Is the integer constant's default type signed or unsigned?

Question

Is the integer constant's default type signed or unsigned? such as 0x80000000, how can I to decide to use it as a signed integer constant or unsigned integer constant without any suffix?

If it is a signed integer constant, how to explain following case?

printf("0x80000000>>3 : %x\n", 0x80000000>>3); 

output:

0x80000000>>3 : 10000000 

The below case can indicate my platform uses arithmetic bitwise shift, not logic bitwise shift:

int n = 0x80000000;  printf("n>>3: %x\n", n>>3); 

output:

n>>3: f0000000 

Answer 1

C has different rules for decimal, octal and hexadecimal constants.

For decimal, it is the first type the value can fit in: int, long, long long

For hexadecimal, it is the first type the value can fit in: int, unsigned int, long, unsigned long, long long, unsigned long long

For example on a system with 32-bit int and unsigned int: 0x80000000 is unsigned int.

Note that for decimal constants, C90 had different rules (but rules didn't change for hexadecimal constants).