Use an undefinied variable

[pierozanetti]

Hi,
I have a question. This statement is accepted by the g++ compiler (v.
4.1.2, under linux):

char c = 'a';
string s = s + c;

But the variable s is used before being defined! Why? It's correct?

The program aborts when the statement is reached.
If we change the above code into:

string s = s + 'a';

the error changes. A segmentation fault is raised.
Can anyone tell me why the code is accepted by the compiler?
Thanks

Piero Zanetti

[Lionel B]

On Fri, 07 Mar 2008 01:16:56 -0800, pierozanetti wrote:

Hi,
I have a question. This statement is accepted by the g++ compiler (v.
4.1.2, under linux):

char c = 'a';
string s = s + c;

But the variable s is used before being defined! Why? It's correct?

It's valid code, but invokes undefined behaviour (actually I was a little
surprised that g++ doesn't emit a warning even with -std=c++98 -pedantic -
Wall).

The program aborts when the statement is reached. If we change the above
code into:

string s = s + 'a';

the error changes. A segmentation fault is raised.

Still undefined behaviour.

Can anyone tell me
why the code is accepted by the compiler?

Because it's valid code.

--
Lionel B

[Lionel B]

On Fri, 07 Mar 2008 02:35:38 -0800, pierozanetti wrote:

On 7 Mar, 11:14, Lionel B <[e-mail]:

>On Fri, 07 Mar 2008 01:16:56 -0800, pierozanetti wrote:

Hi,
I have a question. This statement is accepted by the g++ compiler (v.
4.1.2, under linux):

char c = 'a';
string s = s + c;

But the variable s is used before being defined! Why? It's correct?

It's valid code, but invokes undefined behaviour (actually I was a
little surprised that g++ doesn't emit a warning even with -std=c++98
-pedantic - Wall).

The program aborts when the statement is reached. If we change the
above code into:

string s = s + 'a';

the error changes. A segmentation fault is raised.

Still undefined behaviour.

Can anyone tell me
why the code is accepted by the compiler?

Because it's valid code.

Use an undefined variable is a valid code?!?

Yes. To make it clear:

"valid" == complies with the C++ standard
"valid" != does something sensible

For instance this is valid code:

int main()
{
int x = 0/0;
}

Sure, it's extremely silly code (and my compiler issues a couple of
warnings) but it compiles.

--
Lionel B

[Ron Natalie]

pierozanetti wrote:

Use an undefined variable is a valid code?!?

He means "well formed." That is, the compiler is not required
to detect and issue a diagnositc for it.

[Kai-Uwe Bux]

Lionel B wrote:

On Fri, 07 Mar 2008 02:35:38 -0800, pierozanetti wrote:

[snip]

>Use an undefined variable is a valid code?!?

Yes. To make it clear:

"valid" == complies with the C++ standard
"valid" != does something sensible

For instance this is valid code:

int main()
{
int x = 0/0;
}

Sure, it's extremely silly code (and my compiler issues a couple of
warnings) but it compiles.

I think your example is actually ill-formed according to [5/5]

If during the evaluation of an expression, the result is not
mathematically defined or not in the range of representable values for
its type, the behavior is undefined, unless such an expression is a
constant expression (5.19), in which case the program is ill-formed.

The 0/0 is a constant expression for which the result is not mathematically
defined; and so the program is ill-formed.
Best

Kai-Uwe Bux

[Lionel B]

On Fri, 07 Mar 2008 09:52:13 -0500, Kai-Uwe Bux wrote:

Lionel B wrote:

>On Fri, 07 Mar 2008 02:35:38 -0800, pierozanetti wrote:

[snip]

>>Use an undefined variable is a valid code?!?

Yes. To make it clear:

"valid" == complies with the C++ standard "valid" != does something
sensible

For instance this is valid code:

int main()
{
int x = 0/0;
}

Sure, it's extremely silly code (and my compiler issues a couple of
warnings) but it compiles.

I think your example is actually ill-formed according to [5/5]

If during the evaluation of an expression, the result is not
mathematically defined or not in the range of representable values for
its type, the behavior is undefined, unless such an expression is a
constant expression (5.19), in which case the program is ill-formed.

The 0/0 is a constant expression for which the result is not
mathematically defined; and so the program is ill-formed.

Interesting... in that case shouldn't g++ report an error then rather
than just emit a warning?

$ g++ -std=c++98 main.cpp
main.cpp: In function ‘int main()’:
main.cpp:13: warning: division by zero in ‘0 / 0’
$

--
Lionel B

[Juha Nieminen]

Lionel B wrote:

>string s = s + c;

It's valid code, but invokes undefined behaviour (actually I was a little
surprised that g++ doesn't emit a warning even with -std=c++98 -pedantic -
Wall).

How about this:

struct A { int i; A(): i(7) { std::cout << i << "\n"; } }(A);

Seems to compile and run ok with gcc. Is it valid?

[Lionel B]

On Fri, 07 Mar 2008 18:02:33 +0200, Juha Nieminen wrote:

Lionel B wrote:

>>string s = s + c;

It's valid code, but invokes undefined behaviour (actually I was a
little surprised that g++ doesn't emit a warning even with -std=c++98
-pedantic - Wall).

How about this:

struct A { int i; A(): i(7) { std::cout << i << "\n"; } }(A);

Seems to compile and run ok with gcc. Is it valid?

Perhaps I should clarify that by "valid" I mean conforms to the language
standard; i.e. has no syntax errors, is well-formed, whatever you'd like
to call it. So if a program compiles under a compiler that conforms
perfectly to the standard (probably such a compiler doesn't exist) then
it is by definition valid. Whether it runs "ok" (whatever that means) has
no bearing on its validity.

Your example compiles ok for me, prints out "7" to std output and I can't
see any syntax errors.

What point were you trying to make?

--
Lionel B

[Pete Becker]

On 2008-03-07 10:23:32 -0500, Lionel B <[e-mail]:

>
Interesting... in that case shouldn't g++ report an error then rather
than just emit a warning?

The standard requires "a diagnostic." It doesn't distinguish between a
"warning" and an "error".

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[Default User]

Lionel B wrote:

On Fri, 07 Mar 2008 09:52:13 -0500, Kai-Uwe Bux wrote:

The 0/0 is a constant expression for which the result is not
mathematically defined; and so the program is ill-formed.

Interesting... in that case shouldn't g++ report an error then rather
than just emit a warning?

The standard says nothing about errors or warnings. There are required
diagnostics. The actual output is strictly a QOI issue. A perfectly
comforming implementation could output: "this is a diagnostic" for
every case.

Brian

[James Kanze]

On 7 mar, 17:02, Juha Nieminen <[e-mail]:

Lionel B wrote:

string s = s + c;

It's valid code, but invokes undefined behaviour (actually I was a little
surprised that g++ doesn't emit a warning even with -std=c++98 -pedantic -
Wall).

How about this:

struct A { int i; A(): i(7) { std::cout << i << "\n"; } }(A);

Seems to compile and run ok with gcc. Is it valid?

Yes. What do you think is wrong with it? (In this case,
there's not even any6 undefined behavoir.)

--
James Kanze (GABI Software) email:ja*********@gmail.com
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