C Programming - Functions - Discussion

Well said @Sundar. Thankyou.

In return statement can we return two values at a time?

The output will be 12 12.

I checked in Dev CPP. It takes the right most one in the return statement because according to the comma operator has left to right priority. I hope it will helps for you.

But here the return case we not suppose to do like that it is not possible so it will get error.

RETURN always return the rightmost value.

Since ', ' comma operator is right-left associative right most value will be return at the returning time. This is the simplest explanation of this problem.

Thank you.

Actually it returns a value like that k=(8,12).
if we give multiple values for integers with brackets, the compiler only considers the last value 12.

Note :
if we declare a variable it considers a last value,
int a=(1,2,3) // a=2.

else we initialize a variable that considers the first value,
int a;
a=(1,2,3)
// a=1.

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 is 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".

The last value is returned and all the values before that are simply ignored.

This function is not having a prototype so why isn't it is having compile error.