va_start problem

[gyan]

Why it is that following program behaving like this:

1 #include <stdlib.h>
2 #include <stdarg.h>
3 #include <stdio.h>
4
5 void comd(char *p,int a, ...);
6 void f(void);
7
8 main(){
9 f();
10 }
11
12 void f(void) {
13 char *p = (char *)malloc(10);
14 comd(p,5,'p','a','d','h','j');
15
16 }
17 void comd(char *p,int a, ...)
18 {
19 int i;
20
21 char *q;
22
23 va_list ap;
24 va_start(ap,a);
25 q=p;
26 for(i=0;i<=a;i++)
27 {
28 *p++ = va_arg(ap,char);
29 }
30 printf("\n p = %s\n",q);
31 va_end(ap);
32 }
33
above program give output as
p = padhj
Now if i change line nos 14 to
comd(p,4,'p','a','d','h','j');
then also output is same?
how how va_start exactly locate startingg point
in ap?

[Kenneth Brody]

gyan wrote:

Why it is that following program behaving like this:

1 #include <stdlib.h>
2 #include <stdarg.h>
3 #include <stdio.h>
4
5 void comd(char *p,int a, ...);
6 void f(void);
7
8 main(){
9 f();
10 }
11
12 void f(void) {
13 char *p = (char *)malloc(10);
14 comd(p,5,'p','a','d','h','j');
15
16 }
17 void comd(char *p,int a, ...)
18 {
19 int i;
20
21 char *q;
22
23 va_list ap;
24 va_start(ap,a);
25 q=p;
26 for(i=0;i<=a;i++)
In this case, you will go from 0 though 5, for a total of _6_ items.
That should be "i < a".
27 {
28 *p++ = va_arg(ap,char);
29 }
30 printf("\n p = %s\n",q);
31 va_end(ap);
32 }
33
above program give output as
p = padhj
Actually, the output included an extra character after the 'j', as
you ran off the end of your list. Apparently, whatever happened to
be there ended up being a non-printable character, and you didn't
see it.
Now if i change line nos 14 to
comd(p,4,'p','a','d','h','j');
then also output is same?
how how va_start exactly locate startingg point
in ap?

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+-------------------------+--------------------+-----------------------------+
| Kenneth J. Brody | www.hvcomputer.com | |
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+-------------------------+--------------------+-----------------------------+
Don't e-mail me at: <mailto:Th*************@gmail.com>

[Anand]

gyan wrote:

Why it is that following program behaving like this:

1 #include <stdlib.h>
2 #include <stdarg.h>
3 #include <stdio.h>
4
5 void comd(char *p,int a, ...);
6 void f(void);
7
8 main(){ int main(void) { // Read FAQ to know why 9 f(); return 0; // Read FAQ to know why 10 }
11
12 void f(void) {
13 char *p = (char *)malloc(10); // need to include stdlib.h // Read FAQ to know why
char *p = malloc(10); // Read FAQ to know why 14 comd(p,5,'p','a','d','h','j');
15
16 }
17 void comd(char *p,int a, ...)
18 {
19 int i;
20
21 char *q;
22
23 va_list ap;
24 va_start(ap,a);
25 q=p;
26 for(i=0;i<=a;i++) for(i=0;i<a;i++) // to add a number of elements..
// you were adding a+1 27 {
28 *p++ = va_arg(ap,char);
29 } *p = '\0'; // null terminate the string 30 printf("\n p = %s\n",q);
31 va_end(ap);
32 }
33
above program give output as
p = padhj
Now if i change line nos 14 to
comd(p,4,'p','a','d','h','j');
then also output is same?
how how va_start exactly locate startingg point
in ap?

[Martin Ambuhl]

gyan wrote:

Why it is that following program behaving like this: [....] how how va_start exactly locate startingg point
in ap?
That is not your problem.

No one knows why your program does as it does; you have no reason to
expect it to run at all.

This line might well cause your program to abort: 28 *p++ = va_arg(ap,char); All the arguments are ints.

This line leads to trying to get more arguments than there are 26 for(i=0;i<=a;i++)
and your failure to terminate the string properly means that 30 printf("\n p = %s\n",q); leads to pssibly disastrous outcomes.

