Question

int value = 5; // this type of assignment is called an explicit assignment
int value(5); // this type of assignment is called an implicit assignment

What is the difference between those, if any, and in what cases do explicit and implicit assignment differ and how?


http://weblogs.asp.net/kennykerr/archive/2004/08/31/Explicit-Constructors.aspx

EDIT: I actually just found this article, which makes the whole thing a lot clearer... and it brings up another question, should you (in general) mark constructors taking a single parameter of a primitive type - numeric/bool/string - as explicit and leave the rest as they are (of course keeping watch for gotchas such as constructors like (int, SomeType = SomeType())?

Was it helpful?

Solution

They differ if a class has a constructor marked 'explicit'. Then, one of these does not work.

Otherwise, no difference.

OTHER TIPS

Neither of these is an assignment of any kind -- they're both initialization. The first uses copy initialization, and the second direct initialization. (FWIW, I'm pretty sure I've never heard the terms "explicit assignment" or "implicit assignment" before).

Edit: (Mostly in response to Nathan's comment):

Here's a corrected version of the code from your comment:

#include <iostream>

struct Foo { 
    Foo() { 
        std::cout << "Foo::ctor()" << std::endl; 
    } 
    Foo(Foo const& copy) { 
        std::cout << "Foo::cctor()" << std::endl; 
    } 
    Foo& operator=(Foo const& copy) { 
        std::cout << "foo::assign()" << std::endl; 
        return *this; 
    } 
};

int main(int, const char**) { 
    Foo f; 
    Foo b(f); 
    Foo x = b;
    return 0; 
}

The result from running this should be:

Foo::ctor()
Foo::cctor()
Foo::cctor()

If you run it and get an foo::assign(), throw your compiler away and get one that works (oh, and let us know what compiler it is that's so badly broken)!

Only the first one is an assignment. They are both initialization.

Edit: actually, I'm wrong. Neither are assignment.

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