constructor invocation mechanism

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Case 1)

m is interpreted as a function return my and taking no arguments.
To see the expected output remove () i.e use my m;

Case 2)

This is something better known as the “Most vexing parse”.

n is interpreted as a function returning my that takes an argument of type pointer to function returning my taking no arguments.

To see the expected output in this case try my n((my())); [Instead of treating as an argument specification as in the former case the compiler would now interpret it as an expression because of the extra ()]

My interpretation:

my n((my())) is equivalent to my n = my(). Now the rvalue expression my() creates a temporary[i.e a call to the default constructor] and n is copy initialized to that temporary object[no call to the copy-ctor because of some compiler optimization]

P.S: I am not 100% sure about the last part of my answer. Correct me if I am wrong.

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