C Programming - Functions - Discussion

We can not return more than one value in the function.

Add mult return the value.
return (kk, ll) ;.

In this step, the execution of bracket from left to right and comma (,) operator left to write execution then return value will be ll and the end of file expression then function return this value to calling location.

And answer will be 12, 12.

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\n", k, l); It prints the value of k and l.

Hence the output is "12, 12".

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".

Hi, it is call by value not call by reference so the value should not get change, it should print 3 and 4. Am I right?

How it Returns two value at the same time?

Case-1
return(10,20);
It will return only 20.

Case-2
return 10;
return 20;
It return first statement only.

In return(kk,LL).

The value of LL will be return only commo operation rule.

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\n", k, l); It prints the value of k and l

Hence the output is "12, 12".

Well here the concept of comma operator is used!.

The rightmost value in the return statement will be taken into consideration. Hence, it will be 12 each time.

It is not possible to return multiple values through return keywords. So I think it's a wrong answer.