String matching

String matching
You are encouraged to solve this task according to the task description, using any language you may know.

Given two strings, demonstrate the following 3 types of matchings:

1. Determining if the first string starts with second string
2. Determining if the first string contains the second string at any location
3. Determining if the first string ends with the second string

Optional requirements: A) Print the location of the match for part 2 B) Handle multiple occurrences of a string for part 2.

ALGOL 68

Translation of: python

Works with: ALGOL 68 version Revision 1 - no extensions to language used

Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny

<lang algol68># define some appropriate OPerators # PRIO STARTSWITH = 5, ENDSWITH = 5; OP STARTSWITH = (STRING str, prefix)BOOL: # assuming LWB = 1 #

 IF UPB str < UPB prefix THEN FALSE ELSE str[:UPB prefix]=prefix FI;

OP ENDSWITH = (STRING str, suffix)BOOL: # assuming LWB = 1 #

 IF UPB str < UPB suffix THEN FALSE ELSE str[UPB str-UPB suffix+1:]=suffix FI;

INT loc, loc2;

print((

 "abcd" STARTSWITH "ab", # returns TRUE #
 "abcd" ENDSWITH "zn", # returns FALSE #
 string in string("bb",loc,"abab"), # returns FALSE #
 string in string("ab",loc,"abab"), # returns TRUE #
 (string in string("bb",loc,"abab")|loc|-1), # returns -1 #
 (string in string("ab",loc,"abab")|loc|-1), # returns +1 #
 (string in string("ab",loc2,"abab"[loc+1:])|loc+loc2|-1) # returns +3 #

))</lang> Output:

TFFT         -1         +1         +3

Objective-C

<lang objc> [@"abcd" hasPrefix:@"ab"] //returns true [@"abcd" hasSuffix:@"zn"] //returns false int loc = [@"abab" rangeOfString:@"bb"].location //returns -1 loc = [@"abab" rangeOfString:@"ab"].location //returns 0 loc = [@"abab" rangeOfString:@"ab" options:0 range:NSMakeRange(loc+1, [@"abab" length]-(loc+1))].location //returns 2 </lang>

Python

<lang python> "abcd".startswith("ab") #returns true "abcd".endswith("zn") #returns false "bb" in "abab" #returns false "ab" in "abab" #returns true loc = "abab".find("bb") #returns -1 loc = "abab".find("ab") #returns 0 loc = "abab".find("ab",loc+1) #returns 2 </lang>