I have some c++ source where construction via
Type var(init); // compiles and...
Type var = init; // does not
I thought the two are supposed to be completely equivalent? Why does
type promotion seem to take place on the first line, but not the
second? Is there a way it can be enabled for the second?
In this case, init is an enum constant contained within Type. The
compiler is g++ 4.x. Both lines work with VS8, unless you turn
language extensions off, in which case you get the g++ behavior.
(error: conversion from ‘TestChc::Choices’ to non-scalar type ‘Test’
requested). Here's the complete code:
#include <iostream>
#include <string>
using namespace std;
#define TWOCHARINT(A, B) (A+(B<<8)) // FIXME: wrong for big endian
archs
template<typename EnumStrucTstruct EnumChoices: EnumStrucT
{EnumChoices(): EnumStrucT() {} // only works if EnumStrucT
default constructs
EnumChoices(const EnumStrucT& arg): EnumStrucT(arg) {}
EnumChoices& operator=(const EnumStrucT& arg)
{this->value = arg.value; return *this;} // ^ EnumStrucT handles
derivations
bool operator== (const EnumStrucT& arg) const
{return this->value.asScalar==arg.value.asScalar;}
bool operator!= (const EnumStrucT& arg) const
{return this->value.asScalar!=arg.value.asScalar;}
const string asString() const
{return string(EnumStrucT::value.asChars, sizeof
EnumStrucT::value.asChars);}
const uint16_t asScalar() const {return EnumStrucT::value;}};
struct TestChc
{enum Choices {CN = 0, C1 = '1', CB = TWOCHARINT('A', 'B'), CS = '
', CZ = 'Z'};
union Value
{uint16_t asScalar;
char asChars[2];
Value(const Choices& arg): asScalar(arg) {}} value;
TestChc(const Choices& arg):value(arg) {}};
typedef EnumChoices<TestChcTest;
Test& func() {return *(new Test(TestChc::CZ));}
int main (int, char**)
{Test t(TestChc::CB); // , tf('X'); is invalid
// Test t2 = TestChc::CB; // <- this works in VS
Test t2 = func();
// Test t2; // this only works if TestChc default constructs
// Test t2(TestChc::CS);
cout << "size is " << sizeof t2 << endl;
cout << t.asString() << '\t' << t2.asString() << endl;
t = t2;
cout << (t==t2) << endl;
t = TestChc::C1;
cout << (t!=t2) << endl;
return 0;}
There's a little more than needed there, but the crux of the question
is why does the first declaration in main compile while the second (if
uncommented) does not?
Thanks in advance.
Zachary 3 1423
On Sep 17, 7:49*am, Zak <Unceld...@gmail.comwrote:
I have some c++ source where construction via
* * Type var(init); *// compiles and...
* * Type var = init; *// does not
Because the compiler considers only one user defined conversion.
* * template<typename EnumStrucTstruct EnumChoices: EnumStrucT
* * {EnumChoices(): EnumStrucT() {} // only works if EnumStrucT
default constructs
* * *EnumChoices(const EnumStrucT& arg): EnumStrucT(arg) {}
* * *EnumChoices& operator=(const EnumStrucT& arg)
* * *{this->value = arg.value; return *this;} // ^ EnumStrucT handles
derivations
* * struct TestChc
* * {enum Choices {CN = 0, C1 = '1', CB = TWOCHARINT('A', 'B'),CS = '
', CZ = 'Z'};
* * *union Value
* * *{uint16_t asScalar;
* * * char asChars[2];
* * * Value(const Choices& arg): asScalar(arg) {}} value;
* * *TestChc(const Choices& arg):value(arg) {}};
* * // Test t2 = TestChc::CB; // <- this works in VS
Two conversions needed, which the compiler will not consider:
TestChc::Choices -TestChc
TestChc -Test
Ali
P.S. Your code needs lots of vertical white space to be readable.
In article <ga**********@aioe.org>, Andrey Tarasevich
<an**************@hotmail.comwrote:
Zak wrote:
I have some c++ source where construction via
Type var(init); // compiles and...
Type var = init; // does not
I thought the two are supposed to be completely equivalent?
They are not. They are equivalent only in one particular case when the
source and destination types are the same.
Even in that case, the second form can create a temporary if init is not
Type or something derived from it. That is
Type var = init;
can generate either of the following code-wise
Type var( (Type(init)) );
Type var(init);
but must still check for validity as if the first form were used.
See section 8.5 paragraph 14 of the 2003 C++ standard.
On Sep 17, 9:04 pm, blargg....@gishpuppy.com (blargg) wrote:
In article <gare52$75...@aioe.org>, Andrey Tarasevich
<andreytarasev...@hotmail.comwrote:
Zak wrote:
I have some c++ source where construction via
Type var(init); // compiles and...
Type var = init; // does not
I thought the two are supposed to be completely equivalent?
They are not. They are equivalent only in one particular
case when the source and destination types are the same.
Even in that case, the second form can create a temporary if
init is not Type or something derived from it.
In theory, at least.
That is
Type var = init;
can generate either of the following code-wise
Type var( (Type(init)) );
Type var(init);
but must still check for validity as if the first form were used.
Except that if init has type Type, then if the second is legal,
so is the first. In both cases, you're constructing a Type from
another type.
--
James Kanze (GABI Software) email:ja*********@gmail.com
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
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