Creating a char array in C using variables like in a printf

Question

I'm trying to create a char array in a similar way as using a printf statement.

If I do this:

printf("%d:%d:%.2f", hours, minutes, time);

It will print out exactly how I want it. However I'm now trying to store this as a variable. I've been trying to do something like the line of code below, however I get a "invalid initializer" error for char.

What I'm trying to do:

char temp[] = ("%d:%d:%.2f", hours, minutes, time);

I've also messed with strncat and couldn't figure that out either. Any guidance is appreciated!

You are looking for `sprintf` – Igor Tandetnik Oct 05 '13 at 21:15 — Igor Tandetnik, Oct 05 '13 at 21:15
I just don't understand C++ tag. This seems to be C. – LihO Oct 05 '13 at 21:19 — LihO, Oct 05 '13 at 21:19

score 2 · Answer 1 · edited Oct 08 '13 at 10:04

2

You'll want sprintf, it is the same as printf, but instead outputs to a string, as you wish.

EDIT: snprintf is indeed safer. (Thanks Troy)

edited Oct 08 '13 at 10:04

answered Oct 05 '13 at 21:15

Kninnug

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`snprintf` would be a safer bet. – goji Oct 05 '13 at 21:17

LihO · Answer 2 · 2013-10-05T21:44:29.593

1

You could use snprintf:

char temp[20];
snprintf(temp, sizeof(temp), "%d:%d:%.2f", hours, minutes, time);

edited Oct 05 '13 at 21:44

answered Oct 05 '13 at 21:17

LihO

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`20` => `sizeof temp` – Oct 05 '13 at 21:42

Gangadhar · Accepted Answer · 2013-10-05T21:43:10.790

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you can use sprintf() or snprintf()

int sprintf( char *restrict buffer, const char *restrict format, ... );

int snprintf( char *restrict buffer, int buf_size,const char *restrict format, ... );

   char temp[30];
   sprintf(temp,"%d:%d:%.2f", hours, minutes, time);
   printf("%s\n",temp);

For secure purpose use snprintf() as below

   char temp[30];
   snprintf(temp,sizeof temp,"%d:%d:%.2f", hours, minutes, time);
   printf("%s\n",temp);

edited Oct 05 '13 at 21:43

answered Oct 05 '13 at 21:18

Gangadhar

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I think it's worth pointing out that you should generally _never_ use `sprintf` (especially if any of the arguments are user-supplied or derived from data that's user-supplied). Using `sprintf`, if your formatted data ends up being larger than the buffer you passed it, the result will overflow into space outside the buffer, corrupting data and potentially causing crashes or (worse) a security vulnerability. `snprintf` avoids this by allowing you to specify the size of the buffer you passed it-- `snprintf` will never overflow its buffer. Best to get into good security practices early. – JonathonW Oct 05 '13 at 21:59
@JonathonW `never use sprintf (especially if any of the arguments are user-supplied or derived from data that's user-supplied)` Indeed info.Thanks for adding. – Gangadhar Oct 05 '13 at 22:07
Please leave comment when down voted my answer.some users are misusing down vote. – Gangadhar Oct 05 '13 at 22:26

score 0 · Answer 4 · edited Oct 08 '13 at 10:04

0

You can use snprintf to do that. It is exacly what you need:

#include <cstdio>

char temp[50];
snprintf( temp, sizeof(temp), "%d:%d:%.2f", hours, minutes, time);

edited Oct 08 '13 at 10:04

answered Oct 05 '13 at 21:18

Pierre Fourgeaud

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`50` => `sizeof temp` – Oct 05 '13 at 21:42

Creating a char array in C using variables like in a printf

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