int pointing to char

[Gautam]

this piece of code assigns an int pointer(evident ) to a char,
and when i try to access the ascii value of the char through the
integer pointer(p) ,

what i get is a junk value or not i don't know !
'cause it shows consistent values for diff. chars , for eg.
c = 'a' ans=>-9119

c = 'b' ans=>-9118

c = 'c' ans=>-9117

... and so on
#include <stdio.h>

void main()
{
char c = 'a';
int * p;

p = &c; /*suspicious pointer conversion WARNING !*/
clrscr();
pf("\n%d", *p);
getch();
}

. . . . whats happening ?

Gautam

[James Hu]

On 2004-03-03, Gautam <ga********@yah oo.com> wrote:

this piece of code assigns an int pointer(evident ) to a char,
and when i try to access the ascii value of the char through the
integer pointer(p) ,

what i get is a junk value or not i don't know !
'cause it shows consistent values for diff. chars , for eg.
c = 'a' ans=>-9119

c = 'b' ans=>-9118

c = 'c' ans=>-9117

... and so on
#include <stdio.h>

void main()
{
char c = 'a';
int * p;

p = &c; /*suspicious pointer conversion WARNING !*/
clrscr();
pf("\n%d", *p);
getch();
}
Your compiler seems to be doing its job, according to your comment,
and is warning you that you are doing something wrong. This means
you should fix it. Change the type of c so that p can point to it.

Since you appear to be learning C, you should try to stick to the
standard language features. pf(), clrscr() and getch() are not part of
the standard C language. In standard C in a hosted environment, main
returns an int.

#include <stdio.h>

int main(void)
{
int c = 'a';
int * p;

p = &c;
printf("%d\n", *p);
return 0;
}
. . . . whats happening ?

You are lying to the compiler about what p is pointing to. Assuming
that pf() in your program is not #defined to be an empty statement,
your program results in undefined behavior, so anything could happen.

-- James

[Dan Pop]

In <44************ *************@p osting.google.c om> ga********@yaho o.com (Gautam) writes:

this piece of code assigns an int pointer(evident ) to a char,
Nope, it doesn't.
and when i try to access the ascii value of the char through the
integer pointer(p) ,
An operation devoid of any sense. If you want to access the value of the
char through a pointer, you NEED a pointer to char. No other pointer
type is suitable for the job.
#include <stdio.h>

void main()
So, you didn't bother to read the FAQ *before* posting...
{
char c = 'a';
int * p;

p = &c; /*suspicious pointer conversion WARNING !*/
It's actually a fatal programing error, as far as your program is
concerned. NEVER ignore a compiler warning *before* understanding its
meaning.
clrscr();
What's this nonsense?
pf("\n%d", *p);
What's this nonsense?
getch();
What's this nonsense?
}

. . . . whats happening ?

You're misdefining main(), incorrectly initialising a pointer,
incorrectly using the same pointer and calling a lot functions that
neither your program nor the standard C library defines. What would
you expect to happen?!?

Compare your piece of garbage with the following program that correctly
uses a pointer to figure out the value of 'a':

#include <stdio.h>

int main()
{
char c = 'a';
char *p = &c;

printf("%d\n", *p);
return 0;
}

An equally correct way is by defining c as int and p as pointer to int.

There are cases when you want a pointer to unsigned char to point to the
address of an object of another type, but all the other object pointer
types should point to the address of an object of the corresponding type.

Only experts have a licence to ignore this rule, because they know what
they're doing.

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

[CBFalconer]

Gautam wrote:

this piece of code assigns an int pointer(evident ) to a char,
and when i try to access the ascii value of the char through the
integer pointer(p) ,

what i get is a junk value or not i don't know !
'cause it shows consistent values for diff. chars , for eg.

c = 'a' ans=>-9119
c = 'b' ans=>-9118
c = 'c' ans=>-9117
... and so on

#include <stdio.h>

void main()
Undefined behaviour. Use "int main(void)"
{
char c = 'a';
int * p;

p = &c; /*suspicious pointer conversion WARNING !*/
Well? why didn't you do something about it?
clrscr();
no such function.
pf("\n%d", *p);
no such function.
getch();
no such function.
}

. . . . whats happening ?

Compare to:

#include <stdio.h>
int main(void)
{
int c = 'a';
int *p;

p = &c;
printf("%d\n", *p);
return 0;
}

and compare the resultant values with those you got. That may
give you some clue as to what is going on in your machine even
with all the undefined behavior.

--
Chuck F (cb********@yah oo.com) (cb********@wor ldnet.att.net)
Available for consulting/temporary embedded and systems.
<http://cbfalconer.home .att.net> USE worldnet address!

[gregg]

James Hu wrote:

You are lying to the compiler about what p is pointing to. Assuming
that pf() in your program is not #defined to be an empty statement,
your program results in undefined behavior, so anything could happen.

Is it an undefined behaviour, or result ?

