Programming Forums > Application Development > C++

pegasus001 · Apr 10th, 2007, 11:17 AM

Sorry my misunderstanding. Rushing as always.

Thanx.

Infinite Recursion · Apr 10th, 2007, 11:42 AM

Quote:

Originally Posted by pegasus001

Sorry if this question seems stupid but why do you have to write this :

(Toggle Plain Text)

std::cin >> y;

I put that in there so when you ran the compiled program, the results wouldn't come and go as soon as the "window" opened. I would have put in system("PAUSE"); but that isn't very portable.

Basically, its there to give you time to read the results.

The Dark · Apr 10th, 2007, 8:33 PM

Quote:

Originally Posted by rwm

ok so this works:

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//number of digits
inline int num_digits(int num) {
	return ceil(log10(num));
}

but lots of warnings! :/

thanks guys!

Try it when num = 100

DaWei · Apr 10th, 2007, 10:46 PM

I recommend reading posts #1 for the problem (integers, not strings) and #4 for the solution.

Infinite Recursion · Apr 10th, 2007, 11:55 PM

I suppose the dividing by 10 would be easier to follow... either way would work really, no need to convert to a string and return the length, just one out of a few options.

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int CountDigits (int src)
{
    int count = 0;
    while (src > 0)
    {
          src = src/10;
          count++;
    }
    return count;
}

Jimbo · Apr 11th, 2007, 12:03 AM

Quote:

Originally Posted by Infinite Recursion

I suppose that would be easier...

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int CountDigits (int src)
{
    int count = 0;
    while (src > 0)
    {
          src = src/10;
          count++;
    }
    return count;
}

This code assumes positive values. The OP didn't say anything about negatives, but I'm assuming they're valid input as well.

Infinite Recursion · Apr 11th, 2007, 12:10 AM

Ahhh. True.

If there is a potential for negative values, you could use a call to the abs() function prior to the while block...

...
src = abs(src);
...

Jimbo · Apr 11th, 2007, 12:26 AM

or just use while(src)

rwm · Apr 11th, 2007, 3:00 AM

hey guys! wow lots of replies!

just discovered a problem, it should be:

(Toggle Plain Text)

//number of digits
inline int num_digits(int num) {
	return floor(log10(num) + 1);
}

since log10 of 10,100,1000 etc is one less than the number of digits...

so you guess divide by 10 would be optimal?

i guess i might as well do that

rwm · Apr 11th, 2007, 3:09 AM

hey well both work nicely...

but gonna stick with division...

thanks for help guys!

much appreciated!

Apr 10th, 2007, 11:17 AM	#11
pegasus001 Hobbyist Programmer Join Date: Nov 2006 Location: 163H Posts: 213 Rep Power: 2	Sorry my misunderstanding. Rushing as always. Thanx. __________________ You never test the depth of a river with both feet. The believer is happy. The doubter is wise. Free speech carries with it some freedom to listen. The next generation will always surpass the previous one. It`s one of the never ending cycles of life.

Apr 10th, 2007, 10:46 PM	#14
DaWei Resident Grouch Join Date: Jun 2005 Posts: 6,453 Rep Power: 10	I recommend reading posts #1 for the problem (integers, not strings) and #4 for the solution. __________________ Abstraction doesn't make it impossible to write bad code; it makes it possible to write superior code. Contributor's Corner: Grumpy on C++ Exceptions DaWei on Pointers

Apr 10th, 2007, 11:55 PM	#15
Infinite Recursion Programming Guru Join Date: Jul 2004 Location: United States Posts: 3,467 Rep Power: 8	I suppose the dividing by 10 would be easier to follow... either way would work really, no need to convert to a string and return the length, just one out of a few options. (Toggle Plain Text) int CountDigits (int src) { int count = 0; while (src > 0) { src = src/10; count++; } return count; } int CountDigits (int src) { int count = 0; while (src > 0) { src = src/10; count++; } return count; } __________________ http://jasonpowers.net "There are a thousand hacking at the branches of evil to one who is striking at the root."

Apr 11th, 2007, 12:10 AM	#17
Infinite Recursion Programming Guru Join Date: Jul 2004 Location: United States Posts: 3,467 Rep Power: 8	Ahhh. True. If there is a potential for negative values, you could use a call to the abs() function prior to the while block... ... src = abs(src); ... __________________ http://jasonpowers.net "There are a thousand hacking at the branches of evil to one who is striking at the root."

Apr 11th, 2007, 12:26 AM	#18
Jimbo Battle Programmer Join Date: Feb 2006 Location: Bellevue, WA, USA Posts: 754 Rep Power: 3	or just use while(src) __________________ <insert disclaimer here> <insert shameless plug for Visual Studio here>

Apr 11th, 2007, 3:00 AM	#19
rwm Professional Programmer Join Date: Jan 2007 Location: Cape Town Posts: 291 Rep Power: 2	hey guys! wow lots of replies! just discovered a problem, it should be: (Toggle Plain Text) //number of digits inline int num_digits(int num) { return floor(log10(num) + 1); } //number of digits inline int num_digits(int num) { return floor(log10(num) + 1); } since log10 of 10,100,1000 etc is one less than the number of digits... so you guess divide by 10 would be optimal? i guess i might as well do that

Apr 11th, 2007, 3:09 AM	#20
rwm Professional Programmer Join Date: Jan 2007 Location: Cape Town Posts: 291 Rep Power: 2	hey well both work nicely... but gonna stick with division... thanks for help guys! much appreciated!

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Programming Forums > Application Development > C++

counting digits