Re: Problem with Array of char* - C / C++ / MFC Discussion Boards

Good point!
Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

Luc Pattyn wrote:
...you lack the necessary curiosity.

Eloquently said. Thumbs Up | :thumbsup:

BTW, there's a type-o on your home page.

"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown

"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

Hi David,

DavidCrow wrote:
there's a type-o

there could be several, and they could well be real mistakes, can you be more specific please.
I'm a bit in doubt about "client's IP address" ...

Smile | :)

Luc Pattyn [Forum Guidelines] [My Articles]

- before you ask a question here, search CodeProject, then Google
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get
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Luc Pattyn wrote:
...can you be more specific please.

strengthen

Luc Pattyn wrote:
I'm a bit in doubt about "client's IP address" ...

That's correct.

"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown

"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

Hi David,

DavidCrow wrote:
strengthen

Waw. Thanks for pointing that out. I had no idea, and Googling both shows both appear in similar quantities. I am struggling with th and ht quite often, e.g. height as opposed to width.

is there an easy way to figure it out (apart from Google, Webster and other lookups)?

Smile | :)

Luc Pattyn [Forum Guidelines] [My Articles]

- before you ask a question here, search CodeProject, then Google
- the quality and detail of your question reflects on the effectiveness of the help you are likely to get
- use the code block button (PRE tags) to preserve formatting when showing multi-line code snippets

modified on Wednesday, February 18, 2009 10:33 AM

As far as I understand, you want to pass a function pointer to a thread procedure. In a simplified way it could be done like so:

void MyFuncToPass(int)
{
//..
}

UINT MyThread(PVOID param)
{
typedef void (*F)(int);
F func = (F)param;
func(10); // Call my func passed

return 0;
}

void TheCaller()
{
// run the thread and pass the function to it.
AfxBeginThread(MyThread, (void*)MyFuncToPass);

}

--
Arman

Hi,

I have a function in a class that has the following sig:

int strtoken(char *str, char *separator, char *token[])

for the *token I do this before:

char *token[256];
			for(int i = 0; i < 256; i++)
			{
				token[i] = '\0';
			}

I call the function but at this line it crashes:

token[0] = strtok(str, separator);

Error on writing etc. (German Error, so I don't know exactly what it's called in english)

Can someone help me?

You can simply initialize the array as follows.
char *token[256] = { 0 };

strtok return a char* (array of character) not a single char.

See the example on using strtok taken from msdn. Read the documentation before using it.

#include <string.h>
#include <stdio.h>

	char string[] = "A string\tof ,,tokens\nand some  more tokens";
	char seps[]   = " ,\t\n";
	char *token;

	int main( void )
	{
		printf( "Tokens:\n" );

		// Establish string and get the first token:
		token = strtok( string, seps ); // C4996
		// Note: strtok is deprecated; consider using strtok_s instead
		while( token != NULL )
		{
			// While there are tokens in "string"
			printf( " %s\n", token );

			// Get next token: 
			token = strtok( NULL, seps ); // C4996
		}
	}
</stdio.h></string.h>

-Sarath.
"Great hopes make everything great possible" - Benjamin Franklin

My blog - Sharing My Thoughts

It's very difficult to guess without any error codes or messages.

All I can think of is that you passed a constant string for the str parameter because strtok will modify the str parameter.

«_Superman_»

IMHO there is no apparent reason for the crash, could you please elaborate a bit
(i.e. what are str and separator values?
Could you please post

strtoken

code?

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

Ok look, here is the full function (that I did NOT write myself) which should work correct, but I think >>I<< do something NOT correct.
Its a tokenizer for char-strings

int strtoken(char *str, char *separator, char *token[])
	{
		int i = 0;

		token[0] = strtok(str, separator); //This line crashes

		while ( token[i] ) 
		{
			i++;
			token[i] = strtok(NULL, separator);
		}
		return ( i );
	}

And I call it that way atm:

char *token[256]; //I think that is an array of char*, isn't it?
for(int i = 0; i < 256; i++)
{
    token[i] = '\0';
}
Helper::instance()->strtoken(name, ".", token);

name is a char* e.g. "system.fullscreen"

OK, as already suggested by «_Superman_», probably your code is crashing because you're passing a string literal to the function (strtok cannot access for writing a constant string).
for instance

char * name = "system.fullscreen";
token[0] = strtok(name, "."); // exception here

would crash the application, while

char name[] = "system.fullscreen";
token[0] = strtok(name, ".");

should work fine.
Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

Your right, I get a char* name as a parameter and use it to call the function. Thats the cause of the crash. But how do I solve this problem?
I have this char *name and it has to be tokenized. How can I do that as strtok crashes with it?

