C Programming - Functions - Discussion

It is because of the brackets in the return statement. And brackets have to l precedence, that the reason why only multiplicative part is returned back to the main function.

We can return only one value at the time and in the program given return (ll, kk); Hence the compile take only one value of return (ll, kk); left to right and value gives us of kk = 12.

Guys, a function can return only one value at a time. Thus the statement is invalid. Then how will come this answer 12 12?

Hi @Preethi.

The description which you have given is not understable to me.

Please explain briefly.

Hi, guys.

This will provide and prove the needed answer.

#include <stdio.h>

int func()
{
return (3,14);
}

int main()
{
printf("%d\n", func());
return 0;
}

Thank you very much @Sundar.

Return cannot take two arguments.

Why doesn't the value '12' return to k in the next turn? Why should it go to the l?

Well explained. @Rahul.

Explanation:

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".