What is the difference between an int and a long in C++?

Question

Correct me if I am wrong,

int is 4 bytes, with a range of values from -2,147,483,648 to 2,147,483,647 (2^31)
long is 4 bytes, with a range of values from -2,147,

Answer 1

The C++ Standard says it like this :

3.9.1, §2 :

There are five signed integer types : "signed char", "short int", "int", "long int", and "long long int". In this list, each type provides at least as much storage as those preceding it in the list. Plain ints have the natural size suggested by the architecture of the execution environment (44); the other signed integer types are provided to meet special needs.

(44) that is, large enough to contain any value in the range of INT_MIN and INT_MAX, as defined in the header <climits>.

The conclusion : it depends on which architecture you're working on. Any other assumption is false.

Answer 2

When compiling for x64, the difference between int and long is somewhere between 0 and 4 bytes, depending on what compiler you use.

GCC uses the LP64 model, which means that ints are 32-bits but longs are 64-bits under 64-bit mode.

MSVC for example uses the LLP64 model, which means both ints and longs are 32-bits even in 64-bit mode.

Answer 3

As Kevin Haines points out, ints have the natural size suggested by the execution environment, which has to fit within INT_MIN and INT_MAX.

The C89 standard states that UINT_MAX should be at least 2^16-1, USHRT_MAX 2^16-1 and ULONG_MAX 2^32-1 . That makes a bit-count of at least 16 for short and int, and 32 for long. For char it states explicitly that it should have at least 8 bits (CHAR_BIT). C++ inherits those rules for the limits.h file, so in C++ we have the same fundamental requirements for those values. You should however not derive from that that int is at least 2 byte. Theoretically, char, int and long could all be 1 byte, in which case CHAR_BIT must be at least 32. Just remember that "byte" is always the size of a char, so if char is bigger, a byte is not only 8 bits any more.