Hi
imagine a display where new values are added at the bottom line. If a 
value gets untrue that line disappears and all below move up one line.

So list might be the easiest to use? i need to draw something alike:


if (8value == 1)
    {
XPLMDrawString(white, PanelWindowLeft + 20, PanelWindowTop + 20, 
"Test8", 0, xplmFont_Basic);
}

if (7value == 1)
    {
XPLMDrawString(white, PanelWindowLeft + 20, PanelWindowTop + 10, 
"Test7", 0, xplmFont_Basic);
}

if (9value == 1)
    {
XPLMDrawString(white, PanelWindowLeft + 20, PanelWindowTop + 30, 
"Test9", 0, xplmFont_Basic);
}

etc.
Now how could i have them dynamically moving up/down on true/untrue? ie. 
pack this into a list?
Many thanks
Michael

Ok now I try:

     list<char> charList;

     char* arp = "Test1";

     if (something)
         charList.push_back(*arp);

     // Display the list
     list<char>::iterator theIterator;
     for( theIterator = charList.begin(); theIterator != charList.end(); 
theIterator++ ) {
     XPLMDrawString(white, Panel1WindowLeft + 20, Panel1WindowTop - 20, 
theIterator, 0, xplmFont_Proportional);
     }


I get the error:
cannot convert �std::_List_iterator<char>� to �char*� for argument �4� 
to �void XPLMDrawString(float*, int, int,
  char*, int*, XPLMFontID)�
Many thanks for help

On Jan 2, 1:07=A0pm, Michael <Mich...@notspam.com> wrote:
> Ok now I try:
>
> =A0 =A0 =A0list<char> charList;
>
> =A0 =A0 =A0char* arp =3D "Test1";
>
> =A0 =A0 =A0if (something)
> =A0 =A0 =A0 =A0 =A0charList.push_back(*arp);
>
> =A0 =A0 =A0// Display the list
> =A0 =A0 =A0list<char>::iterator theIterator;
> =A0 =A0 =A0for( theIterator =3D charList.begin(); theIterator !=3D charLi=
st.end();
> theIterator++ ) {
> =A0 =A0 =A0XPLMDrawString(white, Panel1WindowLeft + 20, Panel1WindowTop -=
 20,
> theIterator, 0, xplmFont_Proportional);
> =A0 =A0 =A0}
>
> I get the error:
> cannot convert std::_List_iterator<char> to char* for argument 4
> to void XPLMDrawString(float*, int, int,
> =A0 char*, int*, XPLMFontID)
> Many thanks for help

compilers do their best to help you. The function wants a char* (a
pointer to a char) theIterator isn't a char* it's a
list<char>::iterator. To convert it you use the dereference operator
that is defined for list iterators.

        XPLMDrawString(white, Panel1WindowLeft + 20, Panel1WindowTop -
20,
               *theIterator, 0, xplmFont_Proportional);

Yes, that would be?  * theIterator

And i get:

- invalid conversion from �char� to �char*�

Thanks again

Michael wrote:
> Ok now I try:
> 
>      list<char> charList;

Please consider.

char c = 'a';  // c is a variable of type char
               // it can contain a value of one char.

charList is a std::list of type char, each element (or member or
whatever terminology we'll use) can contain one character.


> 
>      char* arp = "Test1";

     arp is a point to char.  A char* can point to a single character or
a zero terminated string or if you've rolled your own string handling
function perhaps something else. In this case it is pointing to a region
in memory that contains the characters 'T', 'e', 's', 't', '1', '\0'
sequentially.

> 
>      if (something)
>          charList.push_back(*arp);


Assuming that something is true, charList isn't empty anymore and has an
element that contains the value 'T'.  This will become clearer if you
will...
> 
>      // Display the list

Here's a little program to do that. I didn't copy and paste, so be careful.

