Programming Forums > Application Development > C++

rwm · Apr 10th, 2007, 11:16 AM

Hey,

im writing some code for padding files.

Im wondering how i can count the number of digits in an integer...

i.e.

int a = 232;
int b = 9343;
int c = 23;
int d = 1;

a would have 3 digits, b would have 4 digits, c would have 2 and d would have 1 digit.

i've looked all over - all i can find is bitwise stuff - thats not what i want - the number of set bits is not always the same as the number of digits in the integer...

hope someone can help!

thanks!

Eoin · Apr 10th, 2007, 11:42 AM

Well you have one obvious option which would be convert to a string and count the characters (possibly accounting for leading zeros depending on your task).

Another option might be to test the int against being less than 10, 100, 1000, etc. The max size for a int will mean you won't have to perform countless comparisons.

Finally you could look into log's, particularly log10 of the int.

rwm · Apr 10th, 2007, 11:44 AM

Hey Eoin,

Thanks for the quick reply!

Can you explain the logs a bit more please?

Thanks!

pegasus001 · Apr 10th, 2007, 11:46 AM

You can divide it by ten until it reaches 0. Like :

(Toggle Plain Text)

213 / 10 = 21
21 / 10 = 2
2 / 10 = 0

So you have 3 numbers.

rwm · Apr 10th, 2007, 11:53 AM

hey sweet!

ok so obviously using log10 - i would have to use the ceiling function?

for example, say our integer is 235, log10 returns 2.37 - and there are actually 3 numbers in the integer, so use celing(2.37) to get 3...

that right?

rwm · Apr 10th, 2007, 11:57 AM

ok so this works:

(Toggle Plain Text)

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

but lots of warnings! :/

thanks guys!

pegasus001 · Apr 10th, 2007, 12:01 PM

I tried in the windows calculator log555 and the result is 2.744etc. So i guess this was not what eoin ment.

Infinite Recursion · Apr 10th, 2007, 12:03 PM

I have a problem with overkilling the simple things I guess... So I won't kill all of your fun, I'll leave the leading zero issue up to you to solve (if there is even a potential for it to be a problem)

(Toggle Plain Text)

#include <iostream>
#include <sstream>

int CountDigits (int source);
std::string itos (int num);

int main (void)
{
    int x = 0;
    int y = 12345;
    
    x = CountDigits(y);
    
    std::cout << "There are " << x << " digits in " << y << std::endl;
    std::cin >> y;
    
    return 0;
}

int CountDigits (int source)
{
    int count = 0;
    std::string z = itos(source);
    return z.length();
}

std::string itos (int num)
{
   std::ostringstream myStream;
   myStream << num << std::flush;
   return (myStream.str());
}

pegasus001 · Apr 10th, 2007, 12:09 PM

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

(Toggle Plain Text)

std::cin >> y;

Eoin · Apr 10th, 2007, 12:14 PM

Hi pegasus001, that result is perfectly ok. log555 = 2.744 implies 10^2.744 = 555. It's just a convenient fact that after rounding up the answer gives you the number of digits.

So for example log_2(x) {here I'm saying log to the base b of x) gives you the number of binary digits which would be necessary to hold x.

rvm, the warnings are because log10 take and returns a float or double. So you should specify the cast manually and ideally account for when there would be an overflow. Boost.numeric_cast can do this nicely.

P.S. a solution based on pegasus' suggestion of dividing by ten would probably be the most optimal as the log functions are very expensive.

Apr 10th, 2007, 11:16 AM	#1
rwm Professional Programmer Join Date: Jan 2007 Location: Cape Town Posts: 291 Rep Power: 2	counting digits Hey, im writing some code for padding files. Im wondering how i can count the number of digits in an integer... i.e. (Toggle Plain Text) int a = 232; int b = 9343; int c = 23; int d = 1; int a = 232; int b = 9343; int c = 23; int d = 1; a would have 3 digits, b would have 4 digits, c would have 2 and d would have 1 digit. i've looked all over - all i can find is bitwise stuff - thats not what i want - the number of set bits is not always the same as the number of digits in the integer... hope someone can help! thanks!

Apr 10th, 2007, 11:42 AM	#2
Eoin Hobbyist Programmer Join Date: Jun 2006 Location: Ireland Posts: 152 Rep Power: 3	Well you have one obvious option which would be convert to a string and count the characters (possibly accounting for leading zeros depending on your task). Another option might be to test the int against being less than 10, 100, 1000, etc. The max size for a int will mean you won't have to perform countless comparisons. Finally you could look into log's, particularly log10 of the int. __________________ Visit my website BinaryNotions.

