C / C++ / MFC

7-Jul-06 6:18

Jay03 wrote:
class RTI_EXPORT RTIambassador

RTI_EXPORT is a preprocessor macro. When dealing with DLLs, there are some compiler directives you can use (instread of having *.def files) to say you are exporting (or importing) classes, variables, functions, etc. If you do a search for RTI_EXPORT, you should find it defined something like the following:

#ifdef _WINDLL
#define RTI_EXPORT __declspec(dllexport)
#else
#define RTI_EXPORT __declspec(dllimport)
#endif

What that does is allow the you to compile the DLL stating that this class is being exported while you can use the same header file in the application and state that the class is being imported.

Now, when writing DLLs for Windows applications (as opposed to so's for *Nix systems), you have to tell the compiler which symbols (variables, functions, classes, etc) are being exposed to be used by anyone wishing to use the library. At the same time, when you write an application that needs to use a DLL, you need to know which symbols are being imported so the linker knows what to do with them.

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week

Zac

User Input

Harold_Wishes7-Jul-06 5:26

7-Jul-06 5:26

Hello,

I need some help with some data validation as I'm not much of a programmer. I have created some string and character variables in a simple program I am writing for my company. Specifically, the program is suppose to accept user input from the keyboard and check to verify if the user enters data or enters nothing.

At the moment, I would like to give the user the option to press the enter key if they choose to skip data entry for a particular variable and go on to key in data for the next variable using the getline() input function.

Right now, the program forces the user to enter data for all variables. Simply hitting the enter key will not allow the user to move on in the program. The prompt just sits there waiting for your input for the current variable. Is there a way to force the program to move ahead in the program?

My next question is how to represent NULL for data that was never entered.

Help is much appreciated.

Thanks, HRW

Zac Howland7-Jul-06 5:31

7-Jul-06 5:31

What code do you have for this currently? If you are using something like (pseudo-code):

while (INeedData())
{
   string data;
   cin >> data;
   // do somethign with data
}

Then you won't be able to skip. If you do something like:

while (INeedData())
{
   string data;
   cin.getline(data, "\n");
   // do somethign with data
}

Enter will allow you to skip.

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week

Zac

Harold_Wishes7-Jul-06 5:51

7-Jul-06 5:51

Hello,

I am using something like to following:

string data2;

cout << "Please enter modifications if any: ";
cin >> data2;

I am studying the pseudo code but I do not quite understand the test condition while (INeedData()). If the user preforms the entry, then I would like to store that data in the assigned variable. But if the user types in a carriage return, I would like to bypass this variable storage or at least assign it a NULL value it that is at all possible. I tried to use the isspace() function but that has not worked for me.

Thanks for the above reply. HRW Smile | :)

Zac Howland7-Jul-06 6:04

7-Jul-06 6:04

That was just defining a loop to show that this operation would be done multiple times. Judging from your response, I imagine your code looks more like this:

string data1;
cout << "Enter Data1:  ";
cin >> data1;

string data2;
cout << "Enter Data2:  ";
cin >> data2;

string data3;
cout << "Enter Data3:  ";
cin >> data3;

...

cin ignores whitespace. To get the behavior you want, you need to do something like this instead:

char buffer[256] = {0};

string data1;
cout << "Enter Data1:  ";
cin.getline(buffer, 255, '\n');
data1 = buffer;
memset(buffer, 0, 256);

string data2;
cout << "Enter Data2:  ";
cin.getline(buffer, 255, '\n');
data2 = buffer;
memset(buffer, 0, 256);

string data3;
cout << "Enter Data3:  ";
cin.getline(buffer, 255, '\n');
data3 = buffer;
memset(buffer, 0, 256);

...

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week

Zac

Harold_Wishes7-Jul-06 7:06

7-Jul-06 7:06

Hello,

I created a character array as you have above. I assigned it to each of my input statements with no trouble at all. It works perfectly. I am curious as to the purpose of the memset function. I would assume it re-initializes buffer to 0.

Thanks Big Grin | :-D

HRW

Zac Howland7-Jul-06 7:33

7-Jul-06 7:33

Harold_Wishes wrote:
I am curious as to the purpose of the memset function. I would assume it re-initializes buffer to 0.

