in what cases is read is not inverse of show?

[S. Alexander Jacobson]

I've seen people say here that Haskell's implementation of 
read and show are not correct.

I have defined a datastructure in which the following fails:

> foo::MyType
> foo = read $ show myType

What features of myType will cause this usage of read/show to fail?

-Alex-

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S. Alexander Jacobson                   i2x Media  
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