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C Discussion Thread
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Aug 19th, 2007

Access Violation

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Consider the following code:

#include<stdio.h>
#define SET_BIT(buf,n,val)buf[n]=val
int main()
{
char ch='D';
char *name="Yankee Duddle";
int i=5;
SET_BIT(name,i,ch);
printf("%s",name);
}

When compiling with Borland C it worked fine. However when I tried cimpiling it with VS 2005 I got the following error:
"Unhandled exception at 0x00413529 in Learning C.exe: 0xC0000005: Access violation writing location 0x00415645."

When I changed the code to:

(Toggle Plain Text)

char name[]="Yankee Duddle";

Instead of

(Toggle Plain Text)

char *name="Yankee Duddle";

It worked fine. I found this resolution in a forum.

But based on what I learnt in C I don't understand why it did not work in VS. And why by making it an array rather than a pointer it made a difference?

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amishosh

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Aug 19th, 2007

Re: Access Violation

In case char *name="Yankee Duddle"; string literal turns into an unnamed, static array of characters, and this unnamed array may be stored in read-only memory, and which therefore cannot necessarily be modified. Trying to modify it is undefined behaviour.
Whereas in case of char name[]="Yankee Duddle"; you can modify the contents safely.

SpS

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Aug 19th, 2007

Re: Access Violation

Click to Expand / Collapse Quote originally posted by amishosh ...

When compiling with Borland C it worked fine. However when I tried cimpiling it with VS 2005 I got the following error:
"Unhandled exception at 0x00413529 in Learning C.exe: 0xC0000005: Access violation writing location 0x00415645."

But based on what I learnt in C I don't understand why it did not work in VS. And why by making it an array rather than a pointer it made a difference?

char *name="Yankee Duddle"; is a constant, it can not be modified.
char name[]="Yankee Duddle"; the elements can be modified.
calling the SET_BIT in main you are trying to modify the value in char *name.

[EDIT]: Sorry, SPS your posted while I was writing this. Making this post a repetition.

Last edited by Aia; Aug 19th, 2007 at 1:35 pm.

Aia

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Aug 19th, 2007

Re: Access Violation

Why do the two compilers behave differently?

amishosh

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Aug 19th, 2007

Re: Access Violation

I think I found the answer (if anyone else has this problem):
http://msdn2.microsoft.com/en-us/lib...yy(vs.71).aspx

amishosh

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Aug 19th, 2007

Re: Access Violation

> Why do the two compilers behave differently?
Because your code invokes undefined behaviour.
The assumption you made was that string constants can be modified, and they cannot.

If your code is correct, then you'll get the same answer with every compiler.

Salem

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