Apr 10th, 2007, 11:46 AM	#4
pegasus001 Hobbyist Programmer Join Date: Nov 2006 Location: 163H Posts: 215 Rep Power: 3	You can divide it by ten until it reaches 0. Like : (Toggle Plain Text) 213 / 10 = 21 21 / 10 = 2 2 / 10 = 0 213 / 10 = 21 21 / 10 = 2 2 / 10 = 0 So you have 3 numbers. __________________ 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, 11:57 AM	#6
rwm Professional Programmer Join Date: Jan 2007 Location: Cape Town Posts: 291 Rep Power: 2	ok so this works: (Toggle Plain Text) //number of digits inline int num_digits(int num) { return ceil(log10(num)); } //number of digits inline int num_digits(int num) { return ceil(log10(num)); } but lots of warnings! :/ thanks guys!

Apr 10th, 2007, 12:01 PM	#7
pegasus001 Hobbyist Programmer Join Date: Nov 2006 Location: 163H Posts: 215 Rep Power: 3	I tried in the windows calculator log555 and the result is 2.744etc. So i guess this was not what eoin ment. __________________ 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, 11:44 AM	#3
rwm Professional Programmer Join Date: Jan 2007 Location: Cape Town Posts: 291 Rep Power: 2	Hey Eoin, Thanks for the quick reply! Can you explain the logs a bit more please? Thanks!

Apr 10th, 2007, 11:53 AM	#5
rwm Professional Programmer Join Date: Jan 2007 Location: Cape Town Posts: 291 Rep Power: 2	hey sweet! ok so obviously using log10 - i would have to use the ceiling function? for example, say our integer is 235, log10 returns 2.37 - and there are actually 3 numbers in the integer, so use celing(2.37) to get 3... that right?

Apr 10th, 2007, 12:03 PM	#8
Infinite Recursion Programming Guru Join Date: Jul 2004 Location: United States Posts: 3,473 Rep Power: 8	I have a problem with overkilling the simple things I guess... So I won't kill all of your fun, I'll leave the leading zero issue up to you to solve (if there is even a potential for it to be a problem) (Toggle Plain Text) #include <iostream> #include <sstream> int CountDigits (int source); std::string itos (int num); int main (void) { int x = 0; int y = 12345; x = CountDigits(y); std::cout << "There are " << x << " digits in " << y << std::endl; std::cin >> y; return 0; } int CountDigits (int source) { int count = 0; std::string z = itos(source); return z.length(); } std::string itos (int num) { std::ostringstream myStream; myStream << num << std::flush; return (myStream.str()); } #include <iostream> #include <sstream> int CountDigits (int source); std::string itos (int num); int main (void) { int x = 0; int y = 12345; x = CountDigits(y); std::cout << "There are " << x << " digits in " << y << std::endl; std::cin >> y; return 0; } int CountDigits (int source) { int count = 0; std::string z = itos(source); return z.length(); } std::string itos (int num) { std::ostringstream myStream; myStream << num << std::flush; return (myStream.str()); } __________________ http://jasonpowers.net "There are a thousand hacking at the branches of evil to one who is striking at the root."

Apr 10th, 2007, 12:09 PM	#9
pegasus001 Hobbyist Programmer Join Date: Nov 2006 Location: 163H Posts: 215 Rep Power: 3	Sorry if this question seems stupid but why do you have to write this : (Toggle Plain Text) std::cin >> y; std::cin >> y; __________________ 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, 12:14 PM	#10
Eoin Hobbyist Programmer Join Date: Jun 2006 Location: Ireland Posts: 152 Rep Power: 3	Hi pegasus001, that result is perfectly ok. log555 = 2.744 implies 10^2.744 = 555. It's just a convenient fact that after rounding up the answer gives you the number of digits. So for example log_2(x) {here I'm saying log to the base b of x) gives you the number of binary digits which would be necessary to hold x. rvm, the warnings are because log10 take and returns a float or double. So you should specify the cast manually and ideally account for when there would be an overflow. Boost.numeric_cast can do this nicely. P.S. a solution based on pegasus' suggestion of dividing by ten would probably be the most optimal as the log functions are very expensive. __________________ Visit my website BinaryNotions.

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Programming Forums > Application Development > C++

counting digits