Yes. Calling it just goes through and makes sure that every character in the buffer is set to 0 before pulling in more input.

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week

Zac

Harold_Wishes7-Jul-06 10:08

Re: User Input [modified]

7-Jul-06 10:08

Hello again,

I did run my program and caught a runtime error with the cin.getline(buffer, 500, '\n'); I have a condensed working version below of the program compiled using Visual C++. One of my cin input streams is not being read in. It is skipped automatically and just goes to the next input prompt. I have not been able to figure out why this is happening.

I have 2 functions in main(). The problem seems to be when I move from function 1 to function 2 (the runtime error occurs in function 2). Oddly enough, the problem disappears when I comment out function 1 and only execute function 2. If you have a complier and execute the program (at least with Visual C++), you may see what I mean.

Regards Blush | :O

#include <iostream>
#include <fstream>
#include <cstdlib>
#include <string>
#include <cctype>
#include <iomanip>
#include "unsortedlist.h"
#include "sortedlist.h"


using std::string;
using namespace std;


struct node                                                         //
{
     string Length;                   //
     string Sequence;                                               //  
     string N_Terminal;                                             // 
     string C_Terminal;
     node *nxt;// Pointer to next node
};


node *start_ptr = NULL;
node *current;		 // Used to move along the list


void add_Part_one (string& data1, int& data2)
{
      string selection = "A";
	cout << "Please Select the sequence type: " << endl;
	cout << endl;
	cout << "A) Complete  B) Partial  C) UNKNOWN" << endl;
   

      cin >> selection;

      while (selection != "A" && selection != "a" && selection != "B" && selection != "b" 
	 && selection != "C" && selection != "c")
	{
	       cout << "Invalid Entry. Please Try Again: " << endl;  
		 cin >> selection;
	}

     
	if ( selection == "A" || selection == "a" )
      {
	     data1 = "Complete";
	}
	else
	if ( selection == "B" || selection == "b" )
      {
	     data1 = "Partial";
      }
	else
	if ( selection == "C" || selection == "c" )
      {
		 data1 = "UNKNOWN";
      }
	

	cout << "Please enter the number of subunits: ";
      cin >> data2;
}

void add_node_at_end()
{        
    node *temp, *temp2;  // Temporary pointers
	temp = new node; // Reserve space for new node and fill it with data
	 
	

    char buffer[501] = {0};

	cout << "Please enter the length of the protein sequence: ";
	cin.getline(buffer, 500, '\n');
	temp->Length = buffer;
    memset(buffer, 0, 501);


	cout << "Please enter the protein sequence : ";
    cin.getline(buffer, 500, '\n');
    temp->Sequence = buffer;
    memset(buffer, 0, 501);


	cout << "Enter brief status on the presence of a modified N_Terminal: ";
    cin.getline(buffer, 500, '\n');
    temp->N_Terminal = buffer;
    memset(buffer, 0, 501);

	cout << "Enter brief status on the presence of a modified C_Terminal: ";
    cin.getline(buffer, 500, '\n');
    temp->C_Terminal = buffer;
    memset(buffer, 0, 501);


    
	temp->nxt = NULL;

      // Set up link to this node
	if (start_ptr == NULL)
	{ start_ptr = temp;
		 current = start_ptr;
	}
      else
	{ temp2 = start_ptr;
	  // We know this is not NULL - list not empty!
	  while (temp2->nxt != NULL)
	  {  temp2 = temp2->nxt;
				  // Move to next link in chain
	  }
	     temp2->nxt = temp;
	}
      	 cout << endl;
}

int main()
{
   int number_of_subunits = 1;
   string SEQUENCE_TYPE;
   
   add_Part_one(SEQUENCE_TYPE, number_of_subunits);
   for (int counter = 0; counter < number_of_subunits; counter++)
   {  
	 add_node_at_end();

   }


return 0;
}

HRW

Zac Howland7-Jul-06 10:35

Re: User Input [modified]

7-Jul-06 10:35

Try this:

#include <string>
#include <list>
#include <iostream>

using namespace std;

struct data 			// renamed to more meaningful name (granted, not much more)
{
     string Length;
     string Sequence;
     string N_Terminal;
     string C_Terminal;
};				// removed pointer since list class would be better choice

