Simple question about C program

[Hugh Kang]

I am a beginner in C programming. I have a question about C
programming.
In one of our c programs:

char prep_name[18];
char s_time[8];
sprintf(prep_name, "S%s%d%d, s_time, pid, sql_stmt_ctr++);

s_time is alawys 8 characters but pid is 4 ~ 5 and sql_stmt_ctr is
usually
1 ~ 6. We found out that prep_name[] could be over 18 characters when
sql_stmt_ctr is more than 6 digits. We'd like to have just 18
characters
in prep_name even when sql_stmt_ctr is over 6 digits(truncate the rest
digits).
Is there any way to do it?

Regards
Hugh

[James Antill]

On Tue, 22 Jul 2003 04:50:55 +0200, Måns Rullgård wrote:

> [email]skang@leaguedata.com[/email] (Hugh Kang) writes:
>

>> I am a beginner in C programming. I have a question about C
>> programming.
>> In one of our c programs:
>>
>> char prep_name[18];
>> char s_time[8];
>> sprintf(prep_name, "S%s%d%d, s_time, pid, sql_stmt_ctr++);
>>
>> s_time is alawys 8 characters but pid is 4 ~ 5 and sql_stmt_ctr is
>> usually
>> 1 ~ 6. We found out that prep_name[] could be over 18 characters when
>> sql_stmt_ctr is more than 6 digits. We'd like to have just 18
>> characters
>> in prep_name even when sql_stmt_ctr is over 6 digits(truncate the rest
>> digits).
>> Is there any way to do it?

>
> snprintf(prep_name, 18, "S%s%d%d, s_time, pid, sql_stmt_ctr++);

Note that this will only have 17 characters and a terminator. This may or
may not be what the OP wanted.
I'd worry that truncating sql_stmt_ctr could affect the result, so it
might be better to make sure prep_name is big enough to hold all values.

Also, if you want to use snprintf() it would be much more maintainable
as...

snprintf(prep_name, sizeof(prep_name), "%c%s%lu%u",
s_time, (unsigned long) pid, sql_stmt_ctr++);

--
James Antill -- [email]james@and.org[/email]
Need an efficent and powerful string library for C?
[url]http://www.and.org/vstr/[/url]