Furthermore, scalar is optimized away even when it isn't redundant, since scalar only has a compile-time effect.

>perl -MO=Concise -e"@a = scalar @b"
a  <@> leave[1 ref] vKP/REFC ->(end)
1     <0> enter ->2
2     <;> nextstate(main 1 -e:1) v:{ ->3
9     <2> aassign[t5] vKS ->a
-        <1> ex-list lK ->6
3           <0> pushmark s ->4
-           <1> scalar sK/1 ->-          <--- Note leading dash
5              <1> rv2av[t4] sK/1 ->6    <--- "s" means "scalar"
4                 <#> gv[*b] s ->5            did its job.
-        <1> ex-list lK ->9
6           <0> pushmark s ->7
8           <1> rv2av[t2] lKRM*/1 ->9
7              <#> gv[*a] s ->8
-e syntax OK
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It's impossible for scalar to work at run-time, since operands are evaluated before their operator.