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Home/ Questions/Q 6010185
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Editorial Team
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Editorial Team
Asked: May 23, 20262026-05-23T02:04:48+00:00 2026-05-23T02:04:48+00:00

When trying to simulate the evaluation behavior of RuleDelayed I faced unexpected behavior of

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When trying to simulate the evaluation behavior of RuleDelayed I faced unexpected behavior of nested Unevaluated. Consider:

In[1]:= f[Verbatim[Unevaluated][expr_]] := f[expr]
f[Unevaluated[1 + 1]]
f[Unevaluated@Unevaluated[1 + 1]]
f[Unevaluated@Unevaluated@Unevaluated[1 + 1]]
f[Unevaluated@Unevaluated@Unevaluated@Unevaluated[1 + 1]]

Out[2]= f[Unevaluated[1 + 1]]

Out[3]= f[2]

Out[4]= f[Unevaluated[1 + 1]]

Out[5]= f[2]

One can see that only even number of nested Unevaluated wrappers are completely removed. Why?

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-23T02:04:49+00:00Added an answer on May 23, 2026 at 2:04 am

    Use Trace to see why:

    In[1]:= f[Verbatim[Unevaluated][expr_]]:=f[expr]
    
    In[2]:= f[Unevaluated[1+1]]//Trace
    Out[2]= {f[1+1],f[Unevaluated[1+1]]}
    
    1. Due to the defining special property of the Unevaluated language construct, f[Unevaluated[1 + 1]] evaluates just like f[1 + 1] except the 1 + 1 is left unevaluated.
    2. f[1 + 1] does not match the definition you gave for f.
    3. Therefore f[Unevaluated[1 + 1]] remains unevaluated.

    Whereas:

    In[3]:= f[Unevaluated@Unevaluated[1 + 1]] // Trace
    Out[3]= {f[Unevaluated[1+1]],f[1+1],{1+1,2},f[2]}
    
    1. Due to the defining special property of the Unevaluated language construct, f[Unevaluated@Unevaluated[1 + 1]] evaluates just like f[Unevaluated[1 + 1]] except the Unevaluated[1 + 1] is left unevaluated.
    2. f[Unevaluated[1 + 1]] matches the definition you gave for f, and evaluates to f[1 + 1].
    3. Therefore f[Unevaluated@Unevaluated[1 + 1]] evaluates to f[2].
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