How this code work?

Question

how this code work? i break my brain
Can someone explain this?
https://code.sololearn.com/cKeMAOacrloS/?ref=app

Answer 1

it work on the concept of call by reference...

function f() is having arguments which defines reference by & this sign......

now....

f(a, a);

means &a = a; and &b = a;

when you make change to &a and &b it will change to a and a......

now a = 2
and b = 4
that means now a is 4 and b is also 4

return a+b means 4+4

as I told making change to &a and &b will change value of a.....

now cout<<a<<b<<c;

a is 4
b is 7 // because we changed only a not b
c is containing a+b so it will be 8

hope you got it....

Answer 2

cool, that is a great challenge and 2 good explanations, also thank you very much from my side 👍😉

Answer 3

int x(int &a, int &b){
a=3;
b=4;
return a+b;
}

int main() {
int a=2;
int b=7;
int c = x(a, a);
cout << a<< b << c;
return 0;
}

The value of a is 2. Function x receives two parameters by reference, which means that both parameters will refer to the same local variables in main. Variable a is passed into the function through both parameters. In the function, variable a is altered to 3, and then to 4. (Remember that both the function parameters now refer to the same variable) The function the returns 4+4, which is 8, assigned to variable c.

Program prints 478

Answer 4

Thanks guys!

Answer 5

thanks for marking best

Answer 6

@tooselfish welcome