Reputation: 2881
How to count total number of digits from a given positive number without looping in C?
Upvotes: 2
Views: 1838
Reputation: 11694
One possible solutions, assuming 16-bit integer values and logical expressions evaluating to 0 or 1. You could replace the conditions with (u > 99) ? 1 : 0
if you're worried about portability.
int digits( unsigned u)
{
return 1 + (u > 9) + (u > 99) + (u > 999) + (u > 9999);
}
Upvotes: 1
Reputation: 57870
For integers, take the log10 of the number, round down, and add one.
TEST:
#include <math.h>
#include <stdio.h>
int
num_digits(unsigned long number)
{
return (int)(floor(log10((double)number)) + 1.0);
}
int
main(int argc, char **argv)
{
unsigned long test_numbers[] = {
1, 9, 10, 99, 100, 999, 1000, 9999, 10000, 99999, 100000, 999999,
123456789ul,
999999999ul,
0
};
unsigned long *ptr;
for(ptr = test_numbers; *ptr; ptr++)
{
printf("Number of digits in %lu: %d\n", *ptr, num_digits(*ptr));
}
return 0;
}
Output:
Number of digits in 1: 1
Number of digits in 9: 1
Number of digits in 10: 2
Number of digits in 99: 2
Number of digits in 100: 3
Number of digits in 999: 3
Number of digits in 1000: 4
Number of digits in 9999: 4
Number of digits in 10000: 5
Number of digits in 99999: 5
Number of digits in 100000: 6
Number of digits in 999999: 6
Number of digits in 123456789: 9
Number of digits in 999999999: 9
Upvotes: 6
Reputation: 5722
Two solutions for you!
int digit_count(unsigned int n)
{
if (n < 10)
return 1;
else
return (1 + digit_count(n / 10));
}
and
unsigned int n = 50;
int i = 0;
HACK:
i++;
if (n < 10)
{
printf("Digits: %d\n", i);
}
else
{
n /= 10;
goto HACK;
}
Don't hate me for the last one :(
Upvotes: 0