concatenating strings and snprintf in c

Question

I'm wondering if this is the proper way to concatenate and NUL terminate strings including width.

#define FOO "foo"
const char *bar = "bar";
int n = 10;
float f = 10.2;

char *s;
int l;

l = snprintf (NULL, 0, "%-6s %-10s %4d %4f",FOO, bar, n, f);
s = malloc (l + 4); // should it be the number of formats tags?
if (s == null) return 1;
sprintf (s, "%-6s %-10s %4d %4f", FOO, bar, n, f);

It might be a good idea for maintenance purposes to define one format string that is used by both the snprintf and the sprintf. It'd really suck if it got changed in one place and not the other! — Fred Larson
– Fred Larson, Commented Dec 8, 2010 at 3:23
That sounds like a good idea, would it be possible to post a sample code in reference to this? Thanks — Lucas
– Lucas, Commented Dec 8, 2010 at 3:27
Just declare something like const char* fmt_str = "%-6s %-10s %4d %4f"; and use that as your format string in both places. — Fred Larson
– Fred Larson, Commented Dec 8, 2010 at 3:31

ephemient · Accepted Answer · 2010-12-08 03:15:20Z

1

Quite a few systems have a function asprintf in their standard C libraries that does exactly what you do here: allocate and sprintf.

answered Dec 8, 2010 at 3:15

ephemient

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Comments

caf · Accepted Answer · 2010-12-08 03:10:55Z

0

You only need to add 1 to the value returned by snprintf(), since there is only one null terminator added.

However, you do need to check for l == -1 (indicating that snprintf() failed).

answered Dec 8, 2010 at 3:10

caf

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2 Comments

ephemient Over a year ago

snprintf isn't defined in the original C89, and I think that there were a few platforms with non-C99-compliant implementations, but C99 does say that snprintf(NULL, ...) is supposed to work the way OP uses it here.

caf Over a year ago

@ephemient: Quite so, I've removed that bit.

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