Can someone explain how the answer is 9

Question

#include <stdio.h>
int i=10;
void f(void){
int i=5;
i=7;
}
void g()
{
i=9;
}
int main() {
int i=1;
g();
f();
printf ("%d",i);
return 0;
}

Answer 1

The answer is one not nine

https://code.sololearn.com/c8iKSET1FY84/?ref=app

Answer 2

The global i is modified from 10 to 9 by g()

Answer 3

However it is the local i of main() that's output.

Answer 4

Unless you comment out the line `int i=1;` the output will be 1 not 9. The reason, declaration of <i> as local variable in main shadows the global variable <i>, and when you print <i> the local variable <i> having value 1 is used instead of the global <i> whose value is changed in function g().

Answer 5

No Nandini , f() declares its own <i> variable locally, and also changes the local <i> value from 5 to 7. But again if we comment out the local variable declaration `int i = 5;` then the same scenario happens, the global <i> variable will be changed in f() also.

Answer 6

Ipang yeah may be the answer in quiz was wrong

Answer 7

Do u really got the output 9?

it will print 1 not nine, because in general if you are doing some operation on variable, c language will check whether it is present in the local method if it not then only it will go for global variables. one point to note is scope of a local variable is with in that method only.

Answer 8

The printf statement in main function should points to i=1 but how function g() changed its value as it is not pass by reference.

Answer 9

Ipang as u r saying that if we comment i=1 then output is 9
After calling function g(),we are calling f().. Wont f() change the value of i ??

Answer 10

To quote from wikipedia:
"Local variable references in the function or block in which it is declared override the same variable name in the larger scope."
https://en.wikipedia.org/wiki/Local_variable

Answer 11

You are right that printf() call should print the local variable <i> declared in main(). That is if we keep the declaration `int i = 1;`. If we comment that line, then the <i> variable recognized in g() and main() will be the one declared global (`int i = 10;`) on top of the code. This global variable is the one recognizable in both g() and main(). The g() function is the one who changed its value from 10 to 9.

Answer 12

Ipang ok i understood, thank u

Answer 13

I think (studio.h)should be (iostream) sorry if this is wrong because Iam a new coder

Answer 14

The answer would be 1
As when the main runs tge variable i is created in its memory, when g() function is called the i in the memory g() is converted to 9 from junk value, when f() is called the i in the memory of f() is declared to 5 and then changed to 7 but none of the functions returns the value.
The value of i in the main() remains tge same which is equal to 1 and 1 is printed as the output.

If you understood my answer then give me a follow back, I'm new here 😅

Answer 15

Aravind Devara No actually the answer is 1 but in challenge the answer given as 9

Answer 16

Nandini
You're most welcome.

But the code you given above (as it is) will output 1 instead of 9, unless we comment the declaration of local variable <i> in main(). If that is exactly how the code is looking about in the challenge, then I suspect the quiz is a faulty one : )

Answer 17

I understood!!!

Answer 18

Can I see that challenge? Can I have the link Nandini how can it be 9!

Answer 19

The answer is 1 not 9
There is no return in function f and g
So the output will be 1

Answer 20

Actually answer is 1