C Programming - Functions - Discussion

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.

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.

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.

The comma operator evaluates a series of expressions. The value of the comma group is the value of the last element in the list.

int multi_return_args(void)
{
return (44,66);
}

In the example, you show the leading constant expression 44 has no effect, but if the expression had a side effect, it would occur.

For example:

return printf( "we're done" ), 66;

In this case, the program would print "we're done" and then return 66.

so, final answer is based on the last argument that is ll.

As when admult function is called it is gone to admult definition i.e

k=admult(3,4),in calling function.
Now in function def.
int addmult(3,4)
i.e kk=3+4;
ll=3*4;
return(kk,ll)i.e
return(7,12)i.e
return(12).

AS because return statement only one value i.e 12 from right side.
Same is also done when other function is call .i.e return 12.

Why does it return value? I thought the variable "kk" is not equals to variable "k".