printf and cout

**Richard Heathfield** · Mar 26 '08, 08:55 AM

Re: printf and cout

laikon said:

Hello, everyone:
>
this is about overflow in C and C++.

The behaviour of a C program on overflow is undefined.

int c = 400;
printf("%c", c);
>
it print ? on screen,

Not necessarily. C doesn't guarantee you a screen, and it says very little
about the character set supplied by your implementation.

and ascii of '?' is 63.

This is irrelevant, I think. If by some chance you'd seen the character
with code point 144 (which is, of course, *not* an ASCII character), I'd
have been able to explain it in concrete terms. I am a little puzzled by
the 63, actually. Try dumping the character to a file and examining a hex
dump thereof.

but
cout << int(char(400));
>
it print -112 on screen.

You might want to ask that in comp.lang.c++

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

**Bernhard Schauer** · Mar 26 '08, 09:05 AM

Re: printf and cout

int c = 400;

printf("%c", c);
>
it print ? on screen, and ascii of '?' is 63.
>
but
cout << int(char(400));
>
it print -112 on screen.
>
so, my question is why comes 63 and -112, what relations between them,
why printf and cout behavior so differently.

Hi!

both values are the same, except that the '?' is not meant as the
character '?' but as an unprintable character. It also "looks" different
than a normal '?'.

regards

--
Bernhard Schauer
schauer_at_crux y_dot_net

**laikon** · Mar 26 '08, 09:25 AM

Re: printf and cout

On Mar 26, 1:51 am, Richard Heathfield <r...@see.sig.i nvalidwrote:

laikon said:
>

Hello, everyone:

>

this is about overflow in C and C++.

>
The behaviour of a C program on overflow is undefined.
>

int c = 400;
printf("%c", c);

>

it print ? on screen,

>
Not necessarily. C doesn't guarantee you a screen, and it says very little
about the character set supplied by your implementation.
>

and ascii of '?' is 63.

>
This is irrelevant, I think. If by some chance you'd seen the character
with code point 144 (which is, of course, *not* an ASCII character), I'd
have been able to explain it in concrete terms. I am a little puzzled by
the 63, actually. Try dumping the character to a file and examining a hex
dump thereof.
>

but
cout << int(char(400));

>

it print -112 on screen.

>
You might want to ask that in comp.lang.c++
>
--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

thanks very much.

I just know:

the printed '?' does not mean to be the result of the overflow, but a
symbol of unknown ascii character. when a number overflows by char,
and the left binary is negate, the printf will print '?'.

**Richard Heathfield** · Mar 26 '08, 10:05 AM

Re: printf and cout

laikon said:

<snip>

I just know:
>
the printed '?' does not mean to be the result of the overflow, but a
symbol of unknown ascii character.

There aren't any unknown ASCII characters. If the printed '?' /is/ an ASCII
character, then it has code point 63. If it hasn't, then it isn't an ASCII
character.

when a number overflows by char,
and the left binary is negate, the printf will print '?'.

No, printf offers no such guarantee.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

**santosh** · Mar 26 '08, 10:15 AM

Re: printf and cout

laikon wrote:

Hello, everyone:
>
this is about overflow in C and C++.
>
int c = 400;
printf("%c", c);
>
it print ? on screen, and ascii of '?' is 63.

No, actually it prints a ? character, which is not defined either in the
basic C character set or in ASCII.

but
cout << int(char(400));
>
it print -112 on screen.

Try this printf statement:

printf("c = %d\nc (cast to unsigned char) = %d\t%c\n",
c, (unsigned char)c, c);

so, my question is why comes 63 and -112, what relations between them,
why printf and cout behavior so differently.

Actually they don't behave differently at all. In each case the int
value 400 is being interpreted as an unsigned char. The result of this
conversion gives the value 144, which is not a valid ASCII code.
However all modern systems have one or more "extended" ASCII code pages
active and on your system the value 144 happens to map to the ?
character.

<snip>

**Richard Heathfield** · Mar 26 '08, 10:35 AM

Re: printf and cout

santosh said:

laikon wrote:
>

<snip>

>>
>int c = 400;
>printf("%c", c);
>>
>it print ? on screen, and ascii of '?' is 63.

>
No, actually it prints a ? character, which is not defined either in the
basic C character set or in ASCII.

The '?' question mark *is* in fact part of the basic C character set. It is
used in trigraphs, and as part of the ? : conditional operator.

<snip>

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

**santosh** · Mar 26 '08, 10:35 AM

Re: printf and cout

santosh wrote:

laikon wrote:
>

>Hello, everyone:
>>
>this is about overflow in C and C++.
>>
>int c = 400;
>printf("%c", c);
>>
>it print ? on screen, and ascii of '?' is 63.

