pointer int/char ...

[bvb]

Hi,
while declaring/defining pointer to int/char, What is doing the
compiler exactly . i'm able to get output(String) for all pointer to
char definition(with out line # 8&14). But when i include line #8 &
line #14 Its only printing lines 10 & 11 & immediatly its giving seg
fault. Please give suggestion to understand the internal things abt
declaration/definition of pointer varibles.

1#include<stdio .h>
2
3 int main()
4 {
5 char *a="linux";
6 char *b="linux";
7 char *c="linux";
8 int *i=4;
9
10 printf("\n a = %s",a);
11 printf("\n b = %s",b);
12 printf("\n c = %s",c);
13
14 printf("\n i=%d",*i);
15
16 return 0;
17 }

NOTE:
Also i'm getting some strange result when just doing pointer to
interger declaration. (Sometimes i'm able to assign/print pointer to
interger variable values EVEN WITHOUT ALLOCATIG MEMORY)
OS : Linux 2.4.20-8 i686 i686 i386 GNU/Linux
Compiler : gcc version 3.2.2 20030222 (Red Hat Linux 3.2.2-5)

ThankX in Advance.......

Urs
Saravanan.....

[Joona I Palaste]

bvb <bm******@faste m.com> scribbled the following:

Hi,
while declaring/defining pointer to int/char, What is doing the
compiler exactly . i'm able to get output(String) for all pointer to
char definition(with out line # 8&14). But when i include line #8 &
line #14 Its only printing lines 10 & 11 & immediatly its giving seg
fault. Please give suggestion to understand the internal things abt
declaration/definition of pointer varibles.
You are causing undefined behaviour, first by assigning an integer
value to a pointer, then by indirecting through that pointer. Your
implementation probably implements this by trying to access address 4
in the virtual memory table, which very likely is in a part of memory
that does not belong to your processs, therefore causing a
segmentation fault.
What is it exactly that you are trying to do? Perhaps you want this:
int i = 4;
int *ip = &i;
1#include<stdio .h>
2
3 int main()
4 {
5 char *a="linux";
6 char *b="linux";
7 char *c="linux";
8 int *i=4;
9
10 printf("\n a = %s",a);
11 printf("\n b = %s",b);
12 printf("\n c = %s",c);
13
14 printf("\n i=%d",*i);
15
16 return 0;
17 } NOTE:
Also i'm getting some strange result when just doing pointer to
interger declaration. (Sometimes i'm able to assign/print pointer to
interger variable values EVEN WITHOUT ALLOCATIG MEMORY)
This is just sheer dumb luck. Undefined behaviour means that anything
can happen: your program might even appear to work.
OS : Linux 2.4.20-8 i686 i686 i386 GNU/Linux
Compiler : gcc version 3.2.2 20030222 (Red Hat Linux 3.2.2-5)
This is not relevant to comp.lang.c.
ThankX in Advance....... Urs
Saravanan.....

I didn't know you were a bear.

--
/-- Joona Palaste (pa*****@cc.hel sinki.fi) ------------- Finland --------\
\-- http://www.helsinki.fi/~palaste --------------------- rules! --------/
"How come even in my fantasies everyone is a jerk?"
- Daria Morgendorfer

[Dan Pop]

In <d5************ **************@ posting.google. com> bm******@fastem .com (bvb) writes:

Hi,
while declaring/defining pointer to int/char, What is doing the
compiler exactly . i'm able to get output(String) for all pointer to
char definition(with out line # 8&14). But when i include line #8 &
line #14 Its only printing lines 10 & 11 & immediatly its giving seg
fault. Please give suggestion to understand the internal things abt
declaration/definition of pointer varibles.

1#include<stdio .h>
2
3 int main()
4 {
5 char *a="linux";
6 char *b="linux";
7 char *c="linux";
8 int *i=4;
Does your compiler silently accept the above line?!? I can't believe
that!
9
10 printf("\n a = %s",a);
11 printf("\n b = %s",b);
12 printf("\n c = %s",c);
13
14 printf("\n i=%d",*i);
15
16 return 0;
17 }

NOTE:
Also i'm getting some strange result when just doing pointer to
interger declaration. (Sometimes i'm able to assign/print pointer to
interger variable values EVEN WITHOUT ALLOCATIG MEMORY)
OS : Linux 2.4.20-8 i686 i686 i386 GNU/Linux
Compiler : gcc version 3.2.2 20030222 (Red Hat Linux 3.2.2-5)

Then, your compiler is issuing the following diagnostic:

test.c:8: warning: initialization makes pointer from integer without a cast

You got exactly what you deserved for ignoring it. If the compiler is
telling you that your code is incorrect, what's the point in executing it,
anyway, without fixing it?

