lifetime of temp string object

[Gabor Drasny]

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("He llo World").c_str() ;
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).

If that is true, then the memory s points to is invalidated, since the
temporary string is destructed.

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.

Thanks,

Gabor Drasny

[Victor Bazarov]

Gabor Drasny wrote:

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("He llo 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.

It does.

V

[Alf P. Steinbach]

* Gabor Drasny:

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("He llo 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.

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?

[Adrian]

Inline.

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

* Gabor Drasny:

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("He llo 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.

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.
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?

[Mike Wahler]

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

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

Typo! Make that:

const char* s = str.c_str();
-Mike

[Alf P. Steinbach]

* Adrian:

Inline.
What do you mean by that?
* Adrian: "Alf P. Steinbach" <al***@start.no > wrote in message
news:42******** ********@news.i ndividual.net.. .

* Gabor Drasny:

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("He llo 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.

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.

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

You're wrong.

--
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?

[Mike Wahler]

"Adrian" <ba******@xoasi s.com> wrote in message
news:42******** @news.microsoft .com...

Inline.

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

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.

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

C&V please

-Mike

[Alf P. Steinbach]

* Mike Wahler:

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

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

Typo! Make that:

const char* s = str.c_str();

Thanks.

--
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?

[Victor Bazarov]

Alf P. Steinbach wrote:

[...]
But note that

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

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

Yes, but why would anybody do that, when it's SO much easier to do

std::string const str("whatever") ;
const char* s = str.c_str();

(if the string is actually what's needed, I mean)...

:-)

V

[Alf P. Steinbach]

* Victor Bazarov:

Alf P. Steinbach wrote:

[...]
But note that

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

std::cout << s << std::endl;
}

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

Yes, but why would anybody do that, when it's SO much easier to do

std::string const str("whatever") ;
const char* s = str.c_str();

(if the string is actually what's needed, I mean)...

I just thought I should mention it, in case the OPs example code was more
like an impression of the code in question.

:-)

--
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?