#include <iostream>
#include <list>
int main() {
  std::list<char> cl;
  const char *p1 = "Test";
  const char *p2 = "Help";
  cl.push_back(*p1);  // 'T'
  cl.push_back(*p2);  // 'H'
  for(std::list<char>::const_iterator i=cl.begin(); i!=cl.end(); i++) {
    std::cout << *i << std::endl;
  }
}

Ought to output:
T
H

and not
Test
Help




>      list<char>::iterator theIterator;
>      for( theIterator = charList.begin(); theIterator != charList.end();
> theIterator++ ) {

I found this, http://www.xsquawkbox.net/xpsdk/mediawiki/XPLMDrawString

So yes, it looks like you want to pass a char* as the fourth parameter.

>      XPLMDrawString(white, Panel1WindowLeft + 20, Panel1WindowTop - 20, 
> theIterator, 0, xplmFont_Proportional);
>      }



Consider,

#include <iostream>
#include <list>


void someFakeFunction(char *p) {
  std::cout << p << std::endl;
}

int main() {
  std::list<char*> cl;
  char *p1 = "Test";
  char *p2 = "Help";
  cl.push_back(p1);
  cl.push_back(p2);

  for(std::list<char*>::iterator i=cl.begin(); i!=cl.end(); i++) {
    someFakeFunction(*i);
  }
}

Ought to output
Test
Help

I don't particularly care for this for several reasons.  I think you
ought to consider using std::string. You might have to write a wrapper
or some sort of smart pointer to get a non-const pointer to char.

Since you indicated earlier in the thread that each of these "Test?"
strings has an associated flag, you may wish to look into std::pair.
You could make a list like std::list< std::pair<bool, std::string> >.
Although it's not clear to me why you're using a std::list and not a
std::vector. Or maybe even just an array.

Also, your use of magic numbers,
>   XPLMDrawString(white, Panel1WindowLeft + 20, Panel1WindowTop - 20,
seems a bad thing to me. In the case of your loop above, it looks like
you will draw over the same space.  Your original code changed the
offsets for each item, also using magic numbers, but might have left
spaces if a particular text wasn't going to be drawn.


>
> I get the error:
> cannot convert �std::_List_iterator<char>� to �char*� for argument �4�
> to �void XPLMDrawString(float*, int, int,
>   char*, int*, XPLMFontID)�
> Many thanks for help


I think you can get the compilation error to go away if you change
  theIterator
to
  &*theIterator
but you will very likely have problems at run time since I suspect the
function you're calling expects a zero terminated string and your list
is of type char.

LR

     list<char> charList;
     char *arp = "TEST0";
         charList.push_back(*arp);

     // Display the list
     list<char>::iterator theIterator;
     for( theIterator = charList.begin(); theIterator != charList.end(); 
theIterator++ ) {
     XPLMDrawString(white, Panel1WindowLeft + 20, Panel1WindowTop - 20, 
&*theIterator, 0, xplmFont_Proportional);
     }

Many thanks. The above works but draws only T. Any idea on how to draw 
all "TEST0"?

Funny, never seen:  &*theIterator
before in my short C++ life...

Michael wrote:
>      list<char> charList;
>      char *arp = "TEST0";
>          charList.push_back(*arp);
> 
>      // Display the list
>      list<char>::iterator theIterator;
>      for( theIterator = charList.begin(); theIterator != charList.end(); 
> theIterator++ ) {
>      XPLMDrawString(white, Panel1WindowLeft + 20, Panel1WindowTop - 20, 
> &*theIterator, 0, xplmFont_Proportional);
>      }
> 
> Many thanks. The above works but draws only T. Any idea on how to draw 
> all "TEST0"?

You're not storing all of "Test0" in your list. Only the 'T' since
you're using a std::list<char>.  I think you got lucky that you only
showed the 'T' and not something like "Tl939wio923i1-0(**(&&*^".

Please review the post that you're replying to and take a look at the
example with std::list<char*>  I don't like it because I think it will
cause more problems than it will cure.

Consider making a small string class of your own, put it in a namespce
that makes sense for you, give it a method that returns a non-const char
*. Make your list of this class.


> Funny, never seen:  &*theIterator
> before in my short C++ life...

Take the address of the instance of char that theIterator 'points' to.

I'm still not clear on why you're using a list.

You might be better off trying this without the call to XPLMDrawString.
At first write a program that just writes the data to std::cout to see
what you get.


