conversion operator problem

[kowochen]

i encounter a problem as following:

#include <iostream>
#include <string>
using namespace std;

class X
{
public:
X(const string& s, int i):s_(s),i_(i)
{}

operator string ()
{
return s_;
}

operator int ()
{
return i_;
}

private:
string s_;
int i_;
};

int main()
{
string s("abc");
int i(10);

X x("abc", 10);

if (x == i)
{
cout<<"int ok"<<endl;
}

return 0;
}


================================
x can be compared with the integer i and the output is "int ok",
but when i wanna compare with string s, it is a compile error:
if (x == s)
{
...
}

If convert x to string explicitly, it is ok:
if ((string)x == s)
{
...
}

Why couldn't it do a implicit type conversion to string like build in
type?

 

[Victor Bazarov]

i encounter a problem as following:

#include <iostream>
#include <string>
using namespace std;

class X
{
public:
X(const string& s, int i):s_(s),i_(i)
{}

operator string ()
{
return s_;
}

operator int ()
{
return i_;
}

private:
string s_;
int i_;
};

int main()
{
string s("abc");
int i(10);

X x("abc", 10);

if (x == i)
{
cout<<"int ok"<<endl;
}

return 0;
}


================================
x can be compared with the integer i and the output is "int ok",
but when i wanna compare with string s, it is a compile error:
if (x == s)
{
...
}

If convert x to string explicitly, it is ok:
if ((string)x == s)
{
...
}

Why couldn't it do a implicit type conversion to string like build in
type?

Probably because operator== for ints is _built_in_, and the
operator== for strings is a member function of 'std::string'.
Try

if (s == x)

and it should work.

V

 

[Alf P. Steinbach]

* (e-mail address removed):

i encounter a problem as following:

#include <iostream>
#include <string>
using namespace std;

class X
{
public:
X(const string& s, int i):s_(s),i_(i)
{}

operator string ()
{
return s_;
}

operator int ()
{
return i_;
}

private:
string s_;
int i_;
};

int main()
{
string s("abc");
int i(10);

X x("abc", 10);

if (x == i)
{
cout<<"int ok"<<endl;
}

return 0;
}


================================
x can be compared with the integer i and the output is "int ok",
but when i wanna compare with string s, it is a compile error:
if (x == s)
{
...
}

If convert x to string explicitly, it is ok:
if ((string)x == s)
{
...
}

Why couldn't it do a implicit type conversion to string like build in
type?

Comparision with built-in type int uses the built-in "==". This invokes
X:perator int().

Comparision with std::string needs some custom "==". No such is found.

 

[kowochen]

Probably because operator== for ints is _built_in_, and the
operator== for strings is a member function of 'std::string'.
Try

if (s == x)

and it should work.

V
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i have tried this, it also doesn't wrok.

 

[kowochen]

* (e-mail address removed):


















Comparision with built-in type int uses the built-in "==". This invokes
X:perator int().

Comparision with std::string needs some custom "==". No such is found.

--
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A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?- Hide quoted text -

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Do u mean that implicit conversion only can be applied to built-in
"=="?

 

[Bo Persson]

(e-mail address removed) wrote:
::: * (e-mail address removed):
:::
:::
:::
:::
:::
:::: i encounter a problem as following:
:::
:::: #include <iostream>
:::: #include <string>
:::: using namespace std;
:::
:::: class X
:::: {
:::: public:
:::: X(const string& s, int i):s_(s),i_(i)
:::: {}
:::
:::: operator string ()
:::: {
:::: return s_;
:::: }
:::
:::: operator int ()
:::: {
:::: return i_;
:::: }
:::
:::: private:
:::: string s_;
:::: int i_;
:::: };
:::
:::: int main()
:::: {
:::: string s("abc");
:::: int i(10);
:::
:::: X x("abc", 10);
:::
:::: if (x == i)
:::: {
:::: cout<<"int ok"<<endl;
:::: }
:::
:::: return 0;
:::: }
:::
:::: ================================
:::: x can be compared with the integer i and the output is "int ok",
:::: but when i wanna compare with string s, it is a compile error:
:::: if (x == s)
:::: {
:::: ...
:::: }
:::
:::: If convert x to string explicitly, it is ok:
:::: if ((string)x == s)
:::: {
:::: ...
:::: }
:::
:::: Why couldn't it do a implicit type conversion to string like
:::: build in type?
:::
::: Comparision with built-in type int uses the built-in "==". This
::: invokes X:perator int().
:::
::: Comparision with std::string needs some custom "==". No such is
::: found.
:::
::
:: Do u mean that implicit conversion only can be applied to built-in
:: "=="?

Not really.

One difference is that operator== for std::string (there are several)
are templates. None of them can be instantiated for type X, so they
are not considered.

To select an overload, the compiler first selects all possible
candidates, and then tries to choose the best match, only then
considering implicit conversions (and other things). In this case,
there are no candidates!


Bo Persson