getchar() and EOF confusion

[arnuld]

Mostly when I want to take input from stdin I use getchar() but I get this
from man page itself:

"If the integer value returned by getchar() is stored into a variable of
type char and then compared against the integer constant EOF, the
comparison may never succeed, because sign-extension of a variable of type
char on widening to integer is implementation-defined"
while( EOF != (ch = getchar()) ) ....
I use it like that. Can I run into problems with that ?


--
www.lispmachine.wordpress.com
my email is @ the above blog.

[arnuld]

On Tue, 14 Oct 2008 23:39:59 -0700, Keith Thompson wrote:

Note that type char is an integer type. It's important to distinguish
between an integer type (of which there are several, including char,
int, unsigned long, etc.) and the specific integer type called "int".
The name "int" was obviously formed as an abbreviation of the word
"integer", but they mean different things.

Now I am much curious. Whats the different between an "integer" and a
variable of type "int". Do "integer types" are different from "int types"

..SNIP..

You're unlikely to run into this in practice. Machines with this
characteristic are typically DSPs (digital signal processors) which
typically have freestanding C implementations , so stdio.h might not
even be available. But if you want your code to be 100% portable, you
can first check whether the result returned by getchar() is equal to
EOF, and then check whether either feof() or ferror() returns a true
value. In practice, we don't generally bother.

Now I know why some clc lurker told me to distinguish between real end
of file (no more input) and the not so real end of file (error in input)
and suggested me to use feof() and ferror() for that.


--
www.lispmachine.wordpress.com
my email is @ the above blog.
Google Groups is UnBlocked now :)

[Richard Heathfield]

arnuld said:

>On Tue, 14 Oct 2008 23:39:59 -0700, Keith Thompson wrote:

>Note that type char is an integer type. It's important to distinguish
between an integer type (of which there are several, including char,
int, unsigned long, etc.) and the specific integer type called "int".
The name "int" was obviously formed as an abbreviation of the word
"integer", but they mean different things.

Now I am much curious. Whats the different between an "integer" and a
variable of type "int". Do "integer types" are different from "int types"

char is an integer type, but it isn't an int. unsigned char and signed char
are other integer types that are not ints. Furthermore, we can qualify the
bare term "int" with, for example, "short" or "long", resulting in types
that are very similar in behaviour to bare-bones int, but with different
ranges: e.g. int's range is INT_MIN to INT_MAX, but short int's range is
SHRT_MIN to SHRT_MAX and long int's is LONG_MIN to LONG_MAX. And yet these
are integer types. Not ints, but integer types.

The difference isn't just one of ranges and types - it can also be one of
behaviour. The unsigned int type is an integer type that isn't an int. Its
range is 0 to UINT_MAX - but any results outside this range that you try
to store in it are reduced modulo (UINT_MAX + 1) so that they come within
range, and this behaviour is well-defined. This sure beats int's behaviour
when you try to store an out of range result.

--
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

[James Kuyper]

arnuld wrote:

>On Tue, 14 Oct 2008 23:39:59 -0700, Keith Thompson wrote:

>Note that type char is an integer type. It's important to distinguish
between an integer type (of which there are several, including char,
int, unsigned long, etc.) and the specific integer type called "int".
The name "int" was obviously formed as an abbreviation of the word
"integer", but they mean different things.

Now I am much curious. Whats the different between an "integer" and a
variable of type "int". Do "integer types" are different from "int types"

The standard doesn't define any meaning for the phrase "int types". It
does define "integer types". "int" is the name one of one particular
integer type.

Integer types (6.2.5p17):
char

signed integer types (6.2.5p4):
standard signed integer types:
signed char, short int, int, long int, long long int

extended signed integer types (implementation-defined)

unsigned integer types (6.2.5p6):
standard unsigned integer types:
_Bool, and unsigned types corresponding to standard signed
integer types

extended unsigned integer types (implementation-defined)

enumerated types

It's not possible to be specific about the extended integer types. They
are implementation-defined types, such as _int36 for a 36-bit integer
type. In C90, such types were allowed only as an extension to C. This
meant that, in particular, things like size_t that were required to be
integer types could only be typedefs for standard types. In C99, the
concept of "extended integer types" was defined, and size_t is allowed
to refer any unsigned integer type, whether standard or extended.

....

Now I know why some clc lurker told me to distinguish between real end
of file (no more input) and the not so real end of file (error in input)
and suggested me to use feof() and ferror() for that.

EOF is just a macro name; it's clearly named in reference to "End Of
File", but it's also used by the character-oriented I/O functions as a
general-purpose error flag, not exclusively to refer to the end of the file.

[James Kuyper]

arnuld wrote:

>On Tue, 14 Oct 2008 23:39:59 -0700, Keith Thompson wrote:

>Note that type char is an integer type. It's important to distinguish
between an integer type (of which there are several, including char,
int, unsigned long, etc.) and the specific integer type called "int".
The name "int" was obviously formed as an abbreviation of the word
"integer", but they mean different things.

Now I am much curious. Whats the different between an "integer" and a
variable of type "int". Do "integer types" are different from "int types"

The standard doesn't define any meaning for the phrase "int types". It
does define "integer types". "int" is the name one of one particular
integer type.

