Given the following snippet:
#include <stdio.h>
typedef signed long long int64;
typedef signed int int32;
typedef signed char int8;
int main()
{
printf("%i\n", sizeof(int8));
printf("%i\n", sizeof(int32));
printf("%i\n", sizeof(int64));
int8 a = 100;
int8 b = 100;
int32 c = a * b;
printf("%i\n", c);
int32 d = 1000000000;
int32 e = 1000000000;
int64 f = d * e;
printf("%I64d\n", f);
}
The output with MinGW GCC 3.4.5 is (-O0):
1
4
8
10000
-1486618624
The first multiplication is casted to an int32 internally (according to the assembler output). The second multiplication is not casted. I’m not sure if the results differ because the program was running on a IA32, or because it is defined somewhere in the C standard. Nevertheless I’m interested if this exact behavior is defined somewhere (ISO/IEC 9899?), because I like to better understand why and when I’ve to cast manually (I’ve problems porting a program from a different architecture).
The C99 standard does specify that binary operators such as
*do not operate on integer types smaller thanint. Expressions of these types are promoted tointbefore the operator is applied. See 6.3.1.4 paragraph 2 and the numerous occurrences of the words “integer promotion”. But this is somewhat orthogonal to the assembly instructions generated by the compiler, which operate onints because this is faster even when the compiler would be allowed to computed a shorter result (because the result is immediately stored in an l-value of a short type, for instance).Regarding
int64 f = d * e;wheredandeare of typeint, the multiplication is done as anintin accordance with the same promotion rules. The overflow is technically undefined behavior, you are getting two-s-complement result here, but you could get anything according to the standard.Note: the promotion rules distinguish signed and unsigned types when promoting. The rule is to promote smaller types to
intunlessintcannot represent all values of the type, in which caseunsigned intis used.