临时字符串对象的生命周期

Gabor Drasny写道：

Gabor Drasny wrote:

有人能告诉我以下代码是否可以保证有效吗？


不。

#include< string>
#include< iostream>

int main（）
{char /> const char * s = std :: string（" Hello World"）。c_str（）;


这很不寻常。为什么不简单写一下


const char * s =" Hello World";


???

std :: cout<< s<< std :: endl;
返回0;
}

据我了解C ++标准，在main的第一行创建的临时字符串
对象的生命周期（ ）当包含它的表达式结束时结束（除非将对象初始化为对临时的引用）。


是。

如果这是真的，那么指向的内存无效，因为
临时字符串被破坏了。


它是。

我正在开发的项目有很多以上的构造，但是
我无法改变自己所有这些，除非我非常确定它会导致不确定的行为。

Could anyone tell me if the following code is guaranteed to work or not?
Not.
#include <string>
#include <iostream>

int main()
{
const char* s = std::string("Hello World").c_str();
This is rather unusual. Why not simply write

const char* s = "Hello World";

???
std::cout << s << std::endl;
return 0;
}
As I understand the C++ standard the lifetime of the temporary string
object created on the first line of main() ends when the expression
containing it ends (unless an object is initialized as a reference to
the temporary).
Yes.
If that is true, then the memory s points to is invalidated, since the
temporary string is destructed.
It is.
The project I am working on has quite a lot of the above construct, but
I could not bring myself to change all of these, unless I am quite sure
it causes undefined behavior.

确实如此。


V

It does.

V

* Gabor Drasny：

* Gabor Drasny:

大家好，

任何人都可以告诉我，以下代码是否可以保证有效或不是？

#include< string>
#include< iostream>

int main（）
{char /> const char * s = std :: string（" Hello World"）。c_str（）;
std :: cout<< s<< std :: endl;
返回0;
}


不保证可以工作：


*正式你需要还包括< ostream>。 ;-)


*临时std :: string在表达式后不再存在。


我正在进行的项目有很多很多上面的构造，但是我无法让自己改变所有这些，除非我非常确定它会导致不确定的行为。

Hi all,

Could anyone tell me if the following code is guaranteed to work or not?

#include <string>
#include <iostream>

int main()
{
const char* s = std::string("Hello World").c_str();
std::cout << s << std::endl;
return 0;
}
Not guaranteed to work:

* Formally you need to also include <ostream>. ;-)

* The temporary std::string ceases to exist after the expression.

The project I am working on has quite a lot of the above construct, but
I could not bring myself to change all of these, unless I am quite sure
it causes undefined behavior.

这是UB。


但请注意


int main（）

{

std :: string const& str = std :: string（" Hello，world！"）;

const char * s = std.c_str（）;

std :: cout< &LT; s<< std :: endl;

}


不是UB，因为对const的引用保留了临时性，可以这么说。
>
-

答：因为它弄乱了人们通常阅读文字的顺序。

问：为什么这么糟糕？

A：热门发布。

问：usenet和电子邮件中最烦人的事情是什么？

It''s UB.

But note that

int main()
{
std::string const& str = std::string( "Hello, world!" );
const char* s = std.c_str();
std::cout << s << std::endl;
}

is not UB, because that reference to const keeps the temporary, so to speak.

--
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?

内联。


" Alf P. Steinbach" &LT;人*** @ start.no&GT;在消息中写道

news：42 **************** @ news.individual.net ...

Inline.

"Alf P. Steinbach" <al***@start.no> wrote in message
news:42****************@news.individual.net...

* Gabor Drasny ：

* Gabor Drasny:

大家好，

有人可以告诉我，以下代码是否可以保证有效吗？

#include< string>
#include< iostream>


{zh_cn const char * s = std :: string（" Hello World"）。c_str（） ;
std :: cout<< s<< std :: endl;
返回0;
}
不保证可以正常工作：

*正式你还需要包含< ostream>。 ;-)

*临时std :: string在表达式后不再存在。

Hi all,

Could anyone tell me if the following code is guaranteed to work or not?

#include <string>
#include <iostream>

int main()
{
const char* s = std::string("Hello World").c_str();
std::cout << s << std::endl;
return 0;
}
Not guaranteed to work:

* Formally you need to also include <ostream>. ;-)

* The temporary std::string ceases to exist after the expression.

我正在进行的项目有很多很多上面的构造，但是我无法让自己改变所有这些，除非我非常确定它会导致不确定的行为。

The project I am working on has quite a lot of the above construct, but
I could not bring myself to change all of these, unless I am quite sure
it causes undefined behavior.

它' UB。

但请注意

int main（）
{
std :: string const& str = std :: string（" Hello，world！"）;
const char * s = std.c_str（）;
std :: cout<< s<< std :: endl;
}
不是UB，因为对const的引用保留了临时，所以要

It''s UB.

But note that

int main()
{
std::string const& str = std::string( "Hello, world!" );
const char* s = std.c_str();
std::cout << s << std::endl;
}

is not UB, because that reference to const keeps the temporary, so to

说话。


它可以用Java或其他语言用GC实现，但不能用C ++实现，

这仍然是UB。

-
答：因为它弄乱了人们通常阅读文本的顺序。
问：为什么这么糟糕？
答：热门帖子。
问：usenet和电子邮件中最烦人的事情是什么？

speak.
It might do that in Java or some other language with a GC but not in C++,
this is still UB.
--
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?