type promotion for some constructors but not others

[Zak]

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::Choic es’ 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<typena me EnumStrucTstruc t EnumChoices: EnumStrucT
{EnumChoices(): EnumStrucT() {} // only works if EnumStrucT
default constructs
EnumChoices(con st 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.asS calar;}

bool operator!= (const EnumStrucT& arg) const
{return this->value.asScalar !=arg.value.asS calar;}

const string asString() const
{return string(EnumStru cT::value.asCha rs, sizeof
EnumStrucT::val ue.asChars);}

const uint16_t asScalar() const {return EnumStrucT::val ue;}};

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<Tes tChcTest;

Test& func() {return *(new Test(TestChc::C Z));}

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

[acehreli@gmail.com]

Re: type promotion for some constructors but not others

On Sep 17, 7:49 am, Zak <Unceld...@gmai l.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<typena me EnumStrucTstruc t EnumChoices: EnumStrucT
    {EnumChoices(): EnumStrucT() {} // only works if EnumStrucT
default constructs
     EnumChoices(con st 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::Choice s -TestChc
TestChc -Test

Ali

P.S. Your code needs lots of vertical white space to be readable.

[blargg]

Re: type promotion for some constructors but not others

In article <gare52$75t$1@a ioe.org>, Andrey Tarasevich
<andreytarasevi ch@hotmail.comw rote:

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.

[James Kanze]

Re: type promotion for some constructors but not others

On Sep 17, 9:04 pm, blargg....@gish puppy.com (blargg) wrote:

In article <gare52$75...@a ioe.org>, Andrey Tarasevich

<andreytarasev. ..@hotmail.comw rote:

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:james.kan ze@gmail.com
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