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Re: Miscompilation of remainder expressions

Andrew Haley <> writes:

| Roberto Bagnara writes:
|  > Andrew Haley wrote:
|  > > Roberto Bagnara writes:
|  > >  > 
|  > >  > Reading the thread "Autoconf manual's coverage of signed integer
|  > >  > overflow & portability" I was horrified to discover about GCC's
|  > >  > miscompilation of the remainder expression that causes INT_MIN % -1
|  > >  > to cause a SIGFPE on CPUs of the i386 family.  Are there plans to
|  > >  > fix this bug (which, to me, looks quite serious)?
|  > > 
|  > > No, there aren't.  It would make more sense for you to wrap % in some
|  > > code that checks for this, rather than for us to slow down every division
|  > > for this one special case.
|  > 
|  > With all due respect, I must say I am shocked.  I always thought
|  > (and taught) that we, Free Software people, value standard conformance
|  > and getting things right.
| This is a disgreement about interpretation of the langauge in the
| standard, which is:
| "The result of the / operator is the quotient from the division of the
| first operand by the second; the result of the % operator is the
| remainder. In both operations, if the value of the second operand is
| zero, the behavior is undefined. When integers are divided, the result
| of the / operator is the algebraic quotient with any fractional part
| discarded.87) If the quotient a/b is representable, the expression
| (a/b)*b + a%b shall equal a."
| If the quotient a/b is *not* representable, is the behaviour of %
| well-defined or not?  It doesn't say.

Andrew and me exchanged (private) viewpoints of this yesterday.  I would
like to add something I forgot to say.

By definition, the absolute value of "a % b" is always less than the
absolute value of b.  Consequently, "a % b" is always defined.

-- Gaby

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