|  | Can't template operator << be private? | | |
| | | lv |  |
|  Posted: Thu Sep 04, 2008 4:24 am Post subject: Can't template operator << be private? |  |
Hi all!
I have a class with a private template operator << like this:
class MyTest
{
template <typename T>
MyTest & operator << (T const & t)
{
std::cout << t << std::endl;
return *this;
}
};
But the operator is accessible outside the class. However, if I make
it a non-template operator that only accepts an integer, it's truly
private. The code was tested on vc9.0.
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| |
| | | Jiang | |
| | Posted: Thu Sep 04, 2008 9:28 am Post subject: Re: Can't template operator << be private? | |
On Sep 4, 1:24 pm, lv <lv.jc...@gmail.com> wrote:
| Quote: | Hi all!
I have a class with a private template operator << like this:
class MyTest
{
template <typename T
MyTest & operator << (T const & t)
{
std::cout << t << std::endl;
return *this;
}
};
But the operator is accessible outside the class. However, if I make
it a non-template operator that only accepts an integer, it's truly
private.
|
Access specifiers should be respected for both template and
non-template code.
| Quote: | The code was tested on vc9.0.
|
Test more please. Seems vc9 is broken here.
* icl 10.1
* GCC (4.3.2/4.3.1/4.2.2/3.4.4)
* como 4.3.10.1
reject your access code in my mind.
Regards,
Jiang
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| |
| | | Thomas Richter | |
| | Posted: Thu Sep 04, 2008 9:28 am Post subject: Re: Can't template operator << be private? | |
lv schrieb:
| Quote: | Hi all!
I have a class with a private template operator << like this:
class MyTest
{
template <typename T
MyTest & operator << (T const & t)
{
std::cout << t << std::endl;
return *this;
}
};
But the operator is accessible outside the class.
|
For which definition of "accessible"? Could you please post a complete
example that demonstrates the problem?
So long,
Thomas
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| |
| | | lv | |
| | Posted: Thu Sep 04, 2008 7:31 pm Post subject: Re: Can't template operator << be private? | |
On Sep 4, 5:28 pm, Thomas Richter <t...@math.tu-berlin.de> wrote:
| Quote: | lv schrieb:
For which definition of "accessible"? Could you please post a complete
example that demonstrates the problem?
So long,
Thomas
|
Well, I should have made it clearer. I meant the following code
compiles ,
#include <iostream>
class MyTest
{
template <typename T>
MyTest & operator << (T const & t)
{
std::cout << t << std::endl;
return *this;
}
};
int main()
{
MyTest test;
test << 10;
return 0;
}
while this one doesn't compile:
#include <iostream>
class MyTest
{
MyTest & operator << (int const & t)
{
std::cout << t << std::endl;
return *this;
}
};
int main()
{
MyTest test;
test << 10;
return 0;
}
I tested it again with vc8.0 and got the same result.
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| |
| | | Guest | |
| | Posted: Thu Sep 04, 2008 10:36 pm Post subject: Re: Can't template operator << be private? | |
On Sep 4, 12:31 pm, lv <lv.jc...@gmail.com> wrote:
| Quote: | On Sep 4, 5:28 pm, Thomas Richter <t...@math.tu-berlin.de> wrote:
For which definition of "accessible"? Could you please post a complete
example that demonstrates the problem?
Well, I should have made it clearer. I meant the following code
compiles ,
[snip] |
Try your code on LINK
You will see that both samples do not compile. It's a bug in visual
studios apparently.
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