Splitting integers into digits

Author

122

June 19, 2000 09:44 AM

ARGGHHH! This ones driving me insane... i have an integer. Lets say it has the value of 5678. I need to "split" it up so i have 5 and 6 and 7 and 8 all in seperate variables / or an array. I''m pulling my hair out on this... i''ve tried converting the integer to a string then incrementing a pointer to it so i can get each value... but that kinda didn''t work and i dont know why... Any help appreciated. it's hard to remember scripture if you're an atheist

jumble-----------

Muzzafarath

146

June 19, 2000 09:53 AM

Check out the function itoa() (Integer TO Array). If you want to reverse the effect, there's atoi() (Array TO Integer).

    #include <stdio.h>int main(void){int test = 12345;char array[50];itoa(test, array, 10);// The last argument (10) represents the base of the number system (in this case decimal, base 10)// Then you can access the numbers like so:if(array[0] == '1') printf("The first number is 1");// Just remember that the ' ' signs around the numbers are important.return 0;}

/. Muzzafarath
Mad House Software

Edited by - Muzzafarath on June 19, 2000 10:59:40 AM

I'm reminded of the day my daughter came in, looked over my shoulder at some Perl 4 code, and said, "What is that, swearing?" - Larry Wall

Anonymous

June 19, 2000 10:07 AM

I''m pretty sure itoa(...) is a MSVC++ specific function. You could use sprintf(...) if you''re running a different C/C++ system. Then when you access an array element, just subtract the ASCII ''0'' (48 I think) if you want to get the actual number.

Anonymous

June 19, 2000 10:13 AM

Forgot to post how to use sprintf(...) in case you don''t know.

int sprintf( char *buffer, const char *format [, argument] ... );

The first parameter is just a buffer, the second is the printf-esque format description.. for a single integer it would be "%d", and then you have the arguements themselves.

so a code snippet would look be like..
#include
#include

int main()
{
while(true)
{
int num;
char buffer[11];
//do stuff to give num a value
cin>>num;
sprintf(buffer, "%d", num);
int rep=0;
while(buffer[rep]!=0)
cout< cout< }
return 0;
}

Muzzafarath

146

June 19, 2000 10:16 AM

quote:
Original post by Anonymous Poster

I'm pretty sure itoa(...) is a MSVC++ specific function. You could use sprintf(...) if you're running a different C/C++ system. Then when you access an array element, just subtract the ASCII '0' (48 I think) if you want to get the actual number.

My code compiles with DJGPP, which means it's not MSVC++ specific. But itoa() isn't standard C though, but I'm pretty sure it will work with most compilers anyway.

/. Muzzafarath
Mad House Software

Edited by - Muzzafarath on June 19, 2000 11:21:46 AM

I'm reminded of the day my daughter came in, looked over my shoulder at some Perl 4 code, and said, "What is that, swearing?" - Larry Wall

Anonymous

June 19, 2000 11:54 AM

Yeah when i said "i''ve tried converting the integer to a string" i actually meant i had used the itoa() function. It didnt produce the results i wanted. I wrote this console program before i posted:

    void main(){	char newscore[6];	int score = 9876;	itoa(score, newscore, 10);	for (int i=0;i<4;i++) printf("%i\n", atoi( &newscore<i> ));}

and it spurted back at me:

9876876766

which isn''t what i want. BTW i am using MSVC++.

it's hard to remember scripture if you're an atheist

jumble

Author

122

June 19, 2000 11:55 AM

thats me above btw

Also in case you're wondering i need to compare the digits in integer form against some other integers, therefore the atoi function.

Thanks

Edited by - jumble on June 19, 2000 12:58:06 PM

jumble-----------

Buster

100

June 19, 2000 12:43 PM

int x=0
int number = 8562

do {
mydigit[x] = number/(10^x);
x++;
} while (x<10000000);

or something

MartinJ

122

June 19, 2000 12:52 PM

You can use:

    int intoarray(unsigned source, int dest[]){ int tmp=0; {  dest[tmp]=source-(source/10)*10); //element is remainder of division  source/=10; //divide source by 10... drop the remainder  tmp++;  //advance to the next array element  } while source>0; //the while is after so we will always pick up one digit. return tmp;}

if you call it like a=intoarray(1234, array);
a will = 4, the number of digits in the array. And, the array elements will be
array[0]=4, array[1]=3, array[2]=2, array[3]=1. Note that they this is 10^0*4 + 10^1*3 + 10^2*2 + 10^3*1.

This routine doesn't handle negative numbers. That would be an additional test that shouldn't be too hard to add (test at the beginning and resolve the test before the return).

Edited by - MartinJ on June 19, 2000 1:54:18 PM

jumble

Author

122

June 19, 2000 01:07 PM

Hey, thanks everybody... i was just testing out the sprintf statement (turns out it generates the same thing as my atoi() routine above). I think i got this wrapped up now... cheers

oh yeah you could use the abs() function in math.h to find the absolute of a number, thereby getting past the negative number thang in your function Martin.

it's hard to remember scripture if you're an atheist

jumble-----------

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