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Uncovering shadowed variable when outer variable is not global

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2

I saw in this post that we can uncover a global variable where a local variable of the same name is in scope by using extern and a compound statement.

In that post, the example given is like

#include<stdio.h>
int a = 12; 
int main(void) 
 
 int a = 15; 
 printf("Inside a's main local a = : %dn", a);

 
 extern int a;
 printf("In a global a = %dn", a);
 

 return 0; 

and the shadowed variable is a global variable.

But when I tried a program like

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i;
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

it failed.

I tried making the outer i static in case it works only if the outer variable is static but still it didn't work.

Can somebody tell me why the second program doesn't work? And how to get around it if it's possible?

Maybe this method works only if the outer variable is global?

Edit: The comments made me realize that the use of extern will just make a global variable visible. But is there a way around this without changing variable name?

c scope global-variables local-variables

share|improve this question

edited Mar 23 at 11:34

J...S

asked Mar 23 at 10:32

J...SJ...S

3,64911231

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not the same: int i=5; is local scope and you can't make it extern. If you make it static it is still at local scope.
  
  – Weather Vane
  Mar 23 at 11:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  extern doesn't go "one level up". I goes to the global scope, so your i must be declared at the gloal scope. Your compiler should have complained.
  
  – Paul Ogilvie
  Mar 23 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @PaulOgilvie Oh, I see. So there's no way around this if it was a non-global variable?
  
  – J...S
  Mar 23 at 11:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Don't. Just use unique varaible names.
  
  – Weather Vane
  Mar 23 at 11:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Due to the compiler seeing the first instance of 'i' as unused, it is eliminated in the resulting object file. So does not exist when the linker is invoked.
  
  – user3629249
  Mar 23 at 19:49
  

  
  
  
  

 | 
show 4 more comments

2

I saw in this post that we can uncover a global variable where a local variable of the same name is in scope by using extern and a compound statement.

In that post, the example given is like

#include<stdio.h>
int a = 12; 
int main(void) 
 
 int a = 15; 
 printf("Inside a's main local a = : %dn", a);

 
 extern int a;
 printf("In a global a = %dn", a);
 

 return 0; 

and the shadowed variable is a global variable.

But when I tried a program like

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i;
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

it failed.

I tried making the outer i static in case it works only if the outer variable is static but still it didn't work.

Can somebody tell me why the second program doesn't work? And how to get around it if it's possible?

Maybe this method works only if the outer variable is global?

Edit: The comments made me realize that the use of extern will just make a global variable visible. But is there a way around this without changing variable name?

c scope global-variables local-variables

share|improve this question

edited Mar 23 at 11:34

J...S

asked Mar 23 at 10:32

J...SJ...S

3,64911231

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  It's not the same: int i=5; is local scope and you can't make it extern. If you make it static it is still at local scope.
  
  – Weather Vane
  Mar 23 at 11:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  extern doesn't go "one level up". I goes to the global scope, so your i must be declared at the gloal scope. Your compiler should have complained.
  
  – Paul Ogilvie
  Mar 23 at 11:08
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  @PaulOgilvie Oh, I see. So there's no way around this if it was a non-global variable?
  
  – J...S
  Mar 23 at 11:29
  

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  Don't. Just use unique varaible names.
  
  – Weather Vane
  Mar 23 at 11:34
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Due to the compiler seeing the first instance of 'i' as unused, it is eliminated in the resulting object file. So does not exist when the linker is invoked.
  
  – user3629249
  Mar 23 at 19:49
  

  
  
  
  

 | 
show 4 more comments

I saw in this post that we can uncover a global variable where a local variable of the same name is in scope by using extern and a compound statement.

In that post, the example given is like

#include<stdio.h>
int a = 12; 
int main(void) 
 
 int a = 15; 
 printf("Inside a's main local a = : %dn", a);

 
 extern int a;
 printf("In a global a = %dn", a);
 

 return 0; 

and the shadowed variable is a global variable.