Compare your code to the following, paying attention to the lines marked
with /* mha */:

#include <stdlib.h>
#include <stdarg.h>
#include <stdio.h>

void comd(char *p, int a, ...);
#define BIGENOUGH 1024

int /* mha */ main()
{
char p[BIGENOUGH];
comd(p, 5, 'p', 'a', 'd', 'h', 'j');
comd(p, 4, 'p', 'a', 'd', 'h', 'j');
return 0;
}
void comd(char *p, int a, ...)
{
int i;
char *q;
va_list ap;

printf("\n comd called with a = %d\n", a);
va_start(ap, a);
q = p;
for (i = 0; i < /* mha */ a; i++)
*p++ = va_arg(ap, int); /* mha */
*p = 0; /* mha */
printf("p = %s\n", q);
va_end(ap);
}
comd called with a = 5
p = padhj

comd called with a = 4
p = padh

[OP's code follows, with those obnoxious line numbers] 1 #include <stdlib.h>
2 #include <stdarg.h>
3 #include <stdio.h>
4
5 void comd(char *p,int a, ...);
6 void f(void);
7
8 main(){
9 f();
10 }
11
12 void f(void) {
13 char *p = (char *)malloc(10);
14 comd(p,5,'p','a','d','h','j');
15
16 }
17 void comd(char *p,int a, ...)
18 {
19 int i;
20
21 char *q;
22
23 va_list ap;
24 va_start(ap,a);
25 q=p;
26 for(i=0;i<=a;i++)
27 {
29 }
30 printf("\n p = %s\n",q);
31 va_end(ap);
32 }
33

[Martin Ambuhl]

Anand wrote:

gyan wrote:

1 #include <stdlib.h>

[...] // need to include stdlib.h // Read FAQ to know why

An interesting observation. Since he did include <stdlib.h>, what's
your point?

[lawrence.jones]

gyan <si*****@anz.com> wrote:

28 *p++ = va_arg(ap,char);

In addition to the problems others have noted, arguments to varargs
functions undergo the default argument promotions which means that char
and short get promoted to int and float gets promoted to double, which
means that line should be:

*p++ = va_arg(ap, int);

-Larry Jones

Everybody's a slave to routine. -- Calvin

[Christian Bau]

In article
<d1******************************@localhost.talkab outprogramming.com>,
"gyan" <si*****@anz.com> wrote:

Why it is that following program behaving like this:

1 #include <stdlib.h>
2 #include <stdarg.h>
3 #include <stdio.h>
4
5 void comd(char *p,int a, ...);
6 void f(void);
7
8 main(){
9 f();
10 }
11
12 void f(void) {
13 char *p = (char *)malloc(10);
14 comd(p,5,'p','a','d','h','j');
15
16 }
17 void comd(char *p,int a, ...)
18 {
19 int i;
20
21 char *q;
22
23 va_list ap;
24 va_start(ap,a);
25 q=p;
26 for(i=0;i<=a;i++) *****************

I think you meant
for (i = 0; i < a; i++)
27 {
28 *p++ = va_arg(ap,char);
29 }
At this point, you forgot to store a missing 0 character at the end of
the string.
30 printf("\n p = %s\n",q);
31 va_end(ap);
32 }
33
above program give output as
p = padhj
Now if i change line nos 14 to
comd(p,4,'p','a','d','h','j');
then also output is same?
how how va_start exactly locate startingg point
in ap?

[Martin Ambuhl]

la************@ugs.com wrote:

gyan <si*****@anz.com> wrote:

28 *p++ = va_arg(ap,char);

In addition to the problems others have noted, arguments to varargs\

s/In addition to/Among the problems/
See my post an hour and a half before yours.
functions undergo the default argument promotions which means that char
and short get promoted to int and float gets promoted to double, which
means that line should be:

*p++ = va_arg(ap, int);

[those who know me have no need of my name]

in comp.lang.c i read:

See my post an hour and a half before yours.

ahh, someone repealed propagation effects from the usenet? about time!

--
a signature

[Peter Nilsson]

la************@ugs.com wrote:

gyan <si*****@anz.com> wrote:

28 *p++ = va_arg(ap,char);

In addition to the problems others have noted, arguments to varargs
functions undergo the default argument promotions which means that char
and short get promoted to int and float gets promoted to double, which
means that line should be:

*p++ = va_arg(ap, int);

For a free-standing implementation, it may pay to be safer with...

