Cody Smith Cody Smith - 4 years ago 141
C++ Question

Why does string::compare return an int?

Why does

return an int instead of a smaller type like
short
or
char
? My understanding is that this method only returns -1, 0 or 1.

Second part, if I was to design a compare method that compares two objects of type
Foo
and I only wanted to return -1, 0 or 1, would using
short
or
char
generally be a good idea?

EDIT: I've been corrected,
string::compare
does not return -1, 0, or 1, it in fact returns a value >0, <0 or 0. Thanks for keeping me in line guys.

It seems like the answer is roughly, there is no reason to return a type smaller than int because return values are "rvalues" and those "rvalues" don't benefit from being smaller than type int (4 bytes). Also, many people pointed out that the registers of most systems are probably going to be of size int anyway, since these registers are going to be filled whether you give them a 1, 2 or 4 byte value, there is no real advantage to returning a smaller value.

EDIT 2: In fact it looks like there may be extra processing overhead when using smaller datatypes such as alignment, masking, etc. The general consensus is that the smaller datatypes exist to conserve on memory when working with a lot of data, as in the case of an array.

Learned something today, thanks again guys!

Answer Source

First, the specification is that it will return a value less than, equal to or greater than 0, not necessarily -1 or 1. Secondly, return values are rvalues, subject to integral promotion, so there's no point in returning anything smaller.

In C++ (as in C), every expression is either an rvalue or an lvalue. Historically, the terms refer to the fact that lvalues appear on the left of an assignment, where as rvalues can only appear on the right. Today, a simple approximation for non-class types is that an lvalue has an address in memory, an rvalue doesn't. Thus, you cannot take the address of an rvalue, and cv-qualifiers (which condition "access") don't apply. In C++ terms, an rvalue which doesn't have class type is a pure value, not an object. The return value of a function is an rvalue, unless it has reference type. (Non-class types which fit in a register will almost always be returned in a register, for example, rather than in memory.)

For class types, the issues are a bit more complex, due to the fact that you can call member functions on an rvalue. This means that rvalues must in fact have addresses, for the this pointer, and can be cv-qualified, since the cv-qualification plays a role in overload resolution. Finally, C++11 introduces several new distinctions, in order to support rvalue references; these, too, are mainly applicable to class types.

Integral promotion refers to the fact that when integral types smaller than an int are used as rvalues in an expression, in most contexts, they will be promoted to int. So even if I have a variable declared short a, b;, in the expression a + b, both a and b are promoted to int before the addition occurs. Similarly, if I write a < 0, the comparison is done on the value of a, converted to an int. In practice, there are very few cases where this makes a difference, at least on 2's complements machines where integer arithmetic wraps (i.e. all but a very few exotics, today—I think the Unisys mainframes are the only exceptions left). Still, even on the more common machines:

short a = 1;
std::cout << sizeof( a ) << std::endl;
std::cout << sizeof( a + 0 ) << std::endl;

should give different results: the first is the equivalent of sizeof( short ), the second sizeof( int ) (because of integral promotion).

These two issues are formally orthogonal; rvalues and lvalues have nothing to do with integral promotion. Except... integral promotion only applies to rvalues, and most (but not all) of the cases where you would use an rvalue will result in integral promotion. For this reason, there is really no reason to return a numeric value in something smaller than int. There is even a very good reason not to return it as a character type. Overloaded operators, like <<, often behave differently for character types, so you only want to return characters as character types. (You might compare the difference:

char f() { return 'a'; }
std::cout << f() << std::endl;      //  displays "a"
std::cout << f() + 0 << std::endl;  //  displays "97" on my machine

The difference is that in the second case, the addition has caused integral promotion to occur, which results in a different overload of << to be chosen.

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