But when I tried a program like

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i;
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

it failed.

I tried making the outer i static in case it works only if the outer variable is static but still it didn't work.

Can somebody tell me why the second program doesn't work? And how to get around it if it's possible?

Maybe this method works only if the outer variable is global?

Edit: The comments made me realize that the use of extern will just make a global variable visible. But is there a way around this without changing variable name?

c scope global-variables local-variables

share|improve this question

edited Mar 23 at 11:34

J...S

asked Mar 23 at 10:32

J...SJ...S

3,64911231

#include<stdio.h>
int a = 12; 
int main(void) 
 
 int a = 15; 
 printf("Inside a's main local a = : %dn", a);

 
 extern int a;
 printf("In a global a = %dn", a);
 

 return 0; 

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i;
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

share|improve this question

edited Mar 23 at 11:34

J...S

asked Mar 23 at 10:32

J...SJ...S

3,64911231

share|improve this question

edited Mar 23 at 11:34

J...S

asked Mar 23 at 10:32

J...SJ...S

3,64911231

asked Mar 23 at 10:32

J...SJ...S

3,64911231

asked Mar 23 at 10:32

J...SJ...S

3,64911231

1 Answer
1

active

oldest

votes

0

In the below code snapshot

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* just declaration */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

It gets failed because linker is unable to find the definition of i.

One way to overcome this is

#include<stdio.h>
int main()

 int i=5; /* linker will not search this for definition of i */
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* search definition other than this translation unit */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

int i = 100; /* this is the definition of extern i */

Other way is to provide definition of extern variable i in the different file. Ideally extern is meant for external linkage i.e in other file, not in the same file.

share|improve this answer

answered Mar 23 at 12:18

AchalAchal

10.1k2931

add a comment | 

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0

In the below code snapshot

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* just declaration */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

It gets failed because linker is unable to find the definition of i.

One way to overcome this is

#include<stdio.h>
int main()

 int i=5; /* linker will not search this for definition of i */
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* search definition other than this translation unit */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

int i = 100; /* this is the definition of extern i */

Other way is to provide definition of extern variable i in the different file. Ideally extern is meant for external linkage i.e in other file, not in the same file.

share|improve this answer

answered Mar 23 at 12:18

AchalAchal

10.1k2931

add a comment | 

0

In the below code snapshot

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* just declaration */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

It gets failed because linker is unable to find the definition of i.

One way to overcome this is

#include<stdio.h>
int main()

 int i=5; /* linker will not search this for definition of i */
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* search definition other than this translation unit */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

int i = 100; /* this is the definition of extern i */

Other way is to provide definition of extern variable i in the different file. Ideally extern is meant for external linkage i.e in other file, not in the same file.

share|improve this answer

answered Mar 23 at 12:18

AchalAchal

10.1k2931

add a comment | 

In the below code snapshot

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* just declaration */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

It gets failed because linker is unable to find the definition of i.

One way to overcome this is

#include<stdio.h>
int main()

 int i=5; /* linker will not search this for definition of i */
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* search definition other than this translation unit */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

int i = 100; /* this is the definition of extern i */

Other way is to provide definition of extern variable i in the different file. Ideally extern is meant for external linkage i.e in other file, not in the same file.

share|improve this answer

answered Mar 23 at 12:18

AchalAchal

10.1k2931

#include<stdio.h>
int main()

 int i=5;
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* just declaration */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

#include<stdio.h>
int main()

 int i=5; /* linker will not search this for definition of i */
 for(int i=0; i<2; ++i)
 
 
 extern int i; /* search definition other than this translation unit */
 printf("nGlobal: %d", i);
 
 printf("nLocal: %d", i);
 

int i = 100; /* this is the definition of extern i */

share|improve this answer

answered Mar 23 at 12:18

AchalAchal

10.1k2931

answered Mar 23 at 12:18

AchalAchal

10.1k2931

answered Mar 23 at 12:18

AchalAchal

10.1k2931