Could the use of the same variable name inside and outside the loop result in shadow variables ?

Question

https://code.sololearn.com/c68B5v700VAz/?ref=app

Answer 1

Sanjay Kamath
just don't use

for(i =0;i<5;i++)

because it will mess up the outside variable i

use instead

for(int i=0;i<5;i++)

so that the i in the for block is different from the i outside

Answer 2

i was declared before the for loop, so nothing shadows it. if you wrote

for (int i=0; i<5; i++)

the new i would outshadow ths previous one, but only in scope of the curly brackets after for

Answer 3

What do you mean by "shadow variables" ?

Answer 4

//Example for shadowing :
// local variables are always pointed first. so it shadows same named global variables..

#include <stdio.h>
int main(){

int i = 9;

for (int i=0;i<5;i++)
printf("inside i %d\n",i);

printf("outside i %d\n",i);

}

Answer 5

Local variables shadows to their global variables.

Answer 6

Bob_Li Sir, the original problem is as described in the question.., which means that shadowing exists.

Answer 7

Sanjay Kamath i outside braces is undefined, and it is initialized inside the for loop - in the conditions section before the first ;. then it is modified after the last ; in the same section - it gets repeatedly increased by 1, so finally it equals to 5, the result

Answer 8

Sanjay Kamath
it just means you did not redefine the variable i in your for loop, which is what you normally should do. That is why the outside variable i was used, and you get the result you are getting.

Answer 9

Patrick exactly.... then how do you explain the result ?
I outside the braces is 0.

Answer 10

Arsenic : Same variable name inside and outside the loop

Answer 11

https://en.m.wikipedia.org/wiki/Variable_shadowing

As a standard it's not advisable to use this feature....

Answer 12

yes

Answer 13

int i;
int main()
{
for (int i=1;i<=4;i++)
cout<<setw(3)<<i<<endl;
return 0;
}

Answer 14

This was from the test battery 🔋.