//node *start_ptr = NULL;
//node *current;		 // no longer needed

list<data> g_DataList;		// global list of data

char selectionMenu()
{
	char selection = 0;	// your menu allows for only a single char
				// no need to allocate a string for that
	cout << "Please Select the sequence type: " << endl;
	cout << endl;
	cout << "A) Complete  B) Partial  C) UNKNOWN" << endl;
	cin >> selection;
	return selection;
}

bool isValidSelection(char selection)
{
	switch (selection)
	{
	case 'A':
	case 'a':
	case 'B':
	case 'b':
	case 'C':
	case 'c':
		return true;
	default:
		return false;
	}
}

string getSelectionString(char selection)
{
	string result = "UNKNOWN";	// C/c is the default
	switch (selection)
	{
	case 'A':
	case 'a':
		result = "Complete";
		break;
	case 'B':
	case 'b':
		result = "Partial";
		break;
	}

	return result;
}

void add_Part_one (string& data1, int& data2)
{
	char selection = 0;
	do
	{
		selection = selectionMenu();
	} while (!isValidSelection(selection));

	data1 = getSelectionString(selection);

	char buffer[100] = {0};
	cout << "Please enter the number of subunits: ";
	cin >> data2;
	cin.getline(buffer, 99, '\n');	// to get endline character out of the way
}

void add_node_at_end()
{        
	data temp;	// what we are adding to	 
	char buffer[501] = {0};	// buffer for input

	cout << "Please enter the length of the protein sequence: ";
	cin.getline(buffer, 500, '\n');
	temp.Length = buffer;
	memset(buffer, 0, 501);

	cout << "Please enter the protein sequence : ";
	cin.getline(buffer, 500, '\n');
	temp.Sequence = buffer;
	memset(buffer, 0, 501);

	cout << "Enter brief status on the presence of a modified N_Terminal: ";
	cin.getline(buffer, 500, '\n');
	temp.N_Terminal = buffer;
	memset(buffer, 0, 501);

	cout << "Enter brief status on the presence of a modified C_Terminal: ";
	cin.getline(buffer, 500, '\n');
	temp.C_Terminal = buffer;
	memset(buffer, 0, 501);

	g_DataList.push_back(temp);	// add to list

	cout << endl;
}

int main()
{
	int number_of_subunits = 1;
//	string SEQUENCE_TYPE;	// just an FYI:  
				// variable names that are all caps are typically thought to be const
				// it is bad form to create non-const variables with all caps names
   	string sequenceType;

	add_Part_one(sequenceType, number_of_subunits);
   	for (int counter = 0; counter < number_of_subunits; counter++)
	{  
		add_node_at_end();
	}

	return 0;
}

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week

Zac

-- modified at 16:36 Friday 7th July, 2006

Harold_Wishes8-Jul-06 17:36

8-Jul-06 17:36

Thanks for the assistance.

One last question:

Will I need to overload the << operator to output each element of the data object to the screen?

I am having trouble using the iterator function for the <list> class when I attempt to print the Length, Sequence, N_Terminal and C_Terminal.

Regards
HRW

-- modified at 23:37 Saturday 8th July, 2006

Zac Howland9-Jul-06 14:28

Simple MAPI Outlook outbox problem

9-Jul-06 14:28

Harold_Wishes wrote:
Will I need to overload the << operator to output each element of the data object to the screen?

You don't need to per se, but you could. Either way, you should create a print method that looks like either of the following:

// operator<< overload
ostream& operator<<(ostream& os, const data& dta)
{
	os << dta.Length.c_str() << endl
	   << dta.Sequence.c_str() << endl
	   << dta.N_Terminal.c_str() << endl
	   << dta.C_Terminal.c_str() << endl;
	return os;
}

// option 1:  print without overload
void print(const data& dta)
{
	cout << dta.Length.c_str() << endl
	     << dta.Sequence.c_str() << endl
	     << dta.N_Terminal.c_str() << endl
	     << dta.C_Terminal.c_str() << endl << endl;	
}

// option 2:  print with overload
void print(const data& dta)
{
	cout << dta << endl;
}

// use for_each to print entire list of items
for_each(g_DataList.begin(), g_DataList.end(), print);

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week

Zac

chilituna7-Jul-06 4:45

chilituna

7-Jul-06 4:45

Hi,

my email sending with Simple MAPI works, sort of.
But send mail is stuck in the outlook outbox
and has to be mailed from there..

