(-2147483648> 0) returns true in C++?

benyl picture benyl · Feb 4, 2013 · Viewed 26.6k times · Source

-2147483648 is the smallest integer for integer type with 32 bits, but it seems that it will overflow in the if(...) sentence:

if (-2147483648 > 0)
    std::cout << "true";
else
    std::cout << "false";

This will print true in my testing. However, if we cast -2147483648 to integer, the result will be different:

if (int(-2147483648) > 0)
    std::cout << "true";
else
    std::cout << "false";

This will print false.

I'm confused. Can anyone give an explanation on this?


Update 02-05-2012:

Thanks for your comments, in my compiler, the size of int is 4 bytes. I'm using VC for some simple testing. I've changed the description in my question.

That's a lot of very good replys in this post, AndreyT gave a very detailed explanation on how the compiler will behaviour on such input, and how this minimum integer was implemented. qPCR4vir on the other hand gave some related "curiosities" and how integer are represented. So impressive!

Answer

AnT picture AnT · Feb 4, 2013

-2147483648 is not a "number". C++ language does not support negative literal values.

-2147483648 is actually an expression: a positive literal value 2147483648 with unary - operator in front of it. Value 2147483648 is apparently too large for the positive side of int range on your platform. If type long int had greater range on your platform, the compiler would have to automatically assume that 2147483648 has long int type. (In C++11 the compiler would also have to consider long long int type.) This would make the compiler to evaluate -2147483648 in the domain of larger type and the result would be negative, as one would expect.

However, apparently in your case the range of long int is the same as range of int, and in general there's no integer type with greater range than int on your platform. This formally means that positive constant 2147483648 overflows all available signed integer types, which in turn means that the behavior of your program is undefined. (It is a bit strange that the language specification opts for undefined behavior in such cases, instead of requiring a diagnostic message, but that's the way it is.)

In practice, taking into account that the behavior is undefined, 2147483648 might get interpreted as some implementation-dependent negative value which happens to turn positive after having unary - applied to it. Alternatively, some implementations might decide to attempt using unsigned types to represent the value (for example, in C89/90 compilers were required to use unsigned long int, but not in C99 or C++). Implementations are allowed to do anything, since the behavior is undefined anyway.

As a side note, this is the reason why constants like INT_MIN are typically defined as

#define INT_MIN (-2147483647 - 1)

instead of the seemingly more straightforward

#define INT_MIN -2147483648

The latter would not work as intended.