why "++a" is undefined behaviour ?

**gara.matt@gmail.com** · Jul 16 '07, 06:45 AM

Re: why "++a&qu ot; is undefined behaviour ?

On Jul 15, 11:10 pm, arnuld <geek.arn...@gm ail.comwrote:

my friend sent me this programme:
>
#include<iostre am>
int main()
{
int a=5;
std ::cout << ++a
<< "\t"
<< ++a
<<
>
return 0;
>
}
>
its output should be "6 7" and this i what i get:
>
(arnuld@arch ~ )% g++ -ansi -pedantic -Wall -Wextra new2.cpp
new2.cpp: In function 'int main()':
new2.cpp:10: warning: operation on 'a' may be undefined
(arnuld@arch ~ )% ./a.out
7 7
(arnuld@arch ~ )%
>
to not to relay on "undefined behaviuor" i changed this programme:
>
#include<iostre am>
>
int
main()
{
int a =
5;
>
int b = +
+a;
int c = +
+a;
>
std::cout <<
b
<<
"\t"
<<
c
<<
std::endl;
>
return
0;
>
}
>
and now it runs as expected:
>
(arnuld@arch ~ )% g++ -ansi -pedantic -Wall -Wextra new.cpp
(arnuld@arch ~ )% ./a.out
6 7
(arnuld@arch ~ )%
>
i want to know why the direct use of ++a in std::cout is undefiend
behaviour" ?

It isn't. You can use ++a directly in your cout statement. If you take
a closer look, your first post didn't have "<< endl;" or a semicolon
at all which is what caused it.

**szclark@googlemail.com** · Jul 16 '07, 06:45 AM

Re: why "++a&qu ot; is undefined behaviour ?

On Jul 16, 7:10 am, arnuld <geek.arn...@gm ail.comwrote:

my friend sent me this programme:
>
#include<iostre am>
int main()
{
int a=5;
std ::cout << ++a
<< "\t"
<< ++a
<<
>
return 0;
>
}
>
its output should be "6 7" and this i what i get:
>
(arnuld@arch ~ )% g++ -ansi -pedantic -Wall -Wextra new2.cpp
new2.cpp: In function 'int main()':
new2.cpp:10: warning: operation on 'a' may be undefined
(arnuld@arch ~ )% ./a.out
7 7
(arnuld@arch ~ )%
>
to not to relay on "undefined behaviuor" i changed this programme:
>
#include<iostre am>
>
int
main()
{
int a =
5;
>
int b = +
+a;
int c = +
+a;
>
std::cout <<
b
<<
"\t"
<<
c
<<
std::endl;
>
return
0;
>
}
>
and now it runs as expected:
>
(arnuld@arch ~ )% g++ -ansi -pedantic -Wall -Wextra new.cpp
(arnuld@arch ~ )% ./a.out
6 7
(arnuld@arch ~ )%
>
i want to know why the direct use of ++a in std::cout is undefiend
behaviour" ?

Hi, I think it might have been down to the fact that you did not
include an endl; statement or even just a ;

#include<iostre am>
int main()
{
int a = 5;
std ::cout << ++a << "\t;
std ::cout << ++a;
return 0;
}

This returned 6 7.

Hope that helped.

**arnuld** · Jul 16 '07, 07:05 AM

Re: why "++a&qu ot; is undefined behaviour ?

On Jul 16, 11:37 am, szcl...@googlem ail.com wrote:

Hi, I think it might have been down to the fact that you did not
include an endl; statement or even just a ;
>
#include<iostre am>
int main()
{
int a = 5;
std ::cout << ++a << "\t;
std ::cout << ++a;
return 0;
>
}
>
This returned 6 7.

you introduced an extra statement in the programme for producing the
correct result. why i can not get the same output using one single
statement ?

**szclark@googlemail.com** · Jul 16 '07, 07:05 AM

Re: why "++a&qu ot; is undefined behaviour ?

On 16 Jul, 07:57, arnuld <geek.arn...@gm ail.comwrote:

On Jul 16, 11:37 am, szcl...@googlem ail.com wrote:
Hi, I think it might have been down to the fact that you did not
include an endl; statement or even just a ;

>

#include<iostre am>
int main()
{
int a = 5;
std ::cout << ++a << "\t;
std ::cout << ++a;
return 0;

>

}

>

This returned 6 7.

>
you introduced an extra statement in the programme for producing the
correct result. why i can not get the same output using one single
statement ?

I think this is because the compiler will increment 'a' twice before
the line is output to the screen, so the output will then be 7 7.

