hi.
i wanted to know why doesn't the scanf functions check for overflow
when reading number. For example scanf("%d" on 32bit machine considers
"1" and "4294967297 " to be the same.
I tracked to code to where the conversion itself happens. Code in
scanfs just ignores return value from conversion procedures.
More info in case of glibc posted here: http://board.flatassembler.net/topic.php?t=6359
AFAIK, implementation doesn't define behavior in case of overflow, so
glibc could consider this error and return errno=ERANGE 26  9527 
In article <11************ **********@79g2 000cws.googlegr oups.com>, vi****@gmail.co m <vi****@gmail.c omwrote:
>i wanted to know why doesn't the scanf functions check for overflow
when reading number. For example scanf("%d" on 32bit machine considers
"1" and "4294967297 " to be the same.
Because that's how it is spec'd.
"An input item is defined as the longest matching sequence of
characters, unless that exceeds a specified field width, in
which case it is the initial subsequence of that length in
the sequence." [...]
"Except in the case of a % specifier, the input item (or, in the
case of a %n directive, the count of input characters) is
converted to a type appropriate for the conversion specifier. [...]
Unless assignment suppression was indicated by a *, the result
of the conversion is placed in the object pointed to by the first
argument following the format argument that has not
already received a conversion result. If this object does not
have an appropriate type, or if the result of the conversion cannot
be represented in the space provided, the behaviour is undefined."
So there you have it: if you didn't put in a field width, then
the %d is *required* to pull in all the decimal digits there, and
if that's too big for an int, then the result is officially undefined.
This is how fscanf (and hence scanf) are -required- to work according
to the standard.
--
I was very young in those days, but I was also rather dim.
-- Christopher Priest
2006-12-15 <el**********@c anopus.cc.umani toba.ca>,
Walter Roberson wrote:
In article <11************ **********@79g2 000cws.googlegr oups.com>,
vi****@gmail.co m <vi****@gmail.c omwrote:
>>i wanted to know why doesn't the scanf functions check for overflow
when reading number. For example scanf("%d" on 32bit machine considers
"1" and "4294967297 " to be the same.
Because that's how it is spec'd.
"An input item is defined as the longest matching sequence of
characters,
And in what way is "429496729" a matching sequence of characters, if
there is no such integer value?
unless that exceeds a specified field width, in
which case it is the initial subsequence of that length in
the sequence." [...]
"Except in the case of a % specifier, the input item (or, in the
case of a %n directive, the count of input characters) is
converted to a type appropriate for the conversion specifier. [...]
Unless assignment suppression was indicated by a *, the result
of the conversion is placed in the object pointed to by the first
argument following the format argument that has not
already received a conversion result. If this object does not
have an appropriate type, or if the result of the conversion cannot
be represented in the space provided, the behaviour is undefined."
It's undefined. Which means there _are_ no requirements. An
implementation is free to treat it as 1, or as 429496729 with 7 still on
the stream, or as such with 7 _not_ still on the stream, or as
4294967295 (saturation), etc, etc
Anyway, I found a possible situation in which my scanf is
non-conformant:
Numerical strings are truncated to 512 characters; for example, %f
and %d are implicitly %512f and %512d.
So, if I send %f
1.0000000000000 000000000000000 000000000000000 000000000000000 000
000000000000000 000000000000000 000000000000000 000000000000000 000
000000000000000 000000000000000 000000000000000 000000000000000 000
000000000000000 000000000000000 000000000000000 000000000000000 000
000000000000000 000000000000000 000000000000000 000000000000000 000
000000000000000 000000000000000 000000000000000 000000000000000 000
000000000000000 000000000000000 000000000000000 000000000000000 000
000000000000000 000000000000000 000000000000000 000000000000000 000
e1
it converts to 1 instead of 10. Does the standard allow this?
Walter Roberson a écrit :
In article <11************ **********@79g2 000cws.googlegr oups.com>,
vi****@gmail.co m <vi****@gmail.c omwrote:
>>i wanted to know why doesn't the scanf functions check for overflow
when reading number. For example scanf("%d" on 32bit machine considers
"1" and "4294967297 " to be the same.
Because that's how it is spec'd.
"An input item is defined as the longest matching sequence of
characters, unless that exceeds a specified field width, in
which case it is the initial subsequence of that length in
the sequence." [...]
"Except in the case of a % specifier, the input item (or, in the
case of a %n directive, the count of input characters) is
converted to a type appropriate for the conversion specifier. [...]
Unless assignment suppression was indicated by a *, the result
of the conversion is placed in the object pointed to by the first
argument following the format argument that has not
already received a conversion result. If this object does not
have an appropriate type, or if the result of the conversion cannot
be represented in the space provided, the behaviour is undefined."
So there you have it: if you didn't put in a field width, then
the %d is *required* to pull in all the decimal digits there, and
if that's too big for an int, then the result is officially undefined.
This is how fscanf (and hence scanf) are -required- to work according
to the standard.
In general functions like scanf are unusable. They are so
problematic, that it is a wonder when they work at all.
