Again pointer question

[oldnewbie]

While I was surfing for pointer tutorials,I came across an example and couldnt get it.
(This is from a question and an answer to it)
Here it is:

char *pstr = "string literal which can't be altered\n";
char astr[] = "string literal which can be altered\n";
char bstr[] = "string literal which can be altered\n";
cout << pstr << astr << bstr; // 3 strings are printed,OK
*(astr+1) = 'X'; // OK
cout << astr;     // output=sXring literal which can be altered
*(pstr+1) = 'X'; // access violation
cout << pstr;

Why this is so?Is it about stack/heap topics? (which I dont know much about)

Also,I tried this:

char *pstr = "string literal which can't be altered\n";    
char astr[] = "string literal which can be altered\n";
char bstr[] = "string literal which can be altered\n";    
  
cout<<(void*)pstr << "  " << (void*) astr << "  " << (void*) bstr;

And the output is;

0X444000 0X23ff10 0X23fee0

The string that we initialized char pointer is put in a very different location on memory.(at least I think so when I see these numbers).Can you explain this?

[firedraco]

Due to C, conversion from a string literal to a char*, while technically illegal, is allowed. However, if you attempt to modify it, you get an illegal access error.

Anyway, this is basically how it works:
1) Make a char* that points to the constant string literal in memory (no copying)
2/3) Create a new character array and *copy* the string literal in the (modifyable) memory

[hanst99]

Well... first of all, this:
char *pstr = "string literal which can't be alteredn"; is deprecated, your compiler should give you warnings about this. String literals are constant, and you are implicitly casting it to a non-constant char* here. I am not entirely sure, but I think char astr[] = "string literal which can be alteredn"; actually creates a copy of the string here. Anyways: Your char* here points to a constant string here, and the result of trying to access a constant string is undefined. The more correct way to do this (I say more, because using C-strings in C++ is usually unnecessary and evil anyways), would be the following:

#include <iostream>
#include <cstring>

using namespace std;

int main(int argc, char** argv)
{
	char* c= new char[strlen("Can't be altered")+1];
	strcpy(c,"Can be altered");
	cout<<c;
	*(c+1) = 'x';
	cout<<c;
	return 0;
}

[oldnewbie]

I see, since we are trying to modify a constant value,we have access violation error.And this is not given on compile time,but on runtime.

I wonder if there is a read-only memory section (which is included in the part that is given to our small program), which we can NOT modify?

OR

The above situation(int the first post) is similar to below one :

const int a=5;
a=6 // a is constant!

But here compiler warns me :
assignment of read-only variable `a'

In previous char *pstr="A string"; case, I face the error in runtime,with a windows error report.Compiler didnt give a warning.

[hanst99]

Your compiler should give a warning here, though it might only do so if you have the Wall flag on.

I wonder if there is a read-only memory section (which is included in the part that is given to our small program), which we can NOT modify?

maybe, maybe not. Depends on your system and what code your compiler generates, but it's defined in the language standard that you aren't allowed to do that.

[techie07]

When you are assigning string as:
char *pstr = "string literal which can't be alteredn";

The string will be saved in constant area in memory, which you can not modify.


Gorav
http://www.kgsepg.com