Zero Fill Function C++

[Beginner01]

I am trying to write a function that passes in a int and if int is <10 puts a 0 infront of the int to print out. I can't use the manipulator.

Instead of 3, it prints 03.

Does anyone have any ideas?

[Beginner01]

I have to pass mutilple numbers into this function. So I don't think it can be in the out statement. I would think that it would have to be assigned to a variable.

[dush]

Hi

If your function doesnt have to return string composed of numbers you passed to it, in another words If you just need to print out the numbers to screen I would recommend you:

in C use printf :

int i = 5, j = 15;
printf("%02d", i);
printf("%d", j);

in C++ you can use cout in your function:

#include <iostream>
using namespace std;
 
void print(const int& x) {
  cout << (x < 10 ? cout << 0, x : x);
}
 
int main() {
 
  print(5);    
  cout << endl;  
  print(15);
 
  return 0;    
}

[D_C]

For A number, say A, with N decimal digits, we can relate them by N = (int)(log A), where the logarithm is base 10.

If you want all numbers to have 30 digits, then print a string with (30-N) "0"s, followed by your number A.

[dush]

thanks for good idea D_C

I was actually thinkin how to get count of digits of integer number.
(int)log10 works fine, so I upgraded my function:

void print(const int& x, int n) {
while(--n > (int)log10(x)) cout << 0;
cout << x;
}

[etaham]

I found this post on google while looking for a solution to the same problem.
Here's what I got,

cout.fill('0');
cout<<std::setw(N)<<your number<<endl;

Where N is the width of the string you wish to print, and your number is your int.
The fill function tells the stream what to fillt he extra spaces with.