Formatting numbers in loop

I want to list all numbers from 0000-9999 however I am having trouble holding the zero places.

I tried:

for(int i = 0; i <= 9开发者_开发问答999; ++i)
{
cout << i << "\n";
}

but I get: 1,2,3,4..ect How can I make it 0001,0002,0003....0010, etc

---

See setfill for specifying the fill character, and setw for specifying the minimum width.

Your case would look like:

for(int i = 0; i <= 9999; ++i)
{
    cout << setfill('0') << setw(4) << i << "\n";
}

---

You just need to set some flags:

#include <iostream>
#include <iomanip>

using namespace std;
int main()
{
    cout << setfill('0');
    for(int i = 999; i >= 0; --i)
    {
        cout << setw(4) << i << "\n";
    }
    return 0;
}

---

Use ios_base::width() and ios::fill():

cout.width(5);
cout.fill('0');
cout << i << endl;

Alternatively, use the IO manipulators:

#include<iomanip>

// ...
cout << setw(5) << setfill('0') << i << endl;

---

Though not required, but if you want to know how to do this with C, here is an example:

for (int i = 0; i <= 9999; i++)
    printf("%04d\n", i);

Here, '0' in "%04d" works like setfill('0') and '4' works like setw(4).