#if UCHAR_MAX > INT_MAX
*p++ = va_arg(ap, unsigned);
#else
*p++ = va_arg(ap, int);
#endif

--
Peter

[Anand]

Martin Ambuhl wrote:

Anand wrote:

gyan wrote:

1 #include <stdlib.h>

[...]

// need to include stdlib.h // Read FAQ to know why

An interesting observation. Since he did include <stdlib.h>, what's
your point?

Oops.. I missed it (sorry gyan), probably (1%) due to small indentation
and mostly (99%) due to my negligence..

[Anand]

Martin Ambuhl wrote:

Anand wrote:

gyan wrote:

1 #include <stdlib.h>

[...]

// need to include stdlib.h // Read FAQ to know why

An interesting observation. Since he did include <stdlib.h>, what's
your point?

Oops.. I missed it.

I missed it due to small indentation change (1%) and
mostly due to my negligenc (99% .)

sorry gyan.

[Anand]

Martin Ambuhl wrote:

gyan wrote: [...]
This line might well cause your program to abort:

> 28 *p++ = va_arg(ap,char);

All the arguments are ints.

[...]
if the chars are always promoted to int in variadic function then why
there's va_arg(ap,char)?
I mean.. why would it be different from va_arg(ap,int)?
Did I miss something?

[Martin Ambuhl]

Anand wrote:

Martin Ambuhl wrote:

gyan wrote:
[...]

This line might well cause your program to abort:

> 28 *p++ = va_arg(ap,char);

All the arguments are ints.

[...]
if the chars are always promoted to int in variadic function then why
there's va_arg(ap,char)?

Who said that there _is_, in any meaningful sense, 'va_arg(ap,char)'?
va_arg is a macro, so you can make all sorts of nonsensical strings
involving it.
Any function not governed by a prototype (or as an argument in the '...'
part of a prototype) will have its argument expressions converted by the
usual argument conversions. Any argument with rank less than int will
be converted to int if the type is signed or if all the values of that
type can be represented as an int, otherwise to unsigned int. Types of
rank less than int include short, unsigned short, char, unsigned char,
signed char, and _Bool.

Note that C99 retains the float->double conversion here even though it
doesn't in the usual unitary conversions.
I mean.. why would it be different from va_arg(ap,int)?
unless sizeof(int) == sizeof(char), there is an obvious difference. The
other one is that 'va_arg(ap, char)' has no coherent meaning, since the
argument is never a char.
Did I miss something?

Or else added something.

[Anand]

Martin Ambuhl wrote:

Anand wrote:

Martin Ambuhl wrote:

gyan wrote:

[...]

This line might well cause your program to abort:
> 28 *p++ = va_arg(ap,char);
All the arguments are ints.

[...]
if the chars are always promoted to int in variadic function then why
there's va_arg(ap,char)?

Who said that there _is_, in any meaningful sense, 'va_arg(ap,char)'?

[...]
Bad wordings from my side :-(

I wanted to know the behavior/usage of va_arg(ap, char), if at all?
I got the answer now " *NO* meaningful usage."
Thanks for clarifying
I knew that integer promotions in variadic functions.. But somehow I
thought va_arg would handle this.
To add to my confusion my compiler handled the smaller-than-int properly
sunch that if I write va_arg(ap, char) it would work (by making
something like to va_arg(ap, int)).

[Old Wolf]

la************@ugs.com wrote:

gyan <si*****@anz.com> wrote:

14 comd(p,5,'p','a','d','h','j');

17 void comd(char *p,int a, ...)
23 va_list ap;
24 va_start(ap,a);
28 *p++ = va_arg(ap,char);

In addition to the problems others have noted, arguments to varargs
functions undergo the default argument promotions which means that
char and short get promoted to int and float gets promoted to double,
which means that line should be:

*p++ = va_arg(ap, int);

True, but the arguments were all ints in the first place, so no
promotion occurs in this particular example.

[gyan]

Hi all,
thanks for your replies.
I got, what was wrong with my program.
But, no body pointed out how va_start locate starting point of variable
argument list?
in my program, does it depends on value of a or *p.
Or it is like that va_arg will always fetch 'p' first time and then carry
on to 2nd argument and so on?

[those who know me have no need of my name]

in comp.lang.c i read:

But, no body pointed out how va_start locate starting point of variable
argument list?

you tell it.

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