C Programming - Functions - Discussion

Step 1: int i=3, j=4, k, l; The variables i, j, k, l are declared as an integer type and variables 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".

If a function defined the above main, no need of prototype.

There is a prototype, but no definition, That is also fine till you don't call that function.

Now coming to the problem:
Many are asked how return returns the multiple values. That is correct. Return can not return multiple values at a time. But we can achieve this by storing values in an array, finally we can return multiple values. But in the coding problem, here return like (7, 12). Please consider comma operator here.

We know precedence of comma operator is left to right. Because of that the last value only return i.e, ->12.

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.

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".

In the given question, the comma(,) acts as an operator & the comma operator returns the rightmost operand in the expression and simply evaluates the rest of the operands and finally rejects them.