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Found the answer by myself:
"C++ Primer" said:
The copy form of initialization is primarily supported for compatibility with C usage. When it can do so, the compiler is permitted (but not obligated) to skip the copy constructor and create the object directly.
The initialization of the copy form (ie, initialize with the = sign, I note) is mainly for compatibility with C usage. When this usage is supported, the compiler is allowed (but not forced) to skip the copy constructor and create the object directly.
I think this passage has already answered my question.
Thank you all upstairs, especiallykillingme. |
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