CODEWITHSUNDEEP
crounding
M. Ryan
M. Ryan

Reputation: 7192

Rounding integers to nearest ten or hundred in C

I'm trying to think of a function in C that would satisfy the following conditions:

For example:

53 comes out as 60..

197 comes out as 200..

4937 comes out as 5000..

Is there a way to do this so that the requirement is satisfied regardless of the number of trailing zeroes?

For example, I understand how I could do it in any individual case. divide 53 by 10 then ceil(), multiply by 10, but I would like one that can handle any value.

Opinions? Ideas?

Upvotes: 9

Views: 31337

Answers (12)

user7153622
user7153622

Reputation: 1

Some expert advice you get here... To round a number i up to nearest 10: 

if (i%10)
   i += 10 - i%10;

Similarly for 100s, 1000s, etc.

Upvotes: -3

Danny zadok
Danny zadok

Reputation: 1

/*-------------------------------------------------*/
/* Round up a number                               */
/* (Negative numbers go DOWN)                      */
/* 34 -> 40                                        */
/* -34 -> -40                                      */
/*-------------------------------------------------*/
int round_up_10(int num) 
{
    int sign = 1;
    int tens = 1;

    if (num < 0)
    {
        num = (-1) * num;
        sign = -1;
    }
    num = (num + 9) / 10;
    num = num * 10;

    return sign * num;
}
/*-------------------------------------------------*/
/* Round down a number                             */
/* (Negative numbers go UP)                        */
/* 34 -> 30                                        */
/* -34 -> -30                                      */
/*-------------------------------------------------*/
int round_down_10(int num)   
{
    int sign = 1;
    int tens = 1;

    if (num < 0)
    {
        num = (-1) * num;
        sign = -1;
    }
    num = num / 10;
    num = num * 10;

    return sign * num;
}

main()
{
    printf("round_down_10(25450)= %d\n", round_down_10(25450));
    printf("round_down_10(-25450)= %d\n", round_down_10(-25450));
    printf("round_up_10(25450)= %d\n", round_up_10(25450));
    printf("round_up_10(-25450)= %d\n", round_up_10(-25450));

    printf("round_down_10(1347)= %d\n", round_down_10(1347));
    printf("round_down_10(-1347)= %d\n", round_down_10(-1347));
    printf("round_up_10(1347)= %d\n", round_up_10(1347));
    printf("round_up_10(-1347)= %d\n", round_up_10(-1347));

    printf("round_down_10(34)= %d\n", round_down_10(34));
    printf("round_down_10(-34)= %d\n", round_down_10(-34));
    printf("round_up_10(34)= %d\n", round_up_10(34));
    printf("round_up_10(-34)= %d\n", round_up_10(-34));
}
    /*
    The outout:
    round_down_10(25450)= 25450
    round_down_10(-25450)= -25450
    round_up_10(25450)= 25450
    round_up_10(-25450)= -25450
    round_down_10(1347)= 1340
    round_down_10(-1347)= -1340
    round_up_10(1347)= 1350
    round_up_10(-1347)= -1350
    round_down_10(34)= 30
    round_down_10(-34)= -30
    round_up_10(34)= 40
    round_up_10(-34)= -40
    */

Upvotes: 0

user529758
user529758

Reputation:

It's unnecessary to convert the number to a string and back. You can do this using basic modulo arithmetic and multiplication and division.

Here's a pure numeric solution, hopefully somewhat more efficient in terms of running time:

int round_up_to_max_pow10(int n)
{
    int tmp = n;
    int i = 0;
    while ((tmp /= 10) >= 10) {
        i++;
    }

    if (n % (int)(pow(10, i + 1) + 0.5)) {
        tmp++;
    }

    for (; i >= 0; i--) {
        tmp *= 10;
    }

    return tmp;
}

printf("Original: %d; rounded: %d\n", 4937, round_up_to_max_pow10(4937));

Upvotes: 10

Clifford
Clifford

Reputation: 93476

No loops.

#include <math.h>
unsigned roundToNextExp10( unsigned a )
{
    int d = a ;
    if( a >= 10 )
    {
        int m ;
        d-- ;
        m = (int)pow( 10, ((int)log10(d)) ) ;
        d = (int)((d / m) + 1) * m ;
    }        
    return d ;
}

Upvotes: 0

Pavan Yalamanchili
Pavan Yalamanchili

Reputation: 12099

I am not sure if you want round or ceil. But the behavior you show in the question suggests ceil. So I included that.

int my_ceil(int num)
{
    int den = 1;
    int inc = 0;

    while (num >= 10) {
        inc += num % 10;
        num /= 10;
        den *= 10;
    }

    return (num + (inc > 0)) * den;
}

EDIT

Changed the code to remove ceil and other extra operations.

EDIT 2

Fixed for multiples of 10.

Upvotes: 1

Anoop Vaidya
Anoop Vaidya

Reputation: 46543

By Cocoa APIs:

int number=9435;
NSString *string=[NSString stringWithFormat:@"%d",number];
long length=[string length];    
NSString *roundedString=[NSString stringWithFormat:@"%d",([[string substringToIndex:1]intValue]+1)];
while (--length>0) {
    roundedString=[roundedString stringByAppendingString:@"0"];
}
int roundedNumber=[roundedString intValue];
NSLog(@"%d,   %d",number,roundedNumber);

By Typical C style, mathematically:

int i=8517;

int temp=i;
int len,msb;

for (len=0; temp>0; len++) {
    msb=temp;
    temp/=10;
}
msb++;
int multiplier=1;
for (int i=1; i<len; i++) {
    multiplier*=10;
}
 NSLog(@"Rounded : %d",msb*multiplier);

Upvotes: 1

Lee-Man
Lee-Man

Reputation: 414

This should do it:

static int
rnd_up(int val)
{
    double e, r;

    e = exp10(trunc(log10((double)val)));
    r = round(((double)val / e) + 0.5);

    return(r * e);
}

Upvotes: 0

Guffa
Guffa

Reputation: 700372

You can divide the number by ten until there is only one digit left, then multiply it back to size:

int n = 4937;

int m = 1;
while (n >= 10) {
  n = (n + 9) / 10;
  m *= 10;
}
n *= m;

Upvotes: 0

Satish
Satish

Reputation: 721

I would convert the number to string. Get the length of the string.

Then:

// pseudo code:
divisor = pow(10, len of number string)
answer = ceil (number/divisor) * divisor;

Upvotes: 0

rmaddy
rmaddy

Reputation: 318824

Avoid string conversions and loops:

int num = ... // your number
int len = log10(num);
float div = pow(10, len);
int rounded = ceil(num / div) * div;

Upvotes: 20

argentage
argentage

Reputation: 2778

Logarithms are quite helpful here to provide a constant-time answer to the "how many zeros does this have?"

floor(log10(x))= z //the number of zeros

will take the logarithm base 10 and give you the number of zeros that will be in x.

You can then use the C occasional idiom

(A+B-1)/B

to quickly find the ceiling of A/B, which results in the correct leading digit in this way:

zeros = exp10(1,z);
((x+zeros-1)/zeros) * zeros

This is pseudocode but you should get the idea. The key understanding is that logarithms are the way to mathematically determine how many digits a number has.

Upvotes: 3

Brian Salta
Brian Salta

Reputation: 1576

Try taking the first character of the input number, add 1, then append zeros.

    Dim Input = "23568"
    Dim roundUp = Left(Input, 1) + 1

    For x = 1 To Len(Input) - 1
        roundUp &= "0"
    Next

In VB, but hopefully you get the idea.

Upvotes: 0

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