Btw. thanks for your help, I would have never thought of THAT ^^

well, if you really need to use that fucntion, then you may assign the array the way I shown in my previous post, or do someting like:

char * name;
name = _strdup("system.fullscreen");
if ( name )
{
  strtoken(name, ".", token);
  free(name); // don't forget to free memory
}

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

But if I call it with that char* it crashes!

I have a function

doSomething(char *name)
{
  [...]
  strtoken(name, ".", token); //crashes
}

That works but it does not help me:

doSomething(char *name)
{
  [...]
  char foo[] = "hello.world";
  strtoken(foo, ".", token); //does NOT crash
}

But how can I use strtoken with char *name?

Can I make a char foo[] out of char *name?

if you really need to pass a constant string to doSomething, the you may write
(as already suggested)

doSomething(char * name)
{
  //...
  char * name_writable_copy = _strdup( name );
  if ( name_writable_copy )
  {
    strtoken(name_writable_copy, ".", token);
    free( name_writable_copy );
  }
  else
  {
   // handle (unusual) allocation error 
  }
}

The above code shouldn't crash.
Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

Ok thx. I will try it in the evening today and post my results.

Well, good luck!
Smile | :)

If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler.
-- Alfonso the Wise, 13th Century King of Castile.

This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong.
-- Iain Clarke

[My articles]

You missed to tell us how your 'name' variable is declared/defined Smile | :)

. Except for it, other stuff seems OK to me (in the sense that they should not provoke a crash).

--
Arman

error1408 wrote:
Helper::instance()->strtoken(name, ".", token);

How about something like:

char *temp = new char[strlen(name) + 1];
strcpy(temp, name);
Helper::instance()->strtoken(temp, ".", token);

"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown

"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

DavidCrow wrote:
How about something like:

char *temp = new char[strlen(name) + 1];
strcpy(temp, name);
Helper::instance()->strtoken(temp, ".", token);

Ok this works! But I don't understand why...could someone explain it to me? Why does the param char * not work but the char * thats created here?

Btw. I have to DELETE the char *temp afterwards, right?

error1408 wrote:
Why does the param char * not work but the char * thats created here?

Work through this:

void main( void )
{
    char *abc = "First";
    char xyz[] = "Last";
 
    abc[0] = '1';
    xyz[0] = '2';
}

error1408 wrote:
Btw. I have to DELETE the char *temp afterwards, right?

Correct (since it points to heap memory).

"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown

"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

Ok I think I got it. So is it right, that in my case I can not know if the char *name that I get is a literal or not so to be sure it's not I have to create a temp array?

error1408 wrote:
...I can not know if the char *name that I get is a literal or not so to be sure it's not I have to create a temp array?

You'll notice in my example that both variables pointed to a string literal, yet only one of them could be changed. Therein lies the difference between a char* vs. char[]. See here for more.

"Old age is like a bank account. You withdraw later in life what you have deposited along the way." - Unknown

"Fireproof doesn't mean the fire will never come. It means when the fire comes that you will be able to withstand it." - Michael Simmons

i agree with the pallini's solution and also david gave a good description. string literals r located in a const part of memory which is not writable. this is why arrays don't cause crash. they're located in heap or stack based on whether u allocate them or use local variables which r both writable parts of memory.
there's an API function to determine whether a block of memory or a string is const named AfxIsValidAddress. it may also test whether the block is writable. but if i were u, i would use _strdup anyway, and would pass the arg as const for the caller to make sure that the original version of the passed string is not altered.