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Andy
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 Posted: Sun Nov 28, 2004 6:10 pm Post subject: How does C++ infer this in compilation? |
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Hi,
How does C++ infer an expression is an implicit initialization or not?
For example, for the class BinaryNumber, which uses a boolean vector
to represent a binary number.
class BinaryNumber
{
public:
BinaryNumber(int n) { data.resize(n); }
...
private:
vector<bool> data;
}
The implicit initialization
BinaryNumber num = 5;
is allowed. How does C++ compilers figure this out?
Thanks a lot!
Andy
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Alberto Barbati
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| Posted: Sun Nov 28, 2004 11:54 pm Post subject: Re: How does C++ infer this in compilation? |
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Andy wrote:
| Quote: |
The implicit initialization
BinaryNumber num = 5;
is allowed. How does C++ compilers figure this out?
|
I'm sorry... but I can't see the implicit initialization. According to
§12.6.1/1 it's a case of *explicit* initialization, the code above being
equivalent to
BinaryNumber num(BinaryNumber(5));
Moreover, if certain conditions are met, the compiler is allowed to omit
the temporary and the copy.
Alberto
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James Hopkin
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| Posted: Mon Nov 29, 2004 9:47 pm Post subject: Re: How does C++ infer this in compilation? |
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[email]yyu_mail (AT) yahoo (DOT) com[/email] (Andy) wrote in message news:<c0ec8bb4.0411280535.7acd5c21 (AT) posting (DOT) google.com>...
| Quote: | Hi,
How does C++ infer an expression is an implicit initialization or not?
For example, for the class BinaryNumber, which uses a boolean vector
to represent a binary number.
class BinaryNumber
{
public:
BinaryNumber(int n) { data.resize(n); }
...
private:
vector<bool> data;
}
The implicit initialization
BinaryNumber num = 5;
is allowed. How does C++ compilers figure this out?
|
I'm not quite sure I understand what you're asking, but perhaps this is helpful.
If you declare the constructor explicit:
explicit BinaryNumber(int n) { ... }
then the = form of initialisation is not valid, e.g.
BinaryNumber num = 5;
will not compile, but
BinaryNumber num(5);
will.
James
[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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