regexp: match the entire expression

Hi,
Is there any way to do the match between an expression and a pattern perfectly and not partially?
Example:
setStrings{1} = '2*x + 3';
setStrings{2} = '2*x';
first_kind = regexp(setStrings, '(?<A>-?\d+)\s*\*\s*x', 'names')
first_kind = 1×2 cell array
{1×1 struct} {1×1 struct}
second_kind = regexp(setStrings, '(?<A>-?\d+)\s*\*\s*x\s*(?<B>[+-]\s*\d+)', 'names')
second_kind = 1×2 cell array
{1×1 struct} {0×0 struct}
here both strings 1 and 2 are considered to be of first kind. What I want is to exclude that the expression 2 can be counted as the first kind because it does not have B. So probably the option 'names' should be changed into something that guarantees the perfect match, otherwise it excludes it. Here in this example we have a partial match.

 Accepted Answer

setStrings{1} = '2*x + 3';
setStrings{2} = '2*x';
first_kind = regexp(setStrings, '^\s*(?<A>-?\d+)\s*\*\s*x\s*$', 'names')
first_kind = 1×2 cell array
{0×0 struct} {1×1 struct}
second_kind = regexp(setStrings, '(?<A>-?\d+)\s*\*\s*x\s*(?<B>[+-]\s*\d+)', 'names')
second_kind = 1×2 cell array
{1×1 struct} {0×0 struct}
By default, ^ matches "immediately after the beginning of input" and $ matches "immediately before end of input". However if you add the option 'lineanchors' then $ matches immediately after the beginning of each input line and $ matches immediately before the end of each input line (so $ does not match the newline itself.) If you use 'lineanchors' then it is common that you will also end up needing the option 'dotexceptnewline'

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Asked:

on 11 Nov 2022

Commented:

on 12 Nov 2022

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