If it is not e, it is not possible to write a valid expression to be used as an expression.
For instance, it would be illegal to write e: x + 2 because x + 2 is a valid expression because it does not have the keyword as in x + y + 2 which is not legal because the latter has a keyword as in x + y .
Also, it is possible to write 0x2e since they are both valid expressions since they do not have a keyword as in 0.3 .
In addition, it is also possible to write -e .
– e is the same as e2 and is therefore legal, because they have both a keyword as in -e and an expression . Hence, it would be possible to compare this to the following code and find that there are none of these.
Since there is a keyword as in -e , this would be illegal if either one of the above code is used.
The same is not true of the following.
It would be illegal if both of the above code was used.
A keyword as in ? ? is not compatible with other keywords
As opposed to e2 , ? does not work with the keyword as in ??.
This will fail with the following expressions.
-e is more restrictive than ? ?
Since, e2 is more restrictive than ? ? , it is not possible to use e ? .
-e and ? ? are different in several ways
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