LR

Now if i try:


     do{
         charList.push_back(*arp);
     }while(*arp);


I get an endless loop. How should i do that?
Many thanks again
Michael

Michael wrote:
> Now if i try:
> 
> 
>      do{
>          charList.push_back(*arp);
>      }while(*arp);
> 
> 
> I get an endless loop. How should i do that?

I don't know why you would want to do this.

Consider making an array of pointers to char,
    char *a[]
and using a for loop to index through the array.

Try initializing the array and printing out all the values.


Are you using a book?

What compiler are you using?

LR

If I try something alike:

     list<char> charList;
     char displayTxt[50];
     displayTxt[0]= "Test1";

     for (int i =1; i <= 50; i++)
         charList.push_back(displayTxt[i]);

I get: invalid conversion from �const char*� to �char




> I don't know why you would want to do this.
>
I want a display. New values are added at the bottom. Untrue values are 
not displayed anymore and all values below are moved one position up:

Panel1WindowTop - 20  // draw position
etc.


> Consider making an array of pointers to char,
>      char *a[]
> and using a for loop to index through the array.
>
> Try initializing the array and printing out all the values.
>
>
> Are you using a book?
>
Jesse liberty c++ in 21 days.
> What compiler are you using?
GCC Linux and Mac later also VC.
Thanks again
Michael

Michael <Michael@notspam.com> wrote:

> If I try something alike:
> 
>      list<char> charList;
>      char displayTxt[50];
>      displayTxt[0]= "Test1";
> 
>      for (int i =1; i <= 50; i++)
>          charList.push_back(displayTxt[i]);
> 
> I get: invalid conversion from �const char*� to �char

Of course you are, but the error has nothing to do with the charList 
variable. The offending line is:

   displayTxt[0]= "Test1";

'displayTxt[0]' is of type 'char' and "Test1" is of type 'const char*'. 
You can't convert from const char* to char, hence the error.

Michael wrote:
> If I try something alike:
> 
>      list<char> charList;
>      char displayTxt[50];



>      displayTxt[0]= "Test1";

As was pointed out by Daniel T elsethread the above is the cause of the
error message you're getting below.

    displayTxt[0] = 'T';
    displayTxt[1] = 'e';
    displayTxt[2] = 's';
    displayTxt[3] = 't';
    displayTxt[4] = '1';
    displayTxt[5] = '\0';

would do what I think you would like your assignment statement to do.

A char is only a single character. It doesn't contain a string of
characters.  Very loosely speaking, "Test1" is like an array of char
somewhere in memory. That is, a string is sort of like an array of char.

char x[10]; doesn't mean an array of ten strings, but an array of ten char.

Consider trying this snippet,

char s[] = "hello";
std::cout << s << std::endl;

s[2] = 't';
std::cout << s << std::endl;

> 
>      for (int i =1; i <= 50; i++)
>          charList.push_back(displayTxt[i]);

You'll end up with a list of char with one character in each member of
the list.

> 
>> I don't know why you would want to do this.
>>
> I want a display. New values are added at the bottom. Untrue values are 
> not displayed anymore and all values below are moved one position up:
> 
> Panel1WindowTop - 20  // draw position
> etc.

At this point you seem to be doing more C style programming than C++,
even though you want to use std::list.

At some point you'll find the C++ faq useful, but at the moment you
might find more immediate help in the C faq.

http://c-faq.com/
http://c-faq.com/ptrs/index.html
http://c-faq.com/aryptr/index.html

The C++ faq is here
http://www.parashift.com/c++-faq-lite/

Also, IIRC Alf P. Steinbach who posts here frequently used to have a
pretty good tutorial posted somewhere online about C++ pointers.  I
can't find it at the moment. Perhaps you should ask him if he'd be so
kind as to provide a link to it or resurrect it if it's not online at
the moment.

Please read up on std::string, it might save you some trouble.




>> Consider making an array of pointers to char,
>>      char *a[]
>> and using a for loop to index through the array.
>>
>> Try initializing the array and printing out all the values.

This may still be worth trying as an exercise.

LR