String matching: Difference between revisions
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A) Print the location of the match for part 2 |
A) Print the location of the match for part 2 |
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B) Handle multiple occurrences of a string for part 2. |
B) Handle multiple occurrences of a string for part 2. |
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{{trans|C}} - note: This specimen retains the original [[Character_matching.a68#C|C]] coding style. |
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{{works with|ALGOL 68|Revision 1 - no extensions to language used}} |
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{{works with|ALGOL 68G|Any - tested with release [http://sourceforge.net/projects/algol68/files/algol68g/algol68g-1.18.0/algol68g-1.18.0-9h.tiny.el5.centos.fc11.i386.rpm/download 1.18.0-9h.tiny]}} |
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{{wont work with|ELLA ALGOL 68|Any (with appropriate job cards) - tested with release [http://sourceforge.net/projects/algol68/files/algol68toc/algol68toc-1.8.8d/algol68toc-1.8-8d.fc9.i386.rpm/download 1.8-8d] - due to extensive use of '''format'''[ted] ''transput''}} |
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<lang algol68># define some appropriate OPerators # |
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PRIO STARTSWITH = 5, ENDSWITH = 5; |
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OP STARTSWITH = (STRING str, prefix)BOOL: # assuming LWB = 1 # |
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IF UPB str < UPB prefix THEN FALSE ELSE str[:UPB prefix]=prefix FI; |
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OP ENDSWITH = (STRING str, suffix)BOOL: # assuming LWB = 1 # |
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IF UPB str < UPB suffix THEN FALSE ELSE str[UPB str-UPB suffix+1:]=suffix FI; |
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INT loc, loc2; |
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#include <string> |
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using namespace std; |
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string s1="abcd"; |
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string s2="abab"; |
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string s3="ab"; |
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//Beginning |
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s1.compare(0,s3.size(),s3)!=0; |
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//End |
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s1.compare(s1.size()-s3.size(),s3.size(),s3)!=0; |
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//Anywhere |
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s1.find(s2)//returns string::npos |
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int loc=s2.find(s3)//returns 0 |
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loc=s2.find(s3,loc+1)//returns 2 |
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=={{header|Java}}== |
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<lang java> |
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"abab".contains("bb") //returns false |
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"abab".contains("ab") //returns true |
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int loc = "abab".indexOf("bb") //returns -1 |
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loc = "abab".indexOf("ab") //returns 0 |
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loc = "abab".indexOf("ab",loc+1) //returns 2 |
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</lang> |
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print(( |
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string in string("bb",loc,"abab"), # returns FALSE # |
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string in string("ab",loc,"abab"), # returns TRUE # |
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(string in string("bb",loc,"abab")|loc|-1), # returns -1 # |
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(string in string("ab",loc,"abab")|loc|-1), # returns +1 # |
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(string in string("ab",loc2,"abab"[loc+1:])|loc+loc2|-1) # returns +3 # |
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Output: |
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<pre> |
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TFFT -1 +1 +3 |
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=={{header|Objective-C}}== |
=={{header|Objective-C}}== |
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Revision as of 06:59, 15 October 2010
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:
- Determining if the first string starts with second string
- Determining if the first string contains the second string at any location
- 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
- note: This specimen retains the original C coding style.
<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>