C Programming - Functions - Discussion

@Akshay.

k= func1(k=func1(k=func1(k)));

We are simply solving the brackets. As we know a functions requires an argument. In order to get the required argument for k=func1(), we are solving the inside terms of k=func1(........).

In k= func1(k=func1("k=func1(k)")), the last term inside " " sends a function call to func1(k) with argument k=35.

Once the value is returned from the called function it becomes the parameter of next bracket statement as --> k=func1(k=func1(36)) --> k=func1(37).

# include <stdio.h>


int post(int i)
{
int m=i++;
return m;//////return then increment
}

int post3(int k)
{
k++;
return k;//////increment then return
}

int post2(int j)
{
return j++;/////return then increment
}


int main()
{
int i =10;
printf("%d\n", post(i));


int j=20;
printf("%d\n", post2(j));


int k=23;
printf("%d",post3(k));
return 0;
}

The output should be 35.

Because it's call by value can't increase the value in the calling function as its function activation record will be destroyed each time while returning the value. If it will pass by reference then op would be 38.

@priya: No you are wrong.If the return statement would be like return(k++),then answer should be 35.But here,in this case,first k is post incremented and in next step return(k++) it actually incremented and return incremented value.

Here we are taking about post increment meaning ( first initialize later increment).

void main (){
int i=1 j;
j= i++;
printf ("%d%d", i, j);
}

The output will be i=2, j=1 so in above example output should be 35.

The right answer will be 38.

If there will be statement like this,

return k++ then the value will be 35
and 35 will be also in other case if
we declare as the extra variable as follows
int i =k++;
return i;

This is not the correct answer first there is no change in value of variable if it's treated in function and when we print it in main no change is seen it can be done only if it's addresses is sent.

There is a postfix increment operator so the value should not increase before going to the caller function so there should not be any increment in value.

Can anyone explain me the execution of this statement:

k = func1(k=func1(k=func1(k)));

It works by first calling the internal function call or not?

The difference is: if p=12 then
k=++p;then k value first assiened to 13(k=13)
k=p++;then k value asigned to 12(k=12)
Like that we can calculate