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Problem 2159. A SUBSREF variant that accepts the 'end'-operator.

Solution 1685218

Submitted on 3 Dec 2018 by Binbin Qi
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Test Suite

Test Status Code Input and Output
1   Pass
nocheat = isempty(regexp(evalc('type subsrefbetter'),'(eval|regexprep|inline|str2func)')); s.a.b(3).c{2}.d = 'a':'z'; def = substruct('.','a','.','b','()',{3},'.','c','{}',{2},'.','d','()',{1:3}); y = subsrefbetter(s,def); assert(isequal(y,'abc') && nocheat)
2   Pass
nocheat = isempty(regexp(evalc('type subsrefbetter'),'(eval|regexprep|inline|str2func)')); s.a.B(1).c{2}.d = 'a':'z'; def = substruct('.','a','.','B','()',{1},'.','c','{}',{2},'.','d','()',{'end'}); y = subsrefbetter(s,def); assert(isequal(y,'z') && nocheat)
3   Pass
nocheat = isempty(regexp(evalc('type subsrefbetter'),'(eval|regexprep|inline|str2func)')); s.oz = magic(7); def = substruct('.','oz','()',{'end-1' 'end-2:7'}); y = subsrefbetter(s,def); assert(isequal(y,[43 3 12]) && nocheat)
4   Pass
nocheat = isempty(regexp(evalc('type subsrefbetter'),'(eval|regexprep|inline|str2func)')); [s(1:3).TiTa] = deal(magic(3)); def = substruct('()',{2},'.','TiTa','()',{':' ':'}); y = subsrefbetter(s,def); assert(isequal(y,magic(3)) && nocheat)

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