Why the Access violation here ?

Hi all,
why I am getting "access violation" in line 1 below in foo(...)?

void foo( char* str )
{
if( str != NULL && strlen(str) > 0 )
{
str[0] = '9'; // line 1
}
}


int main(int argc, char* argv[])
{

char* str2 = "123456789";

foo( str2);

return 0;
}

Quote:

---

Originally Posted by Dhafir

Hi all,
why I am getting "access violation" in line 1 below in foo(...)?

void foo( char* str )
{
if( str != NULL && strlen(str) > 0 )
{
str[0] = '9'; // line 1
}
}


int main(int argc, char* argv[])
{

char* str2 = "123456789";

foo( str2);

return 0;
}

---

void foo( char* str )
{
if( str != NULL && strlen(str) > 0 )
{
str[0] = '9'; // line 1
}
}


int main(int argc, char* argv[])
{

char str2[] = "123456789";

foo( &str2);

return 0;
}

... try that!!!


.... btw this is a 'C' question not VC++

Thanks ADSOFT, VC++ gives me :
error C2664: cannot convert parameter 1 from 'char (*)[10]' to 'char *'

However, I am not looking for a solution, I am asking "why" I am getting "access violation" in my example?

The answer is a bit tricky!

int main(int argc, char* argv[])
{

char *str2[] = "123456789";

foo( &str2);

return 0;
}

... try that!!!


char *str2[] = "123456789"; // I placed * before str2

When u declare,
char* str2 = "123456789";
compiler creates a character pointer (str2) and initialized to point to a constant unnamed character array "123456789".
Attempting to modify the const array is causing the problem.

you may try the following,
char str2[] = "123456789";

but you should not attempt to assign some than 9 characters(since the initial size of ur array is 9)

Sreedharan

SreeDharan.. you got it..

It is the "successful" assignement of a const char* ( like "123456789") to a non const char* that is confusing in that piece of code. Normally the compiler would pick this up in other cases and throw a compilation error, for example:

std::string str1("123456789");
char* str2 = str1.c_str(); // error C2440:

compiling...
error C2440: 'initializing' : cannot convert from 'const char *' to 'char *'

the compiler will throw a C2440 at you and you would know, however it doesn't do the same for:


char* = "123456789"; // compiles successfully


Thanks to ADSOFT & SreeDharan

Dhafir

Quote:

---

Originally Posted by Dhafir

Hi all,
why I am getting "access violation" in line 1 below in foo(...)?

---

Just to add, it is undefined behavior to change a string-literal. Your program may not have necessarily crashed, depending on the compiler.

A string-literal:

Code:

---

char *p = "string";
p[0] = 'x';

---

This crashes in VC++ 6.0, but I believe that VC++ 5.0 didn't crash when you attempt to do this. That is what is meant by "undefined behavior" -- anything can happen.

Regards,

Paul McKenzie