Anyway, the FAQ is explaining how to initialise pointers properly.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Dan Pop]

In <cc**********@o ravannahka.hels inki.fi> Joona I Palaste <pa*****@cc.hel sinki.fi> writes:

bvb <bm******@faste m.com> scribbled the following:

OS : Linux 2.4.20-8 i686 i686 i386 GNU/Linux
Compiler : gcc version 3.2.2 20030222 (Red Hat Linux 3.2.2-5)

This is not relevant to comp.lang.c.

When the OP has no clue about the source of the problem, it is a *good*
idea to mention such details. For all he knows, the code might be correct
but is hitting a bug in the implementation. And we have seen such cases
in the past, related to bugs in Microsoft compilers.

It is sheer stupidity to discourage people from providing such
information.

Dan
--
Dan Pop
DESY Zeuthen, RZ group
Email: Da*****@ifh.de

[Darrell Grainger]

On Fri, 8 Jul 2004, bvb wrote:

Hi,
while declaring/defining pointer to int/char, What is doing the
compiler exactly . i'm able to get output(String) for all pointer to
char definition(with out line # 8&14). But when i include line #8 &
line #14 Its only printing lines 10 & 11 & immediatly its giving seg
fault. Please give suggestion to understand the internal things abt
declaration/definition of pointer varibles.

1#include<stdio .h>
2
3 int main()
4 {
5 char *a="linux";
6 char *b="linux";
7 char *c="linux";
Line 7 the expression "linux" evaluates to the memory address that the
string "linux" will reside at. So if this string ends up at memory
location 1000, this is the same as:

char *c = (char *)1000;
or
char *c;
c = (char *)1000;
8 int *i=4;
Line 8 is the same as:

int *i;
i = 4;

Thus, you have set the pointer to reference memory address 4.
9
10 printf("\n a = %s",a);
11 printf("\n b = %s",b);
12 printf("\n c = %s",c);
13
14 printf("\n i=%d",*i);
This is telling the compiler to printf the integer at memory location 4.
That is, of i == 4 then *i is the integer at memory location 4. Most
likely, the operating system has sensitive information at memory location
4 so it is blocking your program from peeking at this memory location.
15
16 return 0;
17 }

NOTE:
Also i'm getting some strange result when just doing pointer to
interger declaration. (Sometimes i'm able to assign/print pointer to
interger variable values EVEN WITHOUT ALLOCATIG MEMORY)

This is called undefined behaviour. There are things in C where the
behaviour is undefined. This means that it could work on your computer but
not on might. It could work on Tuesdays at 4pm but never on Wednesdays at
3:17am. You just never know. Because of the uncertainty of undefined
behaviour you should avoid it at all costs.

--
Send e-mail to: darrell at cs dot toronto dot edu
Don't send e-mail to vi************@ whitehouse.gov

[Christopher Benson-Manica]

bvb <bm******@faste m.com> spoke thus:

<mode="nitpick" >

3 int main() int main( void ) /* better */ 4 {
5 char *a="linux"; const char *a="linux"; /* a good idea */ 6 char *b="linux"; /* ditto */ 7 char *c="linux"; /* ditto */ 8 int *i=4;
13
14 printf("\n i=%d",*i); /* some implementations require a newline at the end
of output, no reason not to include one */ 15
16 return 0;
17 }

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cybers pace.org | don't, I need to know. Flames welcome.

[Default User]

Joona I Palaste wrote:

You are causing undefined behaviour, first by assigning an integer
value to a pointer,
Chapter and verse on this, please. My understanding is that it's
implementation-defined.
then by indirecting through that pointer.

This is correct.


