result type of an operation betweem signed and unsigned, c++

[Andrew Haley]

Dima Sorkin writes:
 > Hi.
 >  The following surprised me a lot: I thought that a subtraction of
 > a 'signed' and 'unsigned' integers has the type "signed",
 > but it seems the opposite:
 > 
 > -------------------------------------------------------
 > // u.cc
 > #include<iostream>
 > 
 > int main(){
 >   int i(0);       unsigned u(1);
 > 
 >   bool isPos( i-u > 0 );   int res(i-u);
 > 
 >   std::cout << i-u << ' ' << isPos << ' ' << res <<'\n';
 > }
 > 
 > > g++ -ansi -pedantic -Wall u.cc
 > > ./a.out
 > 4294967295 1 -1        <----------------- See the first two results
 > ----------------------------------------------------------
 > 
 > 1) Is type of 'signed-unsigned' defined in c++, or it is up to compiler ?
 > (is the above behaviour a failure ?)

No, it's not.  See ISO/IEC 14882:1998(E) Section 5, Expressions,
Para. 9.

 > 2) Is there some way to convert a type of an expression into
 >     _readable_ string  ("typeof"), to debug such things ?
 > 
 > 2b) Is it true that any legal expression has a type, 

Yes.  All will become clear if you read the first part of Section 5.

Andrew.