Global variable validation?

Question

I am expecting the output to be 10 But i get 11 as per my knowledge while calling fun function, it should take the global variable. ??

#include <iostream>
using namespace first;
int var = 0;

void fun()
{
    cout << var;
}

int main()
{  
    var = 1;
    cout << var;
    fun();
}

@OliCharlesworth because fun() would take the value of global var =0 ? — user3127386, Dec 22 '13 at 16:32
But you have set the value of the global variable `var` to `1` just before calling `fun()`. — juanchopanza, Dec 22 '13 at 16:37

score 4 · Answer 1 · edited May 23 '17 at 10:32

4

You only declare var once. If you shadow the global variable by a local declaration, the local variable will be used.

For example, see this StackOverflow post for information on how the shadowing works.

edited May 23 '17 at 10:32

Community

1
1

answered Dec 22 '13 at 16:29

Uli Köhler

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score 4 · Answer 2 · answered Dec 22 '13 at 16:31

4

int main()
{  
    int var=1;
    cout<<var;
    fun();
}

At the moment you're just modifying the global rather than creating a new one scoped in main().

Then your output will be

answered Dec 22 '13 at 16:31

splrs

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score 4 · Accepted Answer · answered Dec 22 '13 at 16:31

4

You are reassigning the value of 'var' to 1 and then printing 'var' two times. that's why you get 11 as output.

answered Dec 22 '13 at 16:31

Selva

338
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12

how can the var inside main( ) update the global var ?? – user3127386 Dec 22 '13 at 16:36
1

@user3127386 the `var` inside `main()` *is* the global variable. – juanchopanza Dec 22 '13 at 16:38
@juanchopanza so you mean we can update global variables from any type of local scope ? – user3127386 Dec 22 '13 at 16:40
3

@user3127386: yes, that's what's meant by "global" ;) – Oliver Charlesworth Dec 22 '13 at 16:40
@juanchopanza can we update local variables from a global scope, anyhow? no i am guessing.??? – user3127386 Dec 22 '13 at 16:42
1

@user3127386 no, the reverse isn't true, that's why local variables aren't global. Look up "variable scope" on Google. – Mr Lister Dec 22 '13 at 16:46

score 2 · Answer 4 · answered Dec 22 '13 at 16:31

2

After assigning 1 to var, you are printing 2 times the same variable - note that you aren't creating 2 diffrent variables but you're changing the value of your variable.

answered Dec 22 '13 at 16:31

enedil

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I'm not sure if I unerstand your post correctly, but he is not declaring another var, just changing the value of the global variable. In the specific case above, he won't see 11 because of the specific values used. – Uli Köhler Dec 22 '13 at 16:34
Yes, I haven't see that there wasn't int before var = 1 – enedil Dec 22 '13 at 16:36

score -1 · Answer 5 · answered Dec 22 '13 at 16:58

-1

Var=0 Then var=1 Print var two times Output 11

answered Dec 22 '13 at 16:58

pavan

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I think this is rather a comment than an answer, you could add more information to make it an answer however. – Uli Köhler Feb 27 '14 at 17:19

Global variable validation?

5 Answers5