Is `int` by default `signed long int` in C++?

Question

Is int by default signed long int in C++?

Is it platform and/or compiler dependent? If so, how?

[EDIT]

Are any of the following guaranteed to be duplicate?

signed short int
signed int
signed long int
signed long long int
unsigned short int
unsigned int
unsigned long int
unsigned long long int

Answer 1

All of the integer types are different, i.e. you can safely overload functions for all of them and you won't get any conflict. However, some times use the same number of bits for their representation. Even if they use the same number of bits signed and unsigned types always have a different range. Except for char, using any integer type without signed is equivalent to using it with signed, i.e. signed int and int are equivalent. char is a different type as signed char and unsigned char but char has the same representation and range of either signed char or unsigned char. You can use std::numeric_limits<char>::is_signed to find out which it uses.

On to the more interesting aspects. The following conditions are all true:

• 7 <= std::numeric_limits<signed char>::digits
• sizeof(char) == 1
• sizeof(char) == sizeof(signed char)
• sizeof(char) == sizeof(unsigned char)
• 15 <= std::numeric_limits<short>::digits
• sizeof(char) <= sizeof(short)
• sizeof(short) <= sizeof(int)
• 31 <= std::numeric_limits<long>::digits
• sizeof(int) <= sizeof(long)
• 63 <= std::numeric_limits<long long>::digits
• sizeof(long) <= sizeof(long long)
• sizeof(X) == sizeof(signed X)
• sizeof(signed X) == sizeof(unsigned X)

(where "X" is one of char, short, int, long, and long long).

This means that the size of all integer types can be the same as long as this types hold at least 64 bits (and apparently the Cray X-MP was such a beast). On contemporary machines typically sizeof(int) == sizeof(long) but there are machines where sizeof(int) == sizeof(short). Whether long is 32 or 64 bits depends on the actual architecture and both kinds are currently around.