outputting a NULL pointer to a stream

[Jacek Dziedzic]

Hello!

I have a piece of code that needs to display a formatted
table of pointers using streams, with the pointers represented
as hex values. It looks more or less like this:

#include <iostream>
#include <iomanip>

using namespace std;

// ptr[] is an array of pointers

int main() {
for(unsigned int i=0;i<some_size ;++i) {

cout << hex << setprecision(16 ) << setw(20)
<< (void*)ptr[i] << '\n';

}

}

What annoys me is that if ptr[i] is NULL, then it does
get output as ' 0' instead of
'0x000000000000 0000' as I'd like it to print.

Can anyone suggest a solution to this, other than
checking for NULL manually? Casting to int is not an
option since pointers are 8-byte and ints are 4-byte
on my system.

thanks in advance,
- J.

[mlimber]

Jacek Dziedzic wrote:

Hello!

I have a piece of code that needs to display a formatted
table of pointers using streams, with the pointers represented
as hex values. It looks more or less like this:

#include <iostream>
#include <iomanip>

using namespace std;

// ptr[] is an array of pointers

int main() {
for(unsigned int i=0;i<some_size ;++i) {

cout << hex << setprecision(16 ) << setw(20)
<< (void*)ptr[i] << '\n';

}

}

What annoys me is that if ptr[i] is NULL, then it does
get output as ' 0' instead of
'0x000000000000 0000' as I'd like it to print.

Can anyone suggest a solution to this, other than
checking for NULL manually? Casting to int is not an
option since pointers are 8-byte and ints are 4-byte
on my system.

thanks in advance,
- J.

Use std::cout::fill ( '0' ) or the equivalent I/O manipulator. I'll bet
a pointer to, say, address 0x100 also prints without the zeros (i.e., '
100'). BTW, you omitted the "0x" in your cout
statement.

M

[Jacek Dziedzic]

mlimber wrote:

Use std::cout::fill ( '0' ) or the equivalent I/O manipulator. I'll bet
a pointer to, say, address 0x100 also prints without the zeros (i.e., '
100'). BTW, you omitted the "0x" in your cout
statement.

You've lost your bet :). Since I've used std::hex, the
pointers print out as (for instance)

0x2000000002aff c00
0x2000000007ccd a34
0x2000000009418 afb
0
0x2000000009418 afb
0x2000000009418 afb

etc.

without having to manually specifying "0x" and setting the fill-char
to "0". But not for NULL, alas.

- J.

[Jim Langston]

"Jacek Dziedzic" <jacek@no_spam. tygrys.no_spam. net> wrote in message
news:80******** *************** ***@news.chello .pl...

mlimber wrote:

Use std::cout::fill ( '0' ) or the equivalent I/O manipulator. I'll bet
a pointer to, say, address 0x100 also prints without the zeros (i.e., '
100'). BTW, you omitted the "0x" in your cout
statement.

You've lost your bet :). Since I've used std::hex, the
pointers print out as (for instance)

0x2000000002aff c00
0x2000000007ccd a34
0x2000000009418 afb
0
0x2000000009418 afb
0x2000000009418 afb

etc.

without having to manually specifying "0x" and setting the fill-char
to "0". But not for NULL, alas.

You missed something. Don't you notice the first value is 2? 0x2....
that's why you're not missing any leading zeros, you dont' have any!

So he hasn't lost the bet yet.

[Jacek Dziedzic]

Jim Langston wrote:

You missed something. Don't you notice the first value is 2? 0x2....
that's why you're not missing any leading zeros, you dont' have any!

So he hasn't lost the bet yet.

void *p=(void*)0x100 ;
void *q=(void*)main; // for instance
void *r=NULL;

cout << fixed << setfill('0');
cout << hex << setprecision(16 ) << setw(18) << p << '\n';
cout << hex << setprecision(16 ) << setw(18) << q << '\n';
cout << hex << setprecision(16 ) << setw(18) << r << '\n';

produces

00000000000000x 100
0x4000000000001 990
000000000000000 000

while I'd like to have

0x0000000000000 100
0x4000000000001 990
0x0000000000000 000

.... so while mliber was right about "a pointer to, say, address
0x100 also prints without the zeros" [without setfill('0')]

he missed the 'hex' issue. Therefore, alas, setfill() does not
help me, or I can't see it helping me.

What I can't understand is, why this special value of NULL
would not print with "0x" in front of it like all other
hex values (for pointers).