Integer types (6.2.5p17):
char

signed integer types (6.2.5p4):
standard signed integer types:
signed char, short int, int, long int, long long int

extended signed integer types (implementation-defined)

unsigned integer types (6.2.5p6):
standard unsigned integer types:
_Bool, and unsigned types corresponding to standard signed
integer types

extended unsigned integer types (implementation-defined)

enumerated types

It's not possible to be specific about the extended integer types. They
are implementation-defined types, such as _int36 for a 36-bit integer
type. In C90, such types were allowed only as an extension to C. This
meant that, in particular, things like size_t that were required to be
integer types could only be typedefs for standard types. In C99, the
concept of "extended integer types" was defined, and size_t is allowed
to refer any unsigned integer type, whether standard or extended.

....

Now I know why some clc lurker told me to distinguish between real end
of file (no more input) and the not so real end of file (error in input)
and suggested me to use feof() and ferror() for that.

EOF is just a macro name; it's clearly named in reference to "End Of
File", but it's also used by the character-oriented I/O functions as a
general-purpose error flag, not exclusively to refer to the end of the file.

[Michael]

arnuld wrote:

Mostly when I want to take input from stdin I use getchar() but I get this
from man page itself:

"If the integer value returned by getchar() is stored into a variable of
type char and then compared against the integer constant EOF, the
comparison may never succeed, because sign-extension of a variable of type
char on widening to integer is implementation-defined"
while( EOF != (ch = getchar()) ) ....
I use it like that. Can I run into problems with that ?

the function

int getchar();

reads a byte from the standard input and return it.
If End-of-file is read, it returns EOF (on my machine, it is 0xffffffff)
If ch is an int, there is no problem at all.
A common mistake is assigning getchar() into a char variable.
For example, if ch is a char:

EOF!=(ch=getcha r())

When the byte of 0xff is read:

getchar()=0x000 000ff
ch=0xff

Because EOF is an int, the value of ch is automatically casted to int.

If ch is unsigned, R.H.S of != is 0x000000ff
If ch is signed, R.H.S of != is 0xffffffff which is equal to EOF and
while loop will exit

Therefore, if ch is a char, there will be a problem if the read
character is expanded to EOF (which is implementation-specific) and the
signedness of char (again which is implementation-specific)

[James Kuyper]

Michael wrote:

arnuld wrote:

....

> while( EOF != (ch = getchar()) ) ....

....

If ch is an int, there is no problem at all.

Unless INT_MAX<UCHAR_M AX, which is possible on systems where CHAR_BIT >=
16. On such systems, it's possible for a valid byte, when converted to
'int', to have the same value as EOF. The only work-around for that
possibility is to check feof() and ferror().

[Keith Thompson]

Michael <mi*****@michae ldadmum.no-ip.orgwrites:
[...]

the function

int getchar();

reads a byte from the standard input and return it.
If End-of-file is read, it returns EOF (on my machine, it is 0xffffffff)

[...]

No, EOF cannot be defined as 0xffffffff. It must expand to "an
integer constant expression, with type int and a negative value". A
typical definition is

#define EOF (-1)

If you convert the value of EOF to unsigned int on a 32-bit system,
the result is likely to be 0xffffffff; that's not the value of EOF,
it's the result of the conversion.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
Nokia
"We must do something. This is something. Therefore, we must do this."
-- Antony Jay and Jonathan Lynn, "Yes Minister"

[Michael]

Keith Thompson wrote:

Michael <mi*****@michae ldadmum.no-ip.orgwrites:
[...]

>the function

int getchar();

reads a byte from the standard input and return it.
If End-of-file is read, it returns EOF (on my machine, it is 0xffffffff)

[...]

No, EOF cannot be defined as 0xffffffff. It must expand to "an
integer constant expression, with type int and a negative value". A
typical definition is

#define EOF (-1)

If you convert the value of EOF to unsigned int on a 32-bit system,
the result is likely to be 0xffffffff; that's not the value of EOF,
it's the result of the conversion.

0xffffffff is hexadecimal *is* -1 in decimal on 32-bit int.

[Chris Dollin]

Michael wrote:

Keith Thompson wrote:

>If you convert the value of EOF to unsigned int on a 32-bit system,
the result is likely to be 0xffffffff; that's not the value of EOF,
it's the result of the conversion.

0xffffffff is hexadecimal *is* -1 in decimal on 32-bit int.

Not if it's an /unsigned/ int (see Keith's first sentence above).

--
'It changed the future .. and it changed us.' /Babylon 5/

Hewlett-Packard Limited registered office: Cain Road, Bracknell,
registered no: 690597 England Berks RG12 1HN

[jameskuyper]

Michael wrote:

Keith Thompson wrote:

....

If you convert the value of EOF to unsigned int on a 32-bit system,
the result is likely to be 0xffffffff; that's not the value of EOF,
it's the result of the conversion.

0xffffffff is hexadecimal *is* -1 in decimal on 32-bit int.

Not in C. In C, 0xFFFFFFFF is just a different way of writing the same
value as 2147483647 - the only difference is that 0xFFFFFFFF might
have an unsigned type, while 2147483647 must have a signed type.
0xFFFFFFFF never has the meaning "-1". It can be converted to an int,
and if 'int' is a 32-bit 2's complement type the result of that
conversion will probably be -1, but that doesn't mean that 0xFFFFFFFF
is -1.