j); //function call with paramaters i,j
printf("k=%d\n",k); //printing the value returned by the function function1()
}
return 0;
}

int function1(int x,int y) //function definition
{
int z;
z=x*2+y; //multiplies x value with 2 and adds y value and assigns result to the 'z'
return z; //returns z to the main()
}

CODE SCREENSHOT:

#include<stdio.h> int function1(int x, int y); int main() //function declaration int i, j=2,k; for(i=1;i<=5; i++) //iterates the loop for 5 times k=function(i,j); //function call with paramaters i,j printf("k=%d\n",k); //printing the value returned by the function function1() return 0; int function1(int x, int y) //function definition int z; Z=x*2+y; return z; //multiplies x value with 2 and adds y value and assigns result to the 'z' //returns z to the main()

CODE OUTPUT:

k=4
k=6
k=8
k=10
k=12

SCREENSHOT:

k=4 k=6 k=8 k=10 k=12