临时工和常数 [英] temporaries and const&
问题描述
大家好，请考虑以下功能： -
const int& foo（const double& d）
{
返回d;
}
g ++编译它有警告和solaris CC给出错误。我想
知道代码是否符合标准？
/ P
Hi everyone, please consider the following function:-
const int& foo ( const double& d )
{
return d;
}
g++ compiles it with warnings and solaris CC gives error. I want to
know if the code is correct according to the standard ?
/P
推荐答案
* dragoncoder：
* dragoncoder:
大家好，请考虑以下功能： -
const int& foo（const double& d）
{
返回d;
}
g ++编译它有警告和solaris CC给出错误。我想
知道代码是否符合标准？
Hi everyone, please consider the following function:-
const int& foo ( const double& d )
{
return d;
}
g++ compiles it with warnings and solaris CC gives error. I want to
know if the code is correct according to the standard ?
这在技术上是正确的。
在返回表达式中，int const&引用必然是一个临时的int和
（初始化）。这是用
" double"初始化的，它被隐式转换为int。
然而，使用调用foo的结果（产生未定义的行为，
，因为你正在返回对临时的引用。 C ++就是这样，
默认非常宽松。它高高兴兴地让你自己射击腹股沟，而不是逮捕你[1]，如果这就是你想要的那样。
干杯，
- Alf
注意：
[1]< url：http：//news.bbc.co。 uk / 1 / hi / england / south_yorkshire / 3891311.stm> ;.
-
答：因为它弄乱了人们通常阅读的顺序文字。
问：为什么这么糟糕？
A：热门帖子。
问：什么是最烦人的事情在usenet和电子邮件中？
It''s technically correct.
In the return expression an "int const&" reference is bound to
(initialized with) a temporary "int" that''s initialized with the
"double", which is implicitly converted to int.
However, using the result of a call to foo() yields Undefined Behavior,
because you''re returning a reference to a temporary. C++ is like that,
very permissive by default. It cheerfully lets you shoot yourself in
the groin, instead of arresting you[1], if that''s what you say you want.
Cheers,
- Alf
Notes:
[1] <url: http://news.bbc.co.uk/1/hi/england/south_yorkshire/3891311.stm>.
--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?
dragoncoder写道：
dragoncoder wrote:
大家好，请考虑以下功能： -
const int& foo（const double& d）
{
返回d;
}
g ++编译它有警告和solaris CC给出错误。我想
知道代码是否符合标准？
Hi everyone, please consider the following function:-
const int& foo ( const double& d )
{
return d;
}
g++ compiles it with warnings and solaris CC gives error. I want to
know if the code is correct according to the standard ?
它是正确的，词汇和语法。问题是
整个事情是无用的，因为通过引用返回const
将导致创建一个临时对象，它只能存活
直到结束对''foo''的调用。 IOW函数
返回后临时对象已被销毁。
V
-
请在通过电子邮件回复时删除资金''A'
我没有回复最热门的回复，请不要问
It''s "correct", lexically and syntactically. The problem is that
the whole thing is useless because returning by a reference to const
will cause creation of a temporary object, which will survive only
until the end the call to ''foo''. IOW right after the function
returns the temporary object has already been destroyed.
V
--
Please remove capital ''A''s when replying by e-mail
I do not respond to top-posted replies, please don''t ask
4月30日下午1:54，Alf P. Steinbach < a ... @ start.nowrote：
On Apr 30, 1:54 pm, "Alf P. Steinbach" <a...@start.nowrote:
* dragoncoder：
* dragoncoder:
大家好，请考虑以下函数： -
Hi everyone, please consider the following function:-
const int& foo（const double& d）
{
返回d;
}
const int& foo ( const double& d )
{
return d;
}
g ++用警告编译它，solaris CC给出错误。我想
知道代码是否符合标准？
g++ compiles it with warnings and solaris CC gives error. I want to
know if the code is correct according to the standard ?
这在技术上是正确的。
在返回表达式中，int const&"引用必然是一个临时的int和
（初始化）。这是用
" double"初始化的，它被隐式转换为int。
然而，使用调用foo的结果（产生未定义的行为，
，因为你正在返回对临时的引用。 C ++就是这样，
默认非常宽松。它高高兴兴地让你自己射击腹股沟，而不是逮捕你[1]，如果这就是你想要的那样。
It''s technically correct.
In the return expression an "int const&" reference is bound to
(initialized with) a temporary "int" that''s initialized with the
"double", which is implicitly converted to int.
However, using the result of a call to foo() yields Undefined Behavior,
because you''re returning a reference to a temporary. C++ is like that,
very permissive by default. It cheerfully lets you shoot yourself in
the groin, instead of arresting you[1], if that''s what you say you want.
感谢您的回复。在相同的上下文中，此代码是否调用
未定义的行为？
#include< iostream>
模板<类T1，类T2>
const T1& max（const T1& a，const T2& b）
{
return（a b）？ a：b;
}
int main（）{
int i = 20; double d = 40;
std :: cout<< max（i，d）<< std :: endl;
返回0;
}
Thanks for the response. In the same context, does this code invoke
undefined behaviour ?
#include <iostream>
template <class T1, class T2>
const T1& max ( const T1& a, const T2& b )
{
return ( a b ) ? a : b;
}
int main() {
int i = 20; double d = 40;
std::cout << max ( i, d ) << std::endl;
return 0;
}
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