I have noticed that both int num(5) and int num = 5 both compile and assign 5 to num.

Questions

  1. Could you help me understand why int num(5) work? This syntax looks like initializing an object
  2. What is the difference between the two in terms of speed and stuff that goes behind

Pardon me if it's a silly question. I am a noob at C++ and in process to learn stuff.

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1 Answer

Could you help me understand why int num(5) work? This syntax looks like initializing an object

For the basic case (where the programmer explicitly types in a line of code like int num(5); or int num = 5;, having two different syntaxes that both do the same thing seems unnecessary, and it is.

However, once you get into templates, you'll find cases where you want to write a template that can work with both user-defined classes and with the built-in types. For example, here is a template I wrote the other day to save (and later restore) the state of an arbitrary value:

template<class T> class SaveGuard { public: SaveGuard(T & saveMe) : _saveMe(saveMe), _tempHolder(saveMe) {/* empty */} // save the current value ~SaveGuard() {_saveMe = _tempHolder;} // restore the saved value private: T & _saveMe; T _tempHolder; };

Note that in this case, the _tempHolder(saveMe) clause is syntactically the same as the int(5) syntax, and since the int(5) syntax is supported, I can declare a SaveGuard<int> guard(blah); in my program and it will work as expected. Alternatively I can declare a SaveGuard<MyClassType> guard(blah); and the template will work for that as well, using the copy-constructor of the MyClassType class.

What is the difference between the two in terms of speed and stuff that goes behind

There's no difference in speed in your example; the compiler will generate the same assembly code for either syntax.

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