C / C++ / MFC

Re: Problem With std::getline And C++

16-Dec-01 4:44

Hi,
I find there is a slight inconvenience with the std::getline function in C++. Here is an example:

-----------------------------------------------------------
#include <iostream>
using std::cin;
using std::cout;
using std::endl;
using std::getline;

#include <string>
using std::string;

int main()
{
string strTemp;

cout << "Enter a sentence: ";
getline(cin, strTemp);

cout << "\nYou entered: "
<< strTemp << endl;

return 0;
}

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

If you run the code above, and hopefully I did not leave anything out, out will find that after you entered a sentence and press Enter, the "cursor" will go to the next line, but the program does not move onto "You have entered: " until you press Enter the second time. In other words, I have had problems with the getline function in term of having to press Enter twice to get it to move onto the next algorithm.

I have tried for example:

getline(cin, strTemp, '\n');

It has the same effect.

Is there a way to get around having to press Enter twice?

Thanks,
Kuphryn

Joaquín M López Muñoz16-Dec-01 10:45

Joaquín M López Muñoz

16-Dec-01 10:45

Probably, you're running into a known bug in Dinkumware's implementation of the C++ standard library. This article addresses the problem (look for the "Fix to <istream>" section) and proposes a fix.

Joaquín M López Muñoz
Telefónica, Investigación y Desarrollo

Re: Problem With std::getline And C++

valikac16-Dec-01 16:03

Reading and Writing Data From ifstream to *STL Container* in *Binary*

16-Dec-01 16:03

zen from cprogramming.com posted something about a bug in VC++:

http://www.cprogramming.com/cboard/showthread.php?s=a4253030337e8f0dc4d9e68fb761087c&postid=48172#post48172

That fixed the problem.

Kuphryn

valikac16-Dec-01 4:35

Re: Reading and Writing Data From ifstream to *STL Container* in *Binary*

16-Dec-01 4:35

Hi.
I am working on a simple program that reads data into memory in binary. It then write the data from memory to a second file in memory.

Is it possible to read from ifstream straight to an STL container instead of having to go through dynamic memory allocation?

Thanks,
Kuphryn

Todd Smith16-Dec-01 4:46

Todd Smith

16-Dec-01 4:46

Yes.

You can read a file into a

<br />
vector<char> data;<br />
<br />
or<br />
<br />
string data;<br />

can use istreambuf_iterator to load the file

<br />
ifstream file("c:\\somefile");<br />
vector<char> data((istreambuf_iterator<char>(file)), istreambuf_iterator<char>());<br />

you can make it faster by getting the file size and using dat.reserve() to pre-allocate the required memory.

Todd Smith

Re: Reading and Writing Data From ifstream to *STL Container* in *Binary*

valikac16-Dec-01 5:05

16-Dec-01 5:05

Thanks.

Okay, this is more advanced STL than I am used to. I have no prior knowledge of istreambuf_iterator or ostreambuf_iterator.

First will:

vector data((istreambuf_iterator(file)), istreambuf_iterator());

read data from a file into vector data in *binary* mode?

I am used to something like:

--------------------
char x[5];

istream data;
data.open(x, ios::in | ios::binary);

data.read(reinterpret_cast<char *="">(x), 5);
--------------------

The same goes for data.reserve("size???").

Kuphryn

Color Ramp

Paul Selormey16-Dec-01 1:07

Paul Selormey

16-Dec-01 1:07

Can anyone help with a nice algorithm to create a gradient style color ramp?

What I need is to specific the first and last colors and generate a specified number of colors between them.

Best regards,
Paul.

Paul Selormey, Bsc (Elect Eng), MSc (Mobile Communication) is currently Windows open source developer in Japan, and open for programming contract anywhere!

PJ Arends16-Dec-01 9:37

PJ Arends

16-Dec-01 9:37

This is what I use :

COLORREF *Gradient(COLORREF start, COLORREF end, int steps)
{
    if (steps < 2)
        return NULL;
    COLORREF *output = new COLORREF[steps];
    output[0] = start;
    output[steps - 1] = end;

    double sR = (double)GetRValue(start);
    double sB = (double)GetBValue(start);
    double sG = (double)GetGValue(start);

    double eR = (double)GetRValue(end);
    double eB = (double)GetBValue(end);
    double eG = (double)GetGValue(end);

    double RedStep = (eR - sR) / (double)steps;
    double BlueStep = (eB - sB) / (double)steps;
    double GreenStep = (eG - sG) / (double)steps;

    for (int x = 1; x < steps - 1; x++)
    {
        sR += RedStep;
        sB += BlueStep;
        sG += GreenStep;

        output[x] = RGB((BYTE)sR, (BYTE)sB, (BYTE)sG);
    }
    return output;
}

---
It may be that your sole purpose in life is simply to serve as a warning to others.

Daniel Ferguson16-Dec-01 9:40

Daniel Ferguson

16-Dec-01 9:40

(I previewed this before posting it, but I don't know why the blank lines are missing from the pre section.)

