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Count occurrences of a character in a string
Posted: Tue Jan 03, 2017 7:31 pm
by rbytes
I just count them
Re: Count occurrences of a character in a string
Posted: Tue Jan 03, 2017 9:31 pm
by Dav
Nice. That's the part of your Bulls and cows code that interested me the most.
The only thing I wonder is that perhaps we should not assume option base 1 in our functions when putting together a library for everyone? Maybe we should do the option_base$() call in the function, change option base at the beginning, reset it when leaving? just wondering about it, but I'm not a good one to say that because I always use option base 1 in my code, lol.
Great function by the way. Small code and very useful.
- Dav
Re: Count occurrences of a character in a string
Posted: Wed Jan 04, 2017 12:53 am
by rbytes
Good idea. I have altered the function as suggested and hope others will do the same.
Thanks, Dav.
Re: Count occurrences of a character in a string
Posted: Mon Jan 09, 2017 4:39 pm
by Joel
a general and more efficient solution:
Code: Select all
'option base 1
a$="Now is the time for all good men to come to the aid of their country"
i$="i"
PRINT "String: "&a$
PRINT "Character: "&i$
PRINT
chk=count_instr(a$,i$)
PRINT "Character '"&i$&"' appears "&chk&" times in string."
a$="Now is the time for all good men to come to the aid of their country"
i$="he"
PRINT "String: "&a$
PRINT "Character: "&i$
PRINT
chk=count_instr(a$,i$)
PRINT "Character '"&i$&"' appears "&chk&" times in string."
a$="xxxx"
i$="xx"
PRINT "String: "&a$
PRINT "Character: "&i$
PRINT
chk=count_instr(a$,i$)
PRINT "Character '"&i$&"' appears "&chk&" times in string."
print "option_base(): ";option_base()
DEF count_instr(string$,SUB$)
result=INSTR(string$,SUB$)
IF result=-1 THEN RETURN 0
IF result=LEN(string$)-LEN(SUB$)+OPTION_BASE() THEN RETURN 1
RETURN 1+ count_instr(MID$(string$,result+LEN(SUB$)),SUB$)
END DEF