Order two numerical lists: Difference between revisions
Added Algol 68
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TRUE
TRUE
</pre>
=={{header|ALGOL 68}}==
<lang algol68>BEGIN # compare lists (rows) of integers #
# returns TRUE if there is an element in a that is < the corresponding #
# element in b and all previous elements are equal; FALSE otherwise #
OP < = ( []INT a, b )BOOL:
IF BOOL result := FALSE;
INT a pos := LWB a;
INT b pos := LWB b;
BOOL equal := TRUE;
WHILE a pos <= UPB a AND b pos <= UPB b AND equal DO
equal := a[ a pos ] = b[ b pos ];
IF equal THEN
a pos +:= 1;
b pos +:= 1
FI
OD;
NOT equal
THEN
# there is an element in a and b that is not equal #
a[ a pos ] < b[ b pos ]
ELSE
# all elements are equal or one list is shorter #
# a is < b if a has fewer elements #
a pos > UPB a AND b pos <= UPB b
FI # < # ;
# tests a < b has the expected result #
PROC test = ( STRING a name, []INT a, STRING b name, []INT b, BOOL expected )VOID:
BEGIN
BOOL result = a < b;
print( ( a name, IF result THEN " < " ELSE " >= " FI, b name
, IF result = expected THEN "" ELSE ", NOT as expected" FI
, newline
)
)
END # test # ;
# test cases as in the BBC basic sample #
[]INT list1 = ( 1, 2, 1, 5, 2 );
[]INT list2 = ( 1, 2, 1, 5, 2, 2 );
[]INT list3 = ( 1, 2, 3, 4, 5 );
[]INT list4 = ( 1, 2, 3, 4, 5 );
test( "list1", list1, "list2", list2, TRUE );
test( "list2", list2, "list3", list3, TRUE );
test( "list3", list3, "list4", list4, FALSE );
# additional test cases #
[]INT list5 = ( 9, 0, 2, 1, 0 );
[]INT list6 = ( 4, 0, 7, 7 );
[]INT list7 = ( 4, 0, 7 );
[]INT list8 = ( );
test( "list5", list5, "list6", list6, FALSE );
test( "list6", list6, "list7", list7, FALSE );
test( "list7", list7, "list8", list8, FALSE );
test( "list8", list8, "list7", list7, TRUE )
END</lang>
{{out}}
<pre>
list1 < list2
list2 < list3
list3 >= list4
list5 >= list6
list6 >= list7
list7 >= list8
list8 < list7
</pre>
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