Printing int values with a user determined padding?

Question

I'm pretty sure there's no way to do this, but I was wondering, so here goes...

Say I get several int values from the user, and I know for a fact that the 1st value is the largest. Any way I can use the length of the first int value as the padding when printing the rest of the numbers?

Browsing the forum I found that you can determine the length of an int (n) like this:

l = (int) log10(n) + 1;

Using padding of 5 a print would look like:

printf("%5d", n);

So what I want to know is whether or not there's a way to use l instead of 5...

Thank you!

Actually there is : http://stackoverflow.com/questions/7105890/set-variable-text-column-width-in-printf — Tomer Arazy, May 21 '13 at 07:47

alk · Accepted Answer · 2013-05-21T09:22:15.380

8

Do it like so:

int n = <some value>;
int i = (int) log10(n) + 1;
printf("%*d", i, n);

The * serves as a placeholder for the width being passed as an argument to printf().

edited May 21 '13 at 09:22

answered May 21 '13 at 07:48

alk

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So there IS a way :D Thank you ^^ – kanpeki May 21 '13 at 08:08
@alk What does the `log10(n)` line does? – Kraken May 21 '13 at 08:38
@Kraken: It calculates the number of decimal places used by `n`. Also please see the OP for this. – alk May 21 '13 at 09:16

paxdiablo · Answer 2 · 2013-05-21T07:53:24.563

0

You can use also snprintf() to get the number of characters printed for a value thus:

char buff[100];
int width = snprintf (buff, sizeof(buff), "%d", number1);

But, however you get the desired width, you can just use the * format modifier:

printf( "%*d %*d\n", width, number1, width, number2);

That modifier uses the next argument to specify a dynamic width rather than one fixed in the format string.

edited May 21 '13 at 07:53

answered May 21 '13 at 07:47

paxdiablo

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Printing int values with a user determined padding?

2 Answers2