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try www.planet-source-code.com
sonork: 100:18407
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I am newly learning windows programming. I am trying to create a sort of simple application. When I try and link my project I get an error reading - fatal error RC1004: unexpected end of file found
I am not sure how to deal with this. Any suggections? The error stems from the header file:
// define for the application icon
#define ICON_T3DX 100
// define for the application cursor
#define CURSOR_CROSSHAIR 200
// define for the application music
#define SOUND_ID_MUSIC 1
// defines for the top level menu OPTIONS
#define MENU_OPTIONS_ID_PRINT 1000
#define MENU_OPTIONS_ID_EXIT 1001
// defines for the top level menu HELP
#define MENU_HELP_ABOUT 2000
#define MENU_HELP_INSTRUCTIONS 2001
I am using the header to help with the implementation of resources.
If anyone can help me resolve this error, that would be great.
Thanks a lot!!!
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did you include the header?  i've done that alot 
~SilverShalkin
![Rose | [Rose]](https://www.codeproject.com/script/Forums/Images/rose.gif)
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#include <stdafx.h>
Regards,
Nish
Native CPian.
Born and brought up on CP.
With the CP blood in him.
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I've tried doing that. I get another error, "the file cannot be opened. I checked for it in the IDE file directory and it isn't thete. Any other suggestion?
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Sorry... I did find and include th file in the project afterwards but ut still gives the same error- unexpected end of file found.
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You need to #include stdafx.h for all your cpp files
Regards,
Nish
Native CPian.
Born and brought up on CP.
With the CP blood in him.
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i have seen the samples in msdn, but all of them is written by vb.net. 
c++ : my dream
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There is a ATL-wizard that will create an ATL-based VS.NET Add-In for you... It includes code for creating a Toolbar.
(No more MFC-based Add-ins, it seems...)
Peace!
-=- James.
"Some People Know How To Drive, Others Just Know How To Operate A Car."
(Try Check Favorites Sometime!)
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can anyone please tell me how do i put this to boxes in my dev studio :NET
Picture [New Window]
they had disappeared... 
and i cant get them back 
i cant find any option
tanks
Casa.Sapo.pt
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I haven't used VS.net so I don't know if it's changed from VS6, but try right-clicking in an empty part of the toolbar area, and put a tick next to "WizardBar" in the popup menu
--
Help me! I'm turning into a grapefruit!
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this is my code:
#include <iostream>
using namespace std;
char *pchar[6];
char character[6] = "hello";
int main()
{
strcpy(pchar, &character);
return 0;
}
that is "all" my code, no more no less. the problem accures on line:
strcpy(pchar, &character);
im not very sure on what this'll do, what i am trying to do is, have pchar point to the address of character. i thought that if there was more than one character i should use strcpy(), but im not very sure.
how would i accomplish this? "see any problems, or better way of doing things, just note it "
Thanks!
~SilverShalkin
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Two problems:
1) pchar is an array of char*, not char. So you will not be able to access the elements in the pchar array as a string. Which means that you will not be able to send that array to strcpy.
What you would actually do with an array like pchar is point to 6 different strings. Then when you wanted to use one of the strings you would access the string in the array like so:
strcpy(pchar[0], ...);
2) character is a char array, which is equivalent to one char*. Therefore you do not need to take the address of the character array, you could directly pass it into strcpy like this:
strcpy(..., character);
Good Luck
Build a man a fire, and he will be warm for a day
Light a man on fire, and he will be warm for the rest of his life!
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but i want to point to character, if i dake the address (&) off, it wouldnt be point to character would it?
so i would have to declare my pchar like
char *pchar[6];
pchar[0] = &character[0];
pchar[1] = &character[1];
pchar[2] = &character[2];
pchar[3] = &character[3];
pchar[4] = &character[4];
pchar[5] = &character[5];
?
i hope thats wrong... thats allot to type for somthing so small 
Thank!
~SilverShalkin
![Rose | [Rose]](https://codeproject.global.ssl.fastly.net/script/Forums/Images/rose.gif)
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Lets assume for a minute that you did do that.
You could then do this:
printf("%s\n",pchar[0]);
printf("%s\n",pchar[1]);
printf("%s\n",pchar[2]);
printf("%s\n",pchar[3]);
printf("%s\n",pchar[4]);
printf("%s\n",pchar[5]);
and the results would be:
hello
ello
llo
lo
o
<<there would="" be="" a="" blank="" line="" here="">>
in summary an array of characters IS a pointer to a sequence of characters.
so if we were to do this instead
char characters[6] = "hello";
char* pchar = characters;
char* pchars = &characters[0];
then examine the values of all three variables in the debugger.
they would all be identical.
Hope this helps in your struggle with pointers.
Roger.
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Do this instead :-
char *pchar;
char abc[6];
strcpy(abc,"hello");
pchar = abc;
Nish
Regards,
Nish
Native CPian.
