I need to convert two integers into two arrays of digits, so for example 544 would become arr[0] = 5, arr[1] = 4, arr[2] = 4.
I have found some algorithms doing this, but they create new array, and return this. I would have to allocate this memory for two arrays, so I wanna pass two integers by reference and do this on them directly.
I guess I can do this, because these integers are in fact template types, so they should be changeable. That's why I added C++ tag here.
Just using something like this:
int n = 544; // your number (this value will Change so you might want a copy)
int i = 0; // the array index
char a[256]; // the array
while (n) { // loop till there's nothing left
a[i++] = n % 10; // assign the last digit
n /= 10; // "right shift" the number
}
Note that this will result in returning the numbers in reverse order. This can easily be changed by modifying the initial value of i as well as the increment/decrement based on how you'd like to determine to length of the value.
(Brett Hale) I hope the poster doesn't mind, but I thought I'd add a code snippet I use for this case, since it's not easy to correctly determine the number of decimal digits prior to conversion:
{
char *df = a, *dr = a + i - 1;
int j = i >> 1;
while (j--)
{
char di = *df, dj = *dr;
*df++ = dj, *dr-- = di; /* (exchange) */
}
}
a[i++] = n % 10 should be changed to a[i++] = '0' + (n % 10)A simple solution is:
int i = 12312278;
std::vector<int> digits;
while (i)
{
digits.push_back(i % 10);
i /= 10;
}
std::reverse(digits.begin(), digits.end());
or, string based ( i >= 0 )
for (auto x : to_string(i))
digits.push_back(x-'0');
c. Not sure why the question is tagged for c as well as c++.Call the integer a.
To get the units digit of a, a % 10
To shift a down so the tens is the units digit, a / 10
To know when you're done, a == 0
To know how large your array needs to be in the first place, min(ceil(log(a+1, 10)), 1) (to convince yourself this works, try the logarithm part of it in a calculator. if you don't have multiple argument log, use the identity log(x,y) == log(x)/log(y))
vector<int> for everything and never again have to fret about if your array is the right length or not :)You can do something like this if you are using C++:
#include <iostream>
#include <string>
#include <sstream>
using namespace std;
int main()
{
int a=544;
stringstream str;
str << a;
string arr;
str>>arr;
for(int i=0; i<arr.length(); i++)
{
cout << arr[i];
}
system("pause");
return 0;
}
%. You'd have had it coded faster than it took you to ask the question.