Brian Rodenborn

[j]

"Darrell Grainger" <da*****@NOMORE SPAMcs.utoronto .ca.com> wrote in message
news:Pi******** *************** ********@drj.pf ...

On Fri, 8 Jul 2004, bvb wrote:

Hi,
while declaring/defining pointer to int/char, What is doing the
compiler exactly . i'm able to get output(String) for all pointer to
char definition(with out line # 8&14). But when i include line #8 &
line #14 Its only printing lines 10 & 11 & immediatly its giving seg
fault. Please give suggestion to understand the internal things abt
declaration/definition of pointer varibles.

1#include<stdio .h>
2
3 int main()
4 {
5 char *a="linux";
6 char *b="linux";
7 char *c="linux";
Line 7 the expression "linux" evaluates to the memory address that the
string "linux" will reside at. So if this string ends up at memory
location 1000, this is the same as:

char *c = (char *)1000;
or
char *c;
c = (char *)1000;

8 int *i=4;

Line 8 is the same as:

int *i;
i = 4;

You are mistaken. The value of ``i'' is initially indeterminate.
You then assign the integer ``4'' to ``i''.

int *i = 4;

Here, ``i'' has an initial value which is not indeterminate.

The two are not the same.

Thus, you have set the pointer to reference memory address 4.

9
10 printf("\n a = %s",a);
11 printf("\n b = %s",b);
12 printf("\n c = %s",c);
13
14 printf("\n i=%d",*i);

This is telling the compiler to printf the integer at memory location 4.
That is, of i == 4 then *i is the integer at memory location 4. Most
likely, the operating system has sensitive information at memory location
4 so it is blocking your program from peeking at this memory location.

15
16 return 0;
17 }

NOTE:
Also i'm getting some strange result when just doing pointer to
interger declaration. (Sometimes i'm able to assign/print pointer to
interger variable values EVEN WITHOUT ALLOCATIG MEMORY)

This is called undefined behaviour. There are things in C where the
behaviour is undefined. This means that it could work on your computer but
not on might. It could work on Tuesdays at 4pm but never on Wednesdays at
3:17am. You just never know. Because of the uncertainty of undefined
behaviour you should avoid it at all costs.

--
Send e-mail to: darrell at cs dot toronto dot edu
Don't send e-mail to vi************@ whitehouse.gov

[Jack Klein]

On Fri, 9 Jul 2004 16:21:18 GMT, Default User
<fi********@boe ing.com.invalid > wrote in comp.lang.c:

Joona I Palaste wrote:

You are causing undefined behaviour, first by assigning an integer
value to a pointer,
Chapter and verse on this, please. My understanding is that it's
implementation-defined.

With a cast, the result is implementation-defined. As written by the
op, which I have replaced since you snipped it:
8 int *i=4;

....it is a constraint violation (6.5.4 paragraph 3 and 6.16.1
paragraph 1). So the compiler is required to emit a diagnostic.

Curiously, the standard neither forbids an implementation from
producing an executable in the presence of a constraint violation, so
long as it issues a diagnostic, nor does it state specifically that
executing a program that contained a constraint violation is undefined
behavior. But it surely is.

then by indirecting through that pointer.

This is correct.

Indeed.

--
Jack Klein
Home: http://JK-Technology.Com
FAQs for
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alt.comp.lang.l earn.c-c++
http://www.contrib.andrew.cmu.edu/~a...FAQ-acllc.html

[Peter Nilsson]

"Default User" <fi********@boe ing.com.invalid > wrote in message
news:40******** *******@boeing. com.invalid...

Joona I Palaste wrote:

[re: int *i=4;]

You are causing undefined behaviour, first by assigning an integer
value to a pointer,

Chapter and verse on this, please. My understanding is that it's
implementation-defined.

The assignment of an uncasted integer to a pointer violates the constraint 6.5.16.1p1.
[Whether it's undefined behaviour is apparently a separate issue.]

The conversion of an int to a pointer via cast is implementation-defined, but subject to
various problems...

then by indirecting through that pointer.

This is correct.

Only if the pointer is a not suitably aligned, is a trap representation, or does not point
to an entity of the referenced type. [cf 6.3.2.3p5]

In the code above, there is no guarantee that the exceptions are excluded.

--
Peter