Why printf() is printing 0 instead of 10 in following code?

Question

If i compile and run the following code, it is printing 0 instead of 10.

#include<stdio.h>
main()
{
    int Var=10;
   {
     char Var=Var;
     printf("%d",Var);
   }
}

Why this is printing 0 and why not 10 ?

Now you would be like to read this: [Point of declaration in C++](http://stackoverflow.com/questions/15746271/point-of-declaration-in-c#) but its a C++ link — Grijesh Chauhan, Apr 09 '14 at 06:59
[Point of Declaration](http://msdn.microsoft.com/en-us/library/179x7xb9.aspx) — Grijesh Chauhan, Apr 09 '14 at 07:03

score 19 · Accepted Answer · edited May 23 '17 at 12:10

19

Because in local declaration

 char Var=Var;

the right occurrence of Var refers to the local Var, not the upper one. As Alk commented, this is undefined behavior to assign from an uninitialized variable.

So your declaration does not initialize Var at all, i.e. Var contains garbage. In your particular case, that garbage happens to be 0.

BTW, having two homonyms Var in the same function is really bad taste.

As this answer suggests, you should compile with gcc -Wall -Wshadow then you'll get some warnings on your code. (also add -g to get debug information to be able to debug with gdb)

edited May 23 '17 at 12:10

Community

1
1

answered Apr 09 '14 at 05:52

Basile Starynkevitch

223,805
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So, shouldn't it contain garbage ? – brokenfoot Apr 09 '14 at 05:54
2

So it can print an garbage value including `0`.. right!! – 0xF1 Apr 09 '14 at 05:55
+1! Does compiler or IDE can warn about potential problem. Sorry , I am used to IntelliJ inspections. – Jayan Apr 09 '14 at 05:56
@Basile: makes sense now. – brokenfoot Apr 09 '14 at 05:57
Also it not always print `0` but anything. – Jayesh Bhoi Apr 09 '14 at 05:57
@Jayan : I think user702245's answer is what you are looking for. (`-Wshadow`) – 0xF1 Apr 09 '14 at 05:59
Moreover reading out the inner `Var` invokes *undefined behaviour*, as it hadn't been initialised prior to being read. – alk Apr 09 '14 at 06:29

score 8 · Answer 2 · answered Apr 09 '14 at 05:58

8

Assuming you are using gcc, you would want to turn on -Wshadow (http://gcc.gnu.org/onlinedocs/gcc/Warning-Options.html). This would pop up an error, at the inner variable having same name as outer. The zero is a random value. It could print any garbage in that place.

answered Apr 09 '14 at 05:58

tpb261

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3
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You might consider deleting your answer now, since it has become redundant after [this](http://stackoverflow.com/a/22953927/2235132) was edited. – devnull Apr 09 '14 at 07:08
That answer has a link to this answer. Is it ok to delete? I don't know, is this even the right place to ask this question? – tpb261 Apr 09 '14 at 07:54

score 2 · Answer 3 · answered Apr 09 '14 at 06:06

2

try this

#include<stdio.h>
main()
{
    int Var1=10;
   {
     char Var=Var1;
     printf("%d",Var);
   }
}

answered Apr 09 '14 at 06:06

Arsalan Qaiser

426
2
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Why printf() is printing 0 instead of 10 in following code?

3 Answers3