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src/utils.c: Introduce parse_{time,duration}()

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These helpers can be used in a fashion similar to parse_date(). In
addition to check_time(), parse_time() and parse_duration() support
short forms such as "23:" (instead of "23:00") and ":45" (instead of
"00:45").

Signed-off-by: Lukas Fleischer <calcurse@cryptocrack.de>
This commit is contained in:
Lukas Fleischer 2011-10-05 09:06:18 +02:00
parent d874f7ff97
commit 0acbc06c43
2 changed files with 87 additions and 0 deletions
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2
src/calcurse.h
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@ -904,6 +904,8 @@ const char *get_tempdir (void);
char *new_tempfile (const char *, int);
int parse_date (const char *, enum datefmt, int *, int *, int *,
struct date *);
int parse_time (const char *, unsigned *, unsigned *);
int parse_duration (const char *, unsigned *);
void str_toupper (char *);
void file_close (FILE *, const char *);
void psleep (unsigned);

85
src/utils.c
View File

@ -689,6 +689,91 @@ parse_date (const char *date_string, enum datefmt datefmt, int *year,
return 1;
}
/*
* Converts a time string into hours and minutes. Short forms like "23:"
* (23:00) or ":45" (0:45) are allowed.
*
* Returns 1 on success and 0 on failure.
*/
int
parse_time (const char *string, unsigned *hour, unsigned *minute)
{
const char *p;
unsigned in[2] = {0, 0}, n = 0;
if (!string)
return 0;
/* parse string into in[], read up to two integers */
for (p = string; *p; p++)
{
if (*p == ':')
{
if ((++n) > 1)
return 0;
}
else if ((*p >= '0') && (*p <= '9'))
in[n] = in[n] * 10 + (int)(*p - '0');
else
return 0;
}
if (n != 1 || in[0] >= DAYINHOURS || in[1] >= HOURINMIN)
return 0;
*hour = in[0];
*minute = in[1];
return 1;
}
/*
* Converts a duration string into minutes. Both "hh:mm" (e.g. "23:42") and
* "mmm" (e.g. "100") formats are accepted. Short forms like "23:" (23:00) or
* ":45" (0:45) are allowed as well.
*
* Returns 1 on success and 0 on failure.
*/
int
parse_duration (const char *string, unsigned *duration)
{
const char *p;
unsigned in[2] = {0, 0}, n = 0;
if (!string || *string == '\0')
return 0;
/* parse string into in[], read up to two integers */
for (p = string; *p; p++)
{
if (*p == ':')
{
if ((++n) > 1)
return 0;
}
else if ((*p >= '0') && (*p <= '9'))
in[n] = in[n] * 10 + (int)(*p - '0');
else
return 0;
}
if (n == 0)
{
if (in[0] > 999)
return 0;
*duration = in[0];
}
else if (n == 1)
{
if (in[0] > DAYINHOURS || in[1] >= HOURINMIN)
return 0;
*duration = in[0] * HOURINMIN + in[1];
}
else
return 0;
return 1;
}
void
str_toupper (char *s)
{