Would somebody please tell me why the code below works, but the code below that gives an illegal access message.
Code:char MyString[] = "465.9900"; MyString[0] = '0';Doesn't the pointer method do the same thing?Code:char* MyString = "465.9900"; MyString[0] = '0';
You can assign any array size to mystring however it hasn't beenCode:char* MyString = "465.9900"; MyString[0] = '0';
done yet, so if you add that it should work.
Code:char* MyString = "465.9900"; MyString=new char[10]; MyString[0] = '0';
The first one creates an array of sufficient size and copies that string into it. The array is modifyable.Originally posted by CodeMonkey
Would somebody please tell me why the code below works, but the code below that gives an illegal access message.
Code:char MyString[] = "465.9900"; MyString[0] = '0';Doesn't the pointer method do the same thing?Code:char* MyString = "465.9900"; MyString[0] = '0';
The second one is just a pointer to that string which resides somewhere in the memory. Trying to write to it will result in an access violation error.
MagosX.com
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Teach a man to fish and you feed him for a lifetime.
That will not keep the old data since the pointer is repointed.Originally posted by Travis Dane
Code:char* MyString = "465.9900"; MyString=new char[10]; MyString[0] = '0';
MagosX.com
Give a man a fish and you feed him for a day.
Teach a man to fish and you feed him for a lifetime.
No, no. < = new char[10] > just sets it to a blank string of 10. Putting that before the < = "blahblah" > yields the same error. (in response to the first reply)
"If you tell the truth, you don't have to remember anything"
-Mark Twain
They are 2 very different things
This means allocate space on the stack and copy "465.9900" to it. The compiler sees "465.9900" as a pointer to constant data and copies it byte for byte to memory on the stack that you can alter.Code:char MyString[] = "465.9900"; MyString[0] = '0';
Here you are simply copying the pointer to the cont data......when you try change the array, you trying to change const data and the OS will likely raise an exception.Code:char* MyString = "465.9900"; MyString[0] = '0';
<edit>Magos was too quick for me</edit>
Thank you. Now how do I make this all work? I need to change individual values of char*'s.
"If you tell the truth, you don't have to remember anything"
-Mark Twain
>Now how do I make this all work? I need to change individual values of char*'s.
If you have a char * like so:
Then the only way to make it work is to either copy the string's contents into an array or dynamically allocate the memory. String literals may be placed in read-only memory and are off limits for assignment. Besides, using a char pointer in this manner is bad form for C++, a better way would be to say:Code:char *p = "some string";
Or just use a std::string.Code:char p[] = "some string";
-Prelude
My best code is written with the delete key.
>the OS will likely raise an ?>exception.
you mean only windows will raise an exception. accessing values from a pointer using array positions works fine under every other operating system ive used (linux, unix, solaris, bsd), just not windows.
>Besides, using a char pointer in this manner is bad form for C++
this is not true, the sentance should read.....
"Besides, using a char pointer in this manner is bad form for windows"
codemonkey, its not you. its windows. although accessing memory this way can be dangerous, it is alowed in other OS'
im not sure how to get around it in windows other than to do as suggested before and just use a char array instead of pointer.
No, it's not just windows. You're VERY confused.Originally posted by Perspective
>the OS will likely raise an ?>exception.
you mean only windows will raise an exception. accessing values from a pointer using array positions works fine under every other operating system ive used (linux, unix, solaris, bsd), just not windows.
>Besides, using a char pointer in this manner is bad form for C++
this is not true, the sentance should read.....
"Besides, using a char pointer in this manner is bad form for windows"
codemonkey, its not you. its windows. although accessing memory this way can be dangerous, it is alowed in other OS'
im not sure how to get around it in windows other than to do as suggested before and just use a char array instead of pointer.
String literals can be placed into real-only memory on any implementation. Attempting to write to read only memory will result in undefined behavior.
>this is not true, the sentance should read.....
>"Besides, using a char pointer in this manner is bad form for windows"
On the contrary, it is bad form for C++ because assigning a string literal to a non-const char * is a deprecated feature supported only for compatibility with C. Whether string literals themselves are in read-only memory or not depends on your implementation, but both the C and C++ standards explicitly state that string literals shall be treated as immutable.
-Prelude
My best code is written with the delete key.
ack, woke up still half drunk and responded to this thread which i totally missunderstood. sorry to get so $$$$$y about it (it was a long night...) anywho, what i meant was access rights crashing windows programs but not linux ones. you guys are right about not being able to write to the read only memory allocated.
>Fordy said modifying a constant string literal
i totally missed the entire concept of this post.... i bow my head in shame....
example of what i was trying to say:
srry, bout that.Code:char* p = "string"; cout << *(p+2); // legit ways of accessing data cout << p[2]; // access memory that is not yours, windoze will crash
>cout << *(p+2); // legit ways of accessing data
>cout << p[2]; // access memory that is not yours, windoze will crash
The two methods are equivalent and correct, regardless of the operating system.
-Prelude
My best code is written with the delete key.
They're exactly same thing, when you compile it, the compiler changes from p[2] to *(p+2).char* p = "string";
cout << *(p+2); // legit ways of accessing data
cout << p[2]; // access memory that is not yours, windoze will crash
>The two methods are equivalent and correct, regardless of the operating system.
right, where the difference in operating systems comes into play is when you try to access memory that is beyond the scope of your c-string. it is not correct to or advisable, but different OS' will respond to it differently. windows will just terminate the program while some other systems will allow you read access to this memory.
the sample of code was, as i already said, just to clarify what i had thought the origional question was in this post. if you didnt notice, i already admited to missunderstanding the intent of the question.
