Thread: how could I get each digit in a float into a array of ints?

Lets say I have

Code:

        float a = 12.50f;

and I want to have

Code:

        int b[4] = { 1, 2, 5, 0 };

I already have a way to know how many digits the float has if you are interested:

Code:

        n = 10;
	while (1)
	{
		Digits++;
		if (n > TheFloat)
			break;
		n *= 10;
	}
        Digits += 2;

the last part is for the 2 decimals.
Thanks for any help.

By far the best way is to convert the number to a string and then assign each digit to an int array. Here's a quickie example:

Code:

#include <iomanip>
#include <iostream>
#include <sstream>
#include <string>

using namespace std;

int
main()
{
  float        a = 12.50f;
  int          b[4];
  char         ch;
  stringstream ss;

  ss<< fixed << setprecision(2) << a;
  for (int i = 0; ss>> ch; ) {
    if (ch == '.') {
      continue;
    }
    b[i++] = ch - '0';
  }
  for (int i = 0; i < 4; i++) {
    cout<< b[i] <<endl;
  }

  return 0;
}

thanks but I am using an old compiler and I dont seem to have the sstream library, if anyone knows of a way to just get like the leftmost digit of the float, then the next digit and so on it would be great. But thanks anyway prelude.

this method uses stdlib.h
http://www.cplusplus.com/ref/cstdlib/ecvt.html

if u want free standard compiler go to www.bloodshed.net

>But thanks anyway prelude.
I'm not done yet though. You can do it the old fashioned way too:

Code:

#include <iostream.h>
#include <stdlib.h>

int
main()
{
  float        a = 12.50f;
  int          b[4];
  int          i, j;
  char         buf[256];
  const int    ndigits = 4;

  sprintf(buf, "%f", a);
  for (i = j = 0; j < ndigits; i++) {
    if (buf[i] == '.') {
      continue;
    }
    b[j++] = buf[i] - '0';
  }
  for (i = 0; i < 4; i++) {
    cout<< b[i] <<endl;
  }

  return 0;
}

thanks very very much prelude and alphaoid, I now have what I was looking for and I can sleep in peace