Finding the number of digits

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May 11th, 2007, 03:33 AM

chaostheory

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Finding the number of digits

Im trying to write a function in which you pass an integer as an argument and you get the number of digits that the integer contains back. Here is what I have, but the code is flawed:

int getsize(int x)
{
int n=1;

for(;

{
if(x%pow(10,n)==0)
return n;
n++;
}

}

I havent tested it out, but the most obvious flaw is that if the integer is a multiple of 10, then x%10 will give 0 which will terminate the loop prematurely.

Can someone point out which way I should head to get this function working? I can do the code on my own.

May 11th, 2007, 04:14 AM

requinix

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...

Two paths here:
1) You discover that floor(log10(abs(x)))+1 will give you the number of digits in x, or

2) Make a loop that until integer x==0, you divide x by 10 and increment a counter

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IOI-RLZ agrees: log is indeed good...

May 11th, 2007, 04:28 AM

WebDunce

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Quote:

Originally Posted by requinix

...

2) Make a loop that until integer x==0, you divide x by 10 and increment a counter

i think it should be loop until x < 1, right?

(nevermind, both conditions work...as long as x is of type integer)

Last edited by WebDunce : May 11th, 2007 at 04:35 AM.

May 11th, 2007, 05:25 AM

chaostheory

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Quote:

Originally Posted by requinix

...

Two paths here:
1) You discover that floor(log10(abs(x)))+1 will give you the number of digits in x, or

2) Make a loop that until integer x==0, you divide x by 10 and increment a counter

Whoa! Thank you. The log thing is cool. Thanks man.

May 11th, 2007, 04:05 PM

Scorpions4ever

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The log10() thing won't work for negative numbers or 0 though. You could get around this by using:
(a) Check for 0 and return 1
(b) Check for - sign and then convert the number to positive and find its length and add 1 to it (for the - sign)

Another way would be to use the sprintf() function, which returns the number of characters that were used in the string:

Code:

char buf[30];
int i = 1234;
int len = sprintf(buf, "%d", i);

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May 12th, 2007, 03:04 AM

clifford

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Quote:

Originally Posted by Scorpions4ever

(b) Check for - sign and then convert the number to positive and find its length and add 1 to it (for the - sign)

Or just use abs(x) as the argument to log10() as requinix did in his original suggestion. Your solution (both of them) counts the sign as a digit for negative numbers. It is not a digit, and counting it as such may not be the intention, it is certainly not what was asked.

If performance were an issue sprintf() is probably far slower.

Clifford

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May 12th, 2007, 03:55 AM

salem

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Logarithms are not exactly cheap either.

Code:

#include <stdio.h>

int intlen ( int a ) {
    int len = 0;
    if ( a < 0 ) a = -a;
    while ( a >= 10000 ) { len += 4; a /= 10000; }
    while ( a >= 100   ) { len += 2; a /= 100; }
    while ( a >= 10    ) { len ++;   a /= 10; }
    return len + 1;
}

int main ( ) {
    printf("%d\n", intlen(0) );
    printf("%d\n", intlen(9) );
    printf("%d\n", intlen(42) );
    printf("%d\n", intlen(123456789) );
    return 0;
}


$ ./a.exe
1
1
2
9

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clifford agrees: Smart

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May 12th, 2007, 06:34 AM

clifford

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Quote:

Originally Posted by salem

Logarithms are not exactly cheap either.

Nice solution. I benchmarked them all, and got some interesting results.

I used the following code:

Code:

#include <limits.h>
#include <stdio.h>
#include <math.h>
#include <time.h>
#include <stdlib.h>

int intlen1 ( int a ) 
{
    int len = 0;
    if ( a < 0 ) a = -a;
    while ( a >= 10000 ) { len += 4; a /= 10000; }
    while ( a >= 100   ) { len += 2; a /= 100; }
    while ( a >= 10    ) { len ++;   a /= 10; }
    return len + 1;
}

int intlen2( int a )
{
    if( a == 0 )
    {
        return 1 ;
    }
    else
    {
        if ( a < 0 ) a = -a;
        return (int)log10((double)a) + 1 ;
    }
}

int intlen3( int a )
{
    char buf[30] ;
    int len = sprintf(buf, "%d", a);
    if( a < 0 )
    {
        len-- ;
    }

    return len ;
}

#define ITERATIONS 10000000
int main()
{
    int i ;
    static int test_data[ITERATIONS] ;
    clock_t end ;
    clock_t start ;
    volatile int checksum ;

    for( i = 0; i < ITERATIONS; i++ )
    {
        test_data[i] = rand() << 17 | (rand() & 0xffff8000) << 1 | (rand() & 0x1) ;
    }

    checksum = 0 ;
    start = clock() ;
    for( i = 0; i < ITERATIONS; i++ )
    {
        for( i = 0; i < ITERATIONS; i++ )
        {
            checksum += intlen1( i ) ;
        }
    }
    end = clock() ;
    printf( "Integer algorithm:\t%d, %d\n", end - start, checksum ) ;

    checksum = 0 ;
    start = clock() ;
    for( i = 0; i < ITERATIONS; i++ )
    {
        checksum += intlen2( i ) ;
    }
    end = clock() ;
    printf( "Log10 algorithm:\t%d, %d\n", end - start, checksum ) ;

    checksum = 0 ;
    start = clock() ;
    for( i = 0; i < ITERATIONS; i++ )
    {
        checksum += intlen3( i ) ;
    }
    end = clock() ;
    printf( "String algorithm:\t%d, %d\n", end - start, checksum ) ;
}

And MS VC++ 8, native Win32 code.

Debug build
----------------
Integer algorithm: 1562, 68888890
Log10 algorithm: 1844, 68888890
String algorithm: 8937, 68888890

Release build
----------------
Integer algorithm: 156, 68888890
Log10 algorithm: 719, 68888890
String algorithm: 4640, 68888890

The string algorithm does not benefit much from optimisation. Salem's solution is obviously superior.

Note I changed the floor() call to a simple cast to int in my version. This has a significant beneficial effect on execution time. Note also the cast to double, this is to disambiguate from the single precision overload in C++ compilation. If you cast to float as I originally did, you get the wrong answer for 7 digit values close to 10,000,000 (this is what the checksum was there for - to make sure they produced the same result).

It is interesting because I have in the past and would have in this case, suggested the log10 method. It remains easy to implement (and remember) and is reasonably fast on a system with an FPU at least. Since I generally do embedded systems development and have no FPU, I'll have to remember Salem's solution!)

Clifford

Finding the number of digits

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