**anon** · Jul 16 '07, 08:35 AM

Re: why "++a&qu ot; is undefined behaviour ?

arnuld wrote:

my friend sent me this programme:
>
#include<iostre am>
int main()
{
int a=5;
std ::cout << ++a
<< "\t"
<< ++a
<<
>
return 0;
}

So, this is only one statement:

std::cout << ++a << "\t" << ++a << std::endl;

therefore a is incremented twice before anything is done (ie. something
is printed to standard output).

That means you got the correct output.

I can not back this up with a chapter number from the standard as I do
not have it, but this is how c++ works with prefix increment operator

>
i want to know why the direct use of ++a in std::cout is undefiend
behaviour" ?
>

IMO undefined behavior is something different, not this

**anon** · Jul 16 '07, 08:45 AM

Re: why "++a&qu ot; is undefined behaviour ?

arnuld wrote:

my friend sent me this programme:
>
#include<iostre am>
int main()
{
int a=5;
std ::cout << ++a
<< "\t"
<< ++a
<<
>
return 0;
}
>
its output should be "6 7" and this i what i get:
>
(arnuld@arch ~ )% g++ -ansi -pedantic -Wall -Wextra new2.cpp
new2.cpp: In function 'int main()':
new2.cpp:10: warning: operation on 'a' may be undefined
(arnuld@arch ~ )% ./a.out
7 7
(arnuld@arch ~ )%
>

Basically the rule in C is that if you have an expression, and somewhere
in that expression, you use an increment operator on a variable, you
can't mention that variable elsewhere in the expression.

If we ignore this, then you got correct values (a is incremented twice,
and then printed)

**arnuld** · Jul 16 '07, 09:15 AM

Re: why "++a&qu ot; is undefined behaviour ?

On Jul 16, 1:31 pm, anon <a...@no.nowrot e:

So, this is only one statement:
std::cout << ++a << "\t" << ++a << std::endl;

therefore a is incremented twice before anything is done (ie. something
is printed to standard output).

That means you got the correct output.

and i thought this works like this:

1.) std::cout << ++a << "\t" << ++a << std::endl;

2.) (std::cout << ++a) << "\t" << ++a << std::endl;

3.) (std::cout << ++a) returns std::cout as a result and as a side-
effect it writes the vale of "++a" (== 6) to the the standard output,
which in this case is my terminal. (from C++ Primer 4/e, section
chapter 1, 1.2.2)

4.) (std::cout << "\t") << ++a << std::endl;

same as step 3

5.) (std::cout << ++a) << std::endl;

same as step 3 and this time ++a will be printed (== 7)

6.) std::cout << std::endl;

this will produce a newline and will flush the buffer

where i am wrong ?

I can not back this up with a chapter number from the standard as I do
not have it, but this is how c++ works with prefix increment operator

you can but i trust your information as GCC agrees with you :-)

IMO undefined behavior is something different, not this

ouch!

**anon** · Jul 16 '07, 10:35 AM

Re: why "++a&qu ot; is undefined behaviour ?

arnuld wrote:

>On Jul 16, 1:31 pm, anon <a...@no.nowrot e:

>

>So, this is only one statement:
>std::cout << ++a << "\t" << ++a << std::endl;

>

>therefore a is incremented twice before anything is done (ie. something
>is printed to standard output).

>

>That means you got the correct output.

>
and i thought this works like this:
>
1.) std::cout << ++a << "\t" << ++a << std::endl;
>
2.) (std::cout << ++a) << "\t" << ++a << std::endl;
>
3.) (std::cout << ++a) returns std::cout as a result and as a side-
effect it writes the vale of "++a" (== 6) to the the standard output,
which in this case is my terminal. (from C++ Primer 4/e, section
chapter 1, 1.2.2)
>
>
4.) (std::cout << "\t") << ++a << std::endl;
>
same as step 3
>
5.) (std::cout << ++a) << std::endl;
>
same as step 3 and this time ++a will be printed (== 7)
>
6.) std::cout << std::endl;
>
this will produce a newline and will flush the buffer
>
>
where i am wrong ?
>

Steps 3 and 4 are executed simultaneously, that is is incremented twice,
before it is sent to the standard output

>

>I can not back this up with a chapter number from the standard as I do
>not have it, but this is how c++ works with prefix increment operator

>
you can but i trust your information as GCC agrees with you :-)
>
>

>IMO undefined behavior is something different, not this

>
ouch!
>

I tried to cancel this message, but looks like it did not work ;)

And you get that warning only with -Wall. But I would rather change the
code, then remove that parameter

why "++a" is undefined behaviour ?

why "++a" is undefined behaviour ?

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