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Consider the following C program :

#include<stdio.h>
int jumble(int x, int y){
x = 2*x+y;
return x;
}
int main(){
int x=2, y=5;
y=jumble(y,x);
x=jumble(y,x);
printf("%d \n",x);
return 0;
}

The value printed by the program is ______________.

### 1 comment

26

$x = 2, y = 5$

$y =$ jumble$(5,2)$ //call by value and $y$ will hold return value. After this call $x = 2, \: y = 12$

$x =$ jumble$(12, 2)$ //call by value and $x$ will hold return value. After this call $x = 26, \: y = 12$

$x=26$

Can we tell like this

that "when there is a storage, call by value can work same as call by reference " or

"when there is a storage, call by value can calculate value from other function"

Can we tell like this

that "when there is a storage, call by value can work same as call by reference " or

"when there is a storage, call by value can calculate value from other function"

I think YES because it is returning the value and coming back to main() so it assigns(updates) returned value to the specific variable [for which function is assigned & invoked]

you can try deleting x=jumble(y,x) ,return 0 and any other values to see the effects.

(please correct me if I'm wrong)

if return statement not given then value of x y not changed it works just simple call by value

jumble function is just taking 2 arguments say arg1 and arg2. and returning (2*arg1) + arg2

in main function we have below:

y=jumble(5,2);---main called jumble function with arguments 5,2 initially  and result was stored in variable y

here y becomes 12  i.e: (2*10)+2

x remains same which is 2

x=jumble(12, 2);----main again called jumble function with arguments 12,2  and stored in variable x

so x becomes 26   i.e: (2*12)+2

finally x will be printed
output: 26

by
1st jumble will return 12 that will be stored in y.

this y will be passed as parameter in 2nd jumble that will return 26 in x.
y=jumble(5,2) will return 12 and

x=jumble(12,2) will return 26

so pf(x)-->26