Born and brought up on CP.
With the CP blood in him.
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would that work? - ill insert some of the code replies and try to understand this more... thanks guys!
~SilverShalkin
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SilverShalkin wrote:
would that work?
Of course, it will work
Nish
Regards,
Nish
Native CPian.
Born and brought up on CP.
With the CP blood in him.
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WARNING: This is a long post.
I think your real problem here is a misunderstanding of how C style pointers work and how to use them. Let me start by saying that a pointer is an intrinisic data type just like float, int, char, etc. The pointer data type (on Win32 systems) is a 32 bit value. This means that when you write a statement like
void* pSomeValue;
you are telling the compiler to allocate 32 bits of memory for your program. It also identifies that you intend to populate that variable with a memory address. In the case of a void type pointer you are saying that the memory pointed to by the variable can be of any type you want. (In other words, no type affinity.)
C and C++ allows pointers to have type affinity also. This means that instead of being a void pointer, you can define a pointer like
char* pszSomeValue;
OR
long* plSomeValue;
OR
CSomeClassName* pSomeValue;
ETC.
When you define a pointer like this you are telling the compiler that you want it to allocate 32 bits of storabe for a variable that will eventually point to a specific data type (char, long, CSomeClassName).
The next subject of confusion (I think) is when you are using C style arrays of a particular data type. In your example, you used char* pchar[6] and char character[6] = "hello".
The char* pchar[6] is instructing the compiler to allocate 6 memory locations each having a type affinity for a pointer to a char. This means that the compiler will allocate six 32 bit memory locations whose values it expects to be pointers to a char (char*).
The char character[6] is instructing the compiler to allocate 6 memory locations each having a type affinity for a char (NOT a pointer to a char). This means that the compiler will allocate six 8 bit memory locations (that is the size of a char data type) whose value it expects to be a char.
One of the nuiansces of C style arrays is that when you allocate a C style array the compiler guarantees that the memory it allocates will be continuous. This means that when you do char *pchar[6];, the compiler is going to allocate 32*6 bits in a row. When you do char character[6], the compiler is going to allocate 8*6 bits in a row.
OK. That was a long winded explanation of pointers and arrays. With that understanding, lets look at the line of code where the compiler is giving you a problem.
strcpy(pchar, &character);
The strcpy function is defined as taking 2 paramaters, a char* and a const char*. (Ignore the const for now.) The second parameter is supposed to be a pointer to an array of characters with the last character being NULL (NULL terminated string.) The first parameter is supposed to be a pointer to an array of characters large enough to contain the data pointed to by the second parameter. (BYW strcpy will copy the contents of the second parameter to the first, therefore you will have 2 seperate copies of the data. Changing one will not affect the other.)
For the first parameter, you are passing in a an array of char* type pointers. This is quite different from an array of char type. (Remember char* = 32 bits, char = 8 bits.) As you can see, for the first parameter, the compiler is confused because it is expecting the char array, not the char* array.
You say "what i am trying to do is, have pchar point to the address of character". If this is really what you want, then what you would want to write is
char *pchar;
char character[6] = "hello";
pChar = character;
You may be thinking "how can I assign an array to a pointer?". The answer to this obvious questions goes back to the previous explanation or how C style arrays work. When you specicify the variable name of a C style array without specifying an index into the array (ie. character instead of character[0]), the compiler knows that you want a pointer to the array.
I hope this helps more that it confuses.
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it helps allot, thank you for spending the amount of time that it took to type that long thing I know how pointers and stuff work, its just, there are a couple of things that are in the gray, as in (I dont know if it will do the exact thing i want it)
i guess i should use strcpy() for a pointer, "it doesnt make sense to do so "
i think i should print this out...
Thanks for your time!
~SilverShalkin
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Strange...
(not picking on you redeemer)
Tim Smith
I know what you're thinking punk, you're thinking did he spell check this document? Well, to tell you the truth I kinda forgot myself in all this excitement. But being this here's CodeProject, the most powerful forums in the world and would blow your head clean off, you've got to ask yourself one question, Do I feel lucky? Well do ya punk?
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did you spell check that?
~SilverShalkin
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Tim Smith wrote:
not picking on you redeemer
Then, who else are ya picking on?
Moi???
Nish
Regards,
Nish
Native CPian.
Born and brought up on CP.
With the CP blood in him.
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i have a picture box controls on a dialog box.i want to show bitmap on it, on the click event of another dialog. doing so causes DEBUG ASSERTION FAILED! msg first and then image iz displayed on the top left corner of screen instead of picture box. i m trying to solve the problem for the last two weeks but didnt find any solution.
(if i use the device context of Dialog instead of picture box then it works well but when i try to get the device context of picture box it don't.)
if u know something about this problem or have some helping code plz give it to me. thanx
God Helps those who help others.
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