I mean, since c is declared to be a char (ie 1 byte, according to ISO-C)
and p is an int* (pointer to 2, or 4 bytes), the program reaches the
memory and reads 2 (or 4) bytes instead of just one.
This is defined as far as behaviour is concerned.
The result itself being undefined (we don't know what's in second, third
and fourth byte).

I'm a bit confused with terminology here... :-(

[Clark Cox]

In article <40************ ***********@new s.free.fr>,
gregg <gr******@netJU STSAYNOcourrier .com> wrote:

James Hu wrote:

You are lying to the compiler about what p is pointing to. Assuming
that pf() in your program is not #defined to be an empty statement,
your program results in undefined behavior, so anything could happen.
Is it an undefined behaviour, or result ?

It's undefined behavior. What does pf do? We don't know, it's not
defined anywhere. The only way that that line could be well defined is
if you had a line similar to the following in your code:

#define pf(x,y)

Which would replace your line with a blank one.
I mean, since c is declared to be a char (ie 1 byte, according to ISO-C)
and p is an int* (pointer to 2, or 4 bytes),
There's nothing stopping int from being 1 byte, 3 bytes or 20 bytes.
the program reaches the
memory and reads 2 (or 4) bytes instead of just one.
This is defined as far as behaviour is concerned.
No, it isn't. If sizeof(char) != sizeof(int), you're attempting to
read from memory that you haven't allocated. That's undefined behavior.
Your program might crash, it might give you a random value, or anything
else may happen.
The result itself being undefined (we don't know what's in second, third
and fourth byte).
The result *is* undefined ... because the *behavior* is undefined to
begin with.
I'm a bit confused with terminology here... :-(

[Martin Ambuhl]

Gautam wrote:

this piece of code assigns an int pointer(evident ) to a char,
and when i try to access the ascii value of the char through the
integer pointer(p) ,
what i get is a junk value or not i don't know !
'cause it shows consistent values for diff. chars , for eg.
Every once in a while, someone decides to post the worst possible excuse
for a C program. Yours is in that group, though you forgot to leave off
the #include statement.

Here's your "code": #include <stdio.h>

void main()
{
char c = 'a';
int * p;

p = &c; /*suspicious pointer conversion WARNING !*/
clrscr();
pf("\n%d", *p);
getch();
}

And here's something closer to C:

#include <stdio.h>

int main()
{
char c = 'a';
char *p = &c;
printf("%d\n", (int) *p);
return 0;
}

[gregg]

Clark Cox wrote:

No, it isn't. If sizeof(char) != sizeof(int), you're attempting to
read from memory that you haven't allocated. That's undefined behavior.
Your program might crash, it might give you a random value, or anything
else may happen.

Right, i got it now !
thanks

g.

[Old Wolf]

> > this piece of code assigns an int pointer(evident ) to a char,

and when i try to access the ascii value of the char through the
integer pointer(p) ,
what i get is a junk value or not i don't know !

#include <stdio.h>

void main()
{
char c = 'a';
int * p;

p = &c; /*suspicious pointer conversion WARNING !*/
clrscr();
pf("\n%d", *p);
getch();
}

. . . . whats happening ?

You are lying to the compiler about what p is pointing to. Assuming
that pf() in your program is not #defined to be an empty statement,
your program results in undefined behavior, so anything could happen.

Actually, undefined behaviour is introduced by "void main()",
and also by the line "p = &c;" (for example, it could produce
a hardware exception on a device which requires int pointers
to be correctly aligned). It is UB to point a pointer to an
object of an incompatible type.

[James Hu]

On 2004-03-04, Old Wolf <ol*****@inspir e.net.nz> wrote:

> this piece of code assigns an int pointer(evident ) to a char,
> and when i try to access the ascii value of the char through the
> integer pointer(p) ,
> what i get is a junk value or not i don't know !

> #include <stdio.h>
>
> void main()
> {
> char c = 'a';
> int * p;
>
> p = &c; /*suspicious pointer conversion WARNING !*/
> clrscr();
> pf("\n%d", *p);
> getch();
> }

> . . . . whats happening ?

You are lying to the compiler about what p is pointing to. Assuming
that pf() in your program is not #defined to be an empty statement,
your program results in undefined behavior, so anything could happen.

Actually, undefined behaviour is introduced by "void main()",

I generally ignore this, since it may be valid in a freestanding
environment.
and also by the line "p = &c;" (for example, it could produce
a hardware exception on a device which requires int pointers
to be correctly aligned).
What would such hardware do when presented with:

char c = 'a';
void *q = &c;
int *p = q;

?
It is UB to point a pointer to an object of an incompatible type.

I don't believe a constraint violation results in undefined behavior,
but IANACLL (xref: 6.5.16.1p1 and 4p1).

However, even though p holds a valid character pointer value, it holds
an invalid integer pointer value, and therefore indirection yields UB
when applied on p (xref: 6.5.3.2p4).

-- James