Question
#include<stdio.h>
int addmult(int ii, int jj)
{
int kk, ll;
kk = ii + jj;
ll = ii * jj;
return (kk, ll);
}
int main()
{
int i=3, j=4, k, l;
k = addmult(i, j);
l = addmult(i, j);
printf("%d %dn", k, l);
return 0;
}
What will be the output of the program?
#include<stdio.h>
int addmult(int ii, int jj)
{
int kk, ll;
kk = ii + jj;
ll = ii * jj;
return (kk, ll);
}
int main()
{
int i=3, j=4, k, l;
k = addmult(i, j);
l = addmult(i, j);
printf("%d %dn", k, l);
return 0;
}
Answer: Option A
Was this answer helpful ?
Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variable
i, j are initialized to 3, 4 respectively.
The function addmult(i, j); accept 2 integer parameters.
Step 2: k = addmult(i, j); becomes k = addmult(3, 4)
In the function addmult(). The variable kk, ll are declared as an integer type int kk, ll;
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'k'.
Step 3: l = addmult(i, j); becomes l = addmult(3, 4)
kk = ii + jj; becomes kk = 3 + 4 Now the kk value is '7'.
ll = ii * jj; becomes ll = 3 * 4 Now the ll value is '12'.
return (kk, ll); It returns the value of variable ll only.
The value 12 is stored in variable 'l'.
Step 4: printf("%d, %d`setminus`n", k, l); It prints the value of k and l
Hence the output is "12, 12".
Was this answer helpful ?
More Questions on This Topic :
Latest Videos
Latest Test Papers
Submit Solution