>
No, actually it prints a ? character, which is not defined either in
the basic C character set or in ASCII.
>

>but
>cout << int(char(400));
>>
>it print -112 on screen.

>
Try this printf statement:
>
printf("c = %d\nc (cast to unsigned char) = %d\t%c\n",
c, (unsigned char)c, c);
>

>so, my question is why comes 63 and -112, what relations between
>them, why printf and cout behavior so differently.

>
Actually they don't behave differently at all. In each case the int
value 400 is being interpreted as an unsigned char. The result of this
conversion gives the value 144, which is not a valid ASCII code.
However all modern systems have one or more "extended" ASCII code
pages active and on your system the value 144 happens to map to the ?
character.
>
<snip>

Actually it is used as a placeholder by your system to indicate that an
appropriate conversion could not be done.

**Bernhard Schauer** · Mar 26 '08, 10:35 AM

Re: printf and cout

Richard Heathfield wrote:

There aren't any unknown ASCII characters. If the printed '?' /is/ an
ASCII character, then it has code point 63. If it hasn't, then it isn't an
ASCII character.

Thats ok. But if there is a number as character printed (depending on the
locale, thus not guaranteed to be ASCII) that is "unprintabl e" (control
codes, ...) a special character may be inserted by the terminal - it looks
like a question mark (it may also be one).

On my computer a white question mark on a black background is printed.

regards,

--
Bernhard Schauer
schauer_at_crux y_dot_net

**Bernhard Schauer** · Mar 26 '08, 10:35 AM

Re: printf and cout

santosh wrote:

<snip>

Actually they don't behave differently at all. In each case the int
value 400 is being interpreted as an unsigned char.

<snip>

Code from laikon:

int c = 400;
printf("%c", c);
cout << int(char(400));

Where exactly do you see unsigned characters? (char != unsigned char)

regards,

--
Bernhard Schauer
schauer_at_crux y_dot_net

**santosh** · Mar 26 '08, 10:35 AM

Re: printf and cout

Richard Heathfield wrote:

santosh said:

>laikon wrote:
>>

<snip>

>>>
>>int c = 400;
>>printf("%c" , c);
>>>
>>it print ? on screen, and ascii of '?' is 63.

>>
>No, actually it prints a ? character, which is not defined either in
>the basic C character set or in ASCII.

>
The '?' question mark *is* in fact part of the basic C character set.
It is used in trigraphs, and as part of the ? : conditional operator.
>
<snip>

Well it appears as a question mark on Usenet but it's actually a symbol
(a white question mark inside a black diamond) that corresponds to the
octal sequence '\357' '\277' '\275', used to represent a character code
value that is unknown in the encoding used.

This is for my system. It would probably be different under other
systems.

**santosh** · Mar 26 '08, 11:25 AM

Re: printf and cout

Bernhard Schauer wrote:

santosh wrote:
>
<snip>

>Actually they don't behave differently at all. In each case the int
>value 400 is being interpreted as an unsigned char.

<snip>
>
Code from laikon:

>int c = 400;
>printf("%c", c);
>cout << int(char(400));

>
Where exactly do you see unsigned characters? (char != unsigned char)

The %c format expects (and thus treats the corresponding argument as) an
unsigned char.

**Richard Heathfield** · Mar 26 '08, 11:25 AM

Re: printf and cout

Bernhard Schauer said:

santosh wrote:
>
<snip>

>Actually they don't behave differently at all. In each case the int
>value 400 is being interpreted as an unsigned char.

<snip>
>
Code from laikon:

>int c = 400;
>printf("%c", c);
>cout << int(char(400));

>
Where exactly do you see unsigned characters? (char != unsigned char)

When printf is handed a %c, it converts the corresponding int argument to
an unsigned char. See 4.9.6.1 of C89 or 7.19.6.1(8) of C99.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

**Bernhard Schauer** · Mar 26 '08, 11:25 AM

Re: printf and cout

When printf is handed a %c, it converts the corresponding int argument to

an unsigned char. See 4.9.6.1 of C89 or 7.19.6.1(8) of C99.

Good to know ;-)

--
Bernhard Schauer
schauer_at_crux y_dot_net

**Bernhard Schauer** · Mar 26 '08, 11:25 AM

Re: printf and cout

><snip>

>>Actually they don't behave differently at all. In each case the int
>>value 400 is being interpreted as an unsigned char.

><snip>
>>
>Code from laikon:

>>int c = 400;
>>printf("%c" , c);
>>cout << int(char(400));

>>
>Where exactly do you see unsigned characters? (char != unsigned char)

>
The %c format expects (and thus treats the corresponding argument as) an
unsigned char.

Ok. But I hope char(400) does not convert to an unsigned char?

--
Bernhard Schauer
schauer_at_crux y_dot_net

printf and cout

printf and cout

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