I know that you can get the digits of a number using modulus and division. The following is how I've done it in the past: (Psuedocode so as to make students reading this do some work for their homework assignment):
int pointer getDigits(int number)
initialize int pointer to array of some size
initialize int i to zero
while number is greater than zero
store result of number mod 10 in array at index i
divide number by 10 and store result in number
increment i
return int pointer
Anyway, I was wondering if there is a better, more efficient way to accomplish this task? If not, is there any alternative methods for this task, avoiding the use of strings? C-style or otherwise?
Thanks. I ask because I'm going to be wanting to do this in a personal project of mine, and I would like to do it as efficiently as possible.
Any help and/or insight is greatly appreciated.
The formula will be integer of (log10(number) + 1). For an example, if the number is 1245, then it is above 1000, and below 10000, so the log value will be in range 3 < log10(1245) < 4. Now taking the integer, it will be 3. Then add 1 with it to get number of digits.
The number of digits can be calculated by using log10(num)+1, where log10() is the predefined function in math. h header file.
First, get the size needed to store all the digits in the number -- do a malloc of an array. Next, take the mod of the number and then divide the number by 10. Keep doing this till you exhaust all digits in the number.
The time it takes to extract the digits will be dwarfed by the time required to dynamically allocate the array. Consider returning the result in a struct:
struct extracted_digits
{
int number_of_digits;
char digits[12];
};
You'll want to pick a suitable value for the maximum number of digits (12
here, which is enough for a 32-bit integer). Alternatively, you could return a std::array<char, 12>
and encode the terminal by using an invalid value (so, after the last value, store a 10
or something else that isn't a digit).
Depending on whether you want to handle negative values, you'll also have to decide how to report the unary minus (-
).
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