/* inputs:
	nColourCount -- the number of colours, including the first and last
	crLowColour, crHighColour -- first and last colour

	outputs:
	pColourTable -- an array of colour refs
*/
void RampColours(int nColourCount, COLORREF crLowColour, COLORREF crHighColour, COLORREF* pColourTable)
{
	double red_step, green_step, blue_step;
	int red_low, green_low, blue_low;

	red_low		= GetRValue(crLowColour);
	green_low	= GetGValue(crLowColour);
	blue_low	= GetBValue(crLowColour);
	// divide the difference in colours by the number of steps
	red_step	= (double)(GetRValue(crHighColour) - red_low) / (double)nColourCount;
	green_step	= (double)(GetGValue(crHighColour) - green_low) / (double)nColourCount;
	blue_step	= (double)(GetBValue(crHighColour) - blue_low) / (double)nColourCount;

	// set the first and last colour
	pColourTable[0] = crLowColour;
	pColourTable[nColourCount] = crHighColour;
	// step through each colour, adding the r,g&b step values
	for (int i = 0; i <= nColourCount; i++) {
		pColourTable[i] = RGB( red_low + (red_step * (double)i), green_low + (green_step * (double)i), blue_low + (blue_step * (double)i) );
	} // for each step
}

"Am I talking too fast, or are you just playing dumb? If you want I can write it down." -Jarvis Cocker/Pulp

Rick York16-Dec-01 10:27

Rick York

16-Dec-01 10:27

Here is another one. This is from a little app I hope to be posting here soon :

<br />
BYTE InterpColor( BYTE top, BYTE bottom, int linendx, int maxlines )<br />
	{<br />
	if( top == bottom )<br />
		return top;<br />
<br />
	double cval = (double)linendx * (double)( bottom - top ) / (double)maxlines;<br />
	cval += (double)top;<br />
	return (BYTE)cval;<br />
	}<br />
<br />
<br />
void GetGradientColor( int count, int index, COLORREF top, COLORREF bottom )<br />
	{<br />
	BYTE r = InterpColor( GetRValue( top ), GetRValue( bottom ), index, count );<br />
	BYTE g = InterpColor( GetGValue( top ), GetGValue( bottom ), index, count );<br />
	BYTE b = InterpColor( GetBValue( top ), GetBValue( bottom ), index, count );<br />
	COLORREF color = RGB( r, g, b );<br />
	return color;<br />
	}<br />
<br />

This is not the absolutely most efficient algorithm but it works well for me. There are some easy optimizations that could be made but I have never seen the speed of this to be a problem.

Which ActiveX controls can i use?

Serge Baltic16-Dec-01 21:56

Serge Baltic

16-Dec-01 21:56

Rick York wrote:
void GetGradientColor( int count, int index, COLORREF top, COLORREF bottom )

That's nice, but what about some dithering? Does anyone know how to implement one (at least of 'ordered' style)?

* Origin: Silence Must Be Heard (2:5030/844.235)

Re: Color Ramp-Thanks

Paul Selormey16-Dec-01 13:03

Paul Selormey

16-Dec-01 13:03

Hello All,
Thanks so much for coming quickly to my aid. May God bless you all.

Best regards,
Paul.

Paul Selormey, Bsc (Elect Eng), MSc (Mobile Communication) is currently Windows open source developer in Japan, and open for programming contract anywhere!

15-Dec-01 23:44

Hi, i have a problem, there are a lot of interesting activex controls - but they are loaded dynamicaly (or not?) so i'm not sure if all the users will have them - can you help me - are there some ActiveX controls that are installed on every win machine ? For example there is "Microsoft Forms Checkbox" (or socalled) is it available on every win-box ?

thanks a lot for explanation

john

Are there any News Groups for Linux newbies?

BigMouth15-Dec-01 23:41

BigMouth

15-Dec-01 23:41

- Can you help?

- Regards,
BigMouth

Michael Dunn16-Dec-01 8:15

Michael Dunn

16-Dec-01 8:15

Check out comp.os.linux.*

--Mike--
My really out-of-date homepage
"Not our fault we are intellectually superior to the rest of the office." -- Paul Watson in the Lounge, 12/12/2001
Sonork - 100.10414 AcidHelm
Big fan of Alyson Hannigan.

BigMouth16-Dec-01 17:53

BigMouth

16-Dec-01 17:53

- Thanks, Mike pal!

- I do not know what you mean? Do you mean "http://comp.os.linux.*" ? How to check? I am a newbie. Can you help?

- Have a nice day,
Maer

markkuk17-Dec-01 20:14

markkuk

17-Dec-01 20:14

It means newsgroups with names starting with "comp.os.linux". Newsgroups aren't WWW pages, so there's no "http://". Use your newsreader program (Outlook Express, Forte Agent, etc. under Windows) and subscribe to groups in the comp.os.linux hierarchy. If your ISP doesn't support newsgroups, you can use Google Groups (http://groups.google.com).