Use strtol, or a similar function that will give reasonable
error returns...
In article <sl************ *******@rlaptop .random.yi.org> ,
Random832 <ra*******@gmai l.comwrote:
>2006-12-15 <el**********@c anopus.cc.umani toba.ca>,
Walter Roberson wrote:
>In article <11************ **********@79g2 000cws.googlegr oups.com>,
vi****@gmail.co m <vi****@gmail.c omwrote:
>>>i wanted to know why doesn't the scanf functions check for overflow
>"An input item is defined as the longest matching sequence of
characters,
>And in what way is "429496729" a matching sequence of characters, if
there is no such integer value?
The match is based upon the lexical grammar, and the lexical
grammar does not put limitations on the number or content of the
decimal digits.
--
Okay, buzzwords only. Two syllables, tops. -- Laurie Anderson
2006-12-15 <el**********@c anopus.cc.umani toba.ca>,
Walter Roberson wrote:
In article <sl************ *******@rlaptop .random.yi.org> ,
Random832 <ra*******@gmai l.comwrote:
>>2006-12-15 <el**********@c anopus.cc.umani toba.ca>,
Walter Roberson wrote:
>>In article <11************ **********@79g2 000cws.googlegr oups.com>,
vi****@gmail.co m <vi****@gmail.c omwrote:
>>>>i wanted to know why doesn't the scanf functions check for overflow
>>"An input item is defined as the longest matching sequence of
characters,
>>And in what way is "429496729" a matching sequence of characters, if
there is no such integer value?
The match is based upon the lexical grammar, and the lexical
grammar does not put limitations on the number or content of the
decimal digits.
OK. The rest of my post stands. undefined is undefined, it's not
"required" to do anything in such a case.
so, we agree, it's undefined.
wouldn't it be better to return this overflow as error? 10 digits would
be read off the file/stream/whatever, and function will return as if
number format was invalid, with errno=ERANGE.
i don't think that current behavior is what people await. and scanf
functions are doing lot of "smart" stuff already, just because people
await such behavior.
In article <sl************ *******@rlaptop .random.yi.org> ,
Random832 <ra*******@gmai l.comwrote:
>"Except in the case of a % specifier, the input item (or, in the
case of a %n directive, the count of input characters) is
converted to a type appropriate for the conversion specifier. [...]
Unless assignment suppression was indicated by a *, the result
of the conversion is placed in the object pointed to by the first
argument following the format argument that has not
already received a conversion result. If this object does not
have an appropriate type, or if the result of the conversion cannot
be represented in the space provided, the behaviour is undefined."
>It's undefined. Which means there _are_ no requirements. An
implementati on is free to treat it as 1, or as 429496729 with 7 still on
the stream, or as such with 7 _not_ still on the stream, or as
4294967295 (saturation), etc, etc
No, consumption of the maximum characters is -required-. It cannot
leave the other characters in the stream. The undefined part comes
in the valuation and storage of the overly-long result, not in
how many characters are consumed from input.
--
All is vanity. -- Ecclesiastes
Walter Roberson wrote:
In article <sl************ *******@rlaptop .random.yi.org> ,
Random832 <ra*******@gmai l.comwrote:
>>
It's undefined. Which means there _are_ no requirements. An
implementati on is free to treat it as 1, or as 429496729 with 7 still on
the stream, or as such with 7 _not_ still on the stream, or as
4294967295 (saturation), etc, etc
No, consumption of the maximum characters is -required-. It cannot
leave the other characters in the stream. The undefined part comes
in the valuation and storage of the overly-long result, not in
how many characters are consumed from input.
Once undefined behavior strikes, the program has no way
to tell how many characters were or were not consumed. All
requirements lose their force in the face of U.B.
--
Eric Sosman es*****@acm-dot-org.invalid
2006-12-15 <el**********@c anopus.cc.umani toba.ca>,
Walter Roberson wrote:
In article <sl************ *******@rlaptop .random.yi.org> ,
Random832 <ra*******@gmai l.comwrote:
>>"Except in the case of a % specifier, the input item (or, in the
case of a %n directive, the count of input characters) is
converted to a type appropriate for the conversion specifier. [...]
Unless assignment suppression was indicated by a *, the result
of the conversion is placed in the object pointed to by the first
argument following the format argument that has not
already received a conversion result. If this object does not
have an appropriate type, or if the result of the conversion cannot
be represented in the space provided, the behaviour is undefined."
>>It's undefined. Which means there _are_ no requirements. An
implementatio n is free to treat it as 1, or as 429496729 with 7 still on
the stream, or as such with 7 _not_ still on the stream, or as
4294967295 (saturation), etc, etc
No, consumption of the maximum characters is -required-. It cannot
leave the other characters in the stream. The undefined part comes
in the valuation and storage of the overly-long result, not in
how many characters are consumed from input.
No, I don't think you get it.
In an undefined situation, the standard forbids nothing.
Meaning the implementation gets to do whatever the f*** it wants to,
regarding anything, once anything has happened that has been undefined. This thread has been closed and replies have been disabled. Please start a new discussion. Similar topics | |
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