I have looked the examples in here, haven't
found a clue for this problem.

I have also tried the trick, where I send the
mail in first session and then make a second
session with MAPI_FORCE_DOWNLOAD on, but that
doesn't help either.

Any ideas?

ostream operator link error

mosu'7-Jul-06 4:33

mosu'

7-Jul-06 4:33

Hy, I have this code and this error which I don't understand. Any help will be very apreciated. Thank you.
this is test1.cpp:
#include <iostream.h>

class Address
{
private:
int house;
int block;

public:
Address(int i, int j) {
house=i;
block=j;
}
friend ostream& operator << (ostream&, const Address&);
};

ostream& operator << (ostream& a, const Address& b)
{
a<<"House No "<

Re: ostream operator link error [modified]

Viorel.7-Jul-06 4:45

Viorel.

7-Jul-06 4:45

I suppose you have to move your implementation of operator << to test1.cpp (or other *.cpp) file.

I hope it helps.

-- modified at 10:56 Friday 7th July, 2006

Re: ostream operator link error

Zac Howland7-Jul-06 5:11

Run an application inside another one

7-Jul-06 5:11

mosu` wrote:
this is test1.cpp:

The following code should be moved to a header file (you should not be including *.cpp files!):

In test1.h

#pragma once

#include <ostream>

class Address
{
private:
	int house;
	int block;

public:
	Address(int i, int j) : house(i), block(j)
	{ }

	int GetHouseNumber() const { return house; }
	int GetBlockNumber() const { return block; }

	// add accessors!  No need to make it a friend here
//	friend ostream& operator << (ostream&, const Address&);
};

inline ostream& operator<<(ostream& a, const Address& b)
{
	a << "House No " << b.GetHouseNumber() << endl;
	a << "Block Number " << b.GetBlockNumber() << endl;
	return a;
};

And in main.cpp

#include "test1.h"
int main()
{
	Address a(10,113);
	cout << a;
	return 0;
}

The error is due to the fact that test1.cpp is being compiled twice in your code and thus has multiple definitions in the object code.

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week

Zac

xtof737-Jul-06 4:03

xtof73

7-Jul-06 4:03

Hello

How to run an application within an application : inside a dialog-based application, I want to run another one inside a rectangle ?

Thank you

Sarath C7-Jul-06 4:25

Sarath C

7-Jul-06 4:25

Hi,
You can create a process using CreateProcess and can set it size in STARTUP_INFO parameter.

see msdn for more details.

AFAIK it is not possible to show that exe insde your appliaction, unless it's a COM control (ActiveX or DLL). I'm not sure about this.

SaRath.

"Where I am from, there is no plan B. So, take advantage of today becuase tomorrow is not promised. - 50 Cent"

<blink>My Blog | Understanding State Patte

Zac Howland7-Jul-06 4:30

7-Jul-06 4:30

Sarath.<fon wrote:<="" div="">AFAIK it is not possible to show that exe insde your appliaction, unless it's a COM control (ActiveX or DLL). I'm not sure about this.

If the application is simple enough, you can wrap it in an ActiveX control and have it run inside that. Also, if the application exposes itself via COM (like the office components), you can have it appear on a dialog, but it won't look like the full application in most cases.

If you decide to become a software engineer, you are signing up to have a 1/2" piece of silicon tell you exactly how stupid you really are for 8 hours a day, 5 days a week

Zac

Sarath C7-Jul-06 4:39

Sarath C

7-Jul-06 4:39

In the case of a notepad or wordpad what? what we can do with that?

Office system is completely developed as COM Components and many applications using it very well. that is what I said, we can't make inside as our control or part of our application unless it is an attachable manner.

SaRath.

"Where I am from, there is no plan B. So, take advantage of today becuase tomorrow is not promised. - 50 Cent"

<blink>My Blog | Understanding State Patte