Will I *really* need to cast this pointer to two (or more)
int values and output them as hex, followed by "0x" which
I'll have to add manually?

- J.

[John Harrison]

Jacek Dziedzic wrote:

Jim Langston wrote:

You missed something. Don't you notice the first value is 2? 0x2....
that's why you're not missing any leading zeros, you dont' have any!

So he hasn't lost the bet yet.

void *p=(void*)0x100 ;
void *q=(void*)main; // for instance
void *r=NULL;

cout << fixed << setfill('0');
cout << hex << setprecision(16 ) << setw(18) << p << '\n';
cout << hex << setprecision(16 ) << setw(18) << q << '\n';
cout << hex << setprecision(16 ) << setw(18) << r << '\n';

produces

00000000000000x 100
0x4000000000001 990
000000000000000 000

while I'd like to have

0x0000000000000 100
0x4000000000001 990
0x0000000000000 000

... so while mliber was right about "a pointer to, say, address
0x100 also prints without the zeros" [without setfill('0')]

he missed the 'hex' issue. Therefore, alas, setfill() does not
help me, or I can't see it helping me.

What I can't understand is, why this special value of NULL
would not print with "0x" in front of it like all other
hex values (for pointers).

Will I *really* need to cast this pointer to two (or more)
int values and output them as hex, followed by "0x" which
I'll have to add manually?

- J.

You might find this helps

cout << hex << setprecision(16 ) << internal << setw(18) << p << '\n';

john

[Jacek Dziedzic]

John Harrison wrote:

You might find this helps

cout << hex << setprecision(16 ) << internal << setw(18) << p << '\n';

It doesn't, unfortunately. AFAIK 'internal' is used if you
want to have the number aligned right and the sign to the left,
correct? Unfortunately "0x" does not want to behave like a
sign.

I browsed the whole section on manipulators in my Josuttis
book and it doesn't say a word on outputting pointers. Perhaps
the implementation is rather free as to the way it outputs
them?

I'm lost now and the prospect of outputting them converted
in sizeof(char) pieces looms.

thanks,
- J.

[Jim Langston]

"Jacek Dziedzic" <jacek@no_spam. tygrys.no_spam. net> wrote in message
news:8b******** *************** ****@news.chell o.pl...

John Harrison wrote:

You might find this helps

cout << hex << setprecision(16 ) << internal << setw(18) << p << '\n';

It doesn't, unfortunately. AFAIK 'internal' is used if you
want to have the number aligned right and the sign to the left,
correct? Unfortunately "0x" does not want to behave like a
sign.

I browsed the whole section on manipulators in my Josuttis
book and it doesn't say a word on outputting pointers. Perhaps
the implementation is rather free as to the way it outputs
them?

I'm lost now and the prospect of outputting them converted
in sizeof(char) pieces looms.

Only thing I could think of would be to somehow print them without the 0x
prefix and add it manually. Don't think there is a flag for that though :/

[Jacek Dziedzic]

Jim Langston wrote:

Only thing I could think of would be to somehow print them without the 0x
prefix and add it manually. Don't think there is a flag for that though :/

I gave up and resorted to:

#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <sstream>

using namespace std;

string ptr_to_str(cons t void* const ptr) {

// represent the pointer as unsigned char[]
unsigned char ptr_representat ion[sizeof(void*)];
memcpy(ptr_repr esentation,&ptr ,sizeof(void*)) ;

string result;
stringstream ss;
ss << "0x" << hex << setfill('0');

// iterate over the array in a little-endian manner
for(int i=sizeof(void*)-1; i>=0; --i) {
unsigned int k = ptr_representat ion[i];
ss << setw(2) << k;
}
ss >> result;
return result;
}

int main() {
void *p=NULL;
void *q=(void*)&main ; // for instance
void *r=(void*)0x123 ; // for instance

cout << ptr_to_str(p) << "\n";
cout << ptr_to_str(q) << "\n";
cout << ptr_to_str(r) << "\n";
}

and this produces

0x0000000000000 000
0x4000000000009 e60
0x0000000000000 123

just as desired, only for little-endian platforms, however.

- J.

[Greg Comeau]

In article <8b************ *************** @news.chello.pl >,
Jacek Dziedzic <jacek@no_spam. tygrys.no_spam. net> wrote:

the implementation is rather free as to the way it outputs
[pointers]?

99.99% sure it is implementation-defined.
--
Greg Comeau / Celebrating 20 years of Comeauity!
Comeau C/C++ ONLINE ==> http://www.comeaucomputing.com/tryitout
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