C / C++ / MFC

8-Apr-02 5:48

Hello. I am stuck on my first
external function call.

Here is a short program:

// convert string to long double
#include <iostream>
#include <math.h>
using namespace std;

int main()
{
char str1[] = "123.21";

long double _atold(const char *);

long double d = _atold(str1);

cout << d;

return 0;
}

My problem is using the _atold() library function, which is in the math.h file.
I have tried everything, and
get the link error -
unresolved external symbol "long double __cdecl _atold(char)" (?_atold@@YAOD@Z)

I would REALLY appreciate help
on this one. I could use
other methods, but I want to
get this working. Thanks

David

lucy8-Apr-02 5:50

lucy

8-Apr-02 5:50

do you mean the function which convert a string to a double variable? It's atof().
example:

#include <stdio.h>
#include <strlib.h>

.....
char s[10] = "123.41"; double x;
x = atol(s);
printf("%f\n", x); // "123.41 is printed
....

I didn't find any funciton defined as atold. There is atof, and atoi (to an integer) and atol (to a long value).

davun8-Apr-02 6:12

8-Apr-02 6:12

Hi Lucy,

thanks for the reply. I am learning
from The Waite Group OOP c++
just now. They require a string
be converted to Long Double,
not float, which I thought about
doing as a short cut. But here
is a clip from my math.h file:

#define _LD_COMPLEX_DEFINED
#endif

long double __cdecl acosl(long double);
long double __cdecl asinl(long double);
long double __cdecl atanl(long double);
long double __cdecl atan2l(long double, long double);
long double __cdecl _atold(const char *);

and at the bottom is _atold().
How can I use this?

char s[7] = "123.21"; long double d;
d = _atold(s);
cout << d;

This way, I get the error, undeclared identifier (_atold).

Thanks.

Ernest Laurentin8-Apr-02 6:31

Ernest Laurentin

8-Apr-02 6:31

The function _atold() is defined only if you are compiling for _M_M68K (Mot. 68K)
is defined. Use atof()!

"Dirty hands lead to important discovery..." - Thomas Edison

davun8-Apr-02 8:34

8-Apr-02 8:34

Thanks Ernest,

but can I ask, how would one
convert a string to long double?

Would I need to use atof(), then
use a static cast to ld? I would
think there is a direct method.

Thanks for your help.

David

lucy8-Apr-02 8:46

lucy

8-Apr-02 8:46

according to MS document, long double is equal to double. and atof() will convert a string to double. I think this is what you want for the assignment?

8-Apr-02 8:47

atof function does return a double value.

Try this ...

#include <stdlib.h>
#include <stdio.h>
#include <isostream.h>

using namespace std;

int main()
{
char str1[] = "123.21";

double d = atof(str1);

cout << d;

return 0;
}

davun8-Apr-02 9:24

Re: ScrollWindowEx Problem!

8-Apr-02 9:24

thanks all. Very much appreciated.

ScrollWindowEx Problem!

G.Richard8-Apr-02 5:43

G.Richard

8-Apr-02 5:43

does anyone knows how to use ScrollWindowEx?

c++ : my dream

alex.barylski8-Apr-02 7:22

alex.barylski

8-Apr-02 7:22

Nope..no one uses ScrollWindowEx anymore... Poke tongue | ;-P

I'm afraid your going to have to be more specific than that...

Cheers

"An expert is someone who has made all the mistakes in his or her field" - Niels Bohr

Re: ScrollWindowEx Problem!

G.Richard9-Apr-02 3:36

G.Richard

9-Apr-02 3:36

I want to use the ScrollWindowEx to animate window, when scrolling.
i mean I can't use ScrollwindowEx do the same thing that scrollwindow do, if i change the last parameter to SW_SMOOTHSCROLL. it can't update the region scrolled.why not have a try? you can test the following code inyou application.

...
CRect rcClient;
GetClientRect(&rcClient);
ScrollwindowEx(100,0, &rcClient, &rcClient, NULL,NULL, SW_SMOOTHSCROLL);
...

c++ : my dream

Help with multiple classes

Xds8-Apr-02 5:32

Xds

8-Apr-02 5:32

Hi all,

This is my first post; I'm new to C++ so go easy on me.
I have two classes; and their functions include code that refer to the other:
i.e: Class X has code that uses a variable/function from Class Y and vice-versa. The classes are completely different, so it would not help to derive one off the other.

In Header.h:

------------------------------------------------------------
class X {
public:
	int  myint;
	void doit(void);
};

void X::doit(void) {
	printf("CY.myint = %i", CY.myint); // Print Class Y's myint
}


class Y {
public:
	int  myint;
	void doit(void);
};

void Y::doit(void) {
	printf("CX.myint = %i", CX.myint); // Print Class X's myint
}
------------------------------------------------------------

In Main.CPP:

------------------------------------------------------------
#include  "stdio.h"
#include "Header.h"

int main(void) {

	class X CX;
	class Y CY;

	CX.myint = 100;
	CY.myint = 200;
	CX.doit();
	CY.doit();

	return 1;

}
------------------------------------------------------------

How can I get these classes to work together?
I can't put an 'extern class Y' before the 'class X' declaration, so how can I let the compiler know that the other class exists and will be declared?

Thanks!

Mike

Rickard Andersson²⁰8-Apr-02 5:39

Rickard Andersson²⁰

8-Apr-02 5:39

Xds wrote:
int main(void)
{
class X CX;
class Y CY;

CX.myint = 100;
CY.myint = 200;

CX.doit();
CY.doit();

return 1;
}

I'm not sure about this but isn't it "wrong" to write 'class X CX'?
Isn't 'X CX' the right way!?

I can be wrong!!!!!!

------------------------------------
Rickard Andersson, Suza Computing
ICQ#: 50302279
I'm from the winter country SWEDEN!

------------------------------------

Xds8-Apr-02 5:49

Xds

8-Apr-02 5:49

Pretty sure it works either way, but in any case, that's not the problem.
Thanks!

Mike

Roger Allen8-Apr-02 6:09

Roger Allen

8-Apr-02 6:09

You need to use a forward declaration of the class.

class Y ; // forward class declaration

class X
{
 void DoIt(Y& y) ; // pass in a reference to a Y object
} ;

class Y
{
} ;

You also cannot print out the content of CY in CX unless you pass a reference / pointer / copy of the object you want to print in as a parameter. The CX in main does not know anything about the CY:

#include  "stdio.h"
#include "Header.h"
int main(void) 
{
	X CX;
	Y CY;
	CX.myint = 100;
	CY.myint = 200;
	CX.doit(CY); // see class declaration above
	CY.doit();
	return 1;
}

Roger Allen
Sonork 100.10016

If I'm not breathing, I'm either dead or holding my breath.
A fool jabbers, while a wise man listens. But is he so wise to listen to the fool?