C 是否在内部使用 2 的补码来评估 5-4 之类的无符号数算术？

Question

I have C code as

#include<stdio.h>

int main()
{
    unsigned int a = 5;
    unsigned int b = 4;
    printf("%u",a-b);
}

Output of above code is 1, I am thinking that C has calculated internally the result as taking 2's compliment of -4 and then using compliment arithmetic to evaluate the result. Please correct me if anything I am interpreting wrong. (Here, I am talking about how C actually calculates result using binary)

Answer 1

Does C uses 2's complement internally to evaluate unsigned numbers arithmetic like 5-4?

No, for two reasons.

unsigned int a = 5, b = 4;
printf("%u",a-b);

C guarantees that arithmetic on unsigned integer types is performed modulo the size of the type. So if you computed ba , you'd get -1 which would wrap around to UINT_MAX , which is probably either 65535 or 4294967295 on your machine. But if you compute ab , that's just an ordinary subtraction that doesn't overflow or underflow in any way, so the result is an uncomplicated 1 without worrying about 2's complement or modulo arithmetic or anything.

If your compiler, or your CPU architecture, chooses to implement a - b as a + -b , that's their choice, but it's an implementation detail way beneath the visibility of the C Standard, or ordinary programmers like you and me, and it won't affect the observable results of a C program at all.

Where things get interesting, of course, is with addition and subtraction of signed quantities. Up above I said that under unsigned arithmetic, 4 - 5 is -1 which wraps around to UINT_MAX . Using signed arithmetic, of course, 4 - 5 is -1 which is -1. Under 2's complement arithmetic, it Just So Happens that the bit patterns for -1 and UINT_MAX are identical (typically 0xffff or 0xffffffff ), and this is why 2's complement arithmetic is popular, because your processor gets to define, and your C compiler gets to use, just one set of add and sub instructions, that work equally well for doing signed and unsigned arithmetic. But (today, at least), C does not mandate 2's complement arithmetic, and that's the other reason why the answer to your original question is "no".

---

But to be clear (and to go back to your question): Just about any C compiler, for just about any architecture, is going to implement a - b by emitting some kind of a sub instruction. Whether the processor then chooses to implement sub as a two's complement negate-and-add, or some other kind of negate-and-add, or via dedicated bitwise subtraction-with-borrow logic, is entirely up to that processor, and it doesn't matter (and is probably invisible) as long as it always returns a mathematically appropriate result.

Answer 2

The arithmetic method generally whatever is most natural on the target hardware. It is not defined by the C language.

When a processor's ALU has at least int sized registers, ab will not doubt be implemented as a single SUB instruction (or whatever the target's op-code mnemonic for subtraction might be). In the hardware logic it may well be that the logic is equivalent to a + (~b + 1) (ie 2's complement the RHS and add) - but that is a hardware logic/micro-code implementation issue, not a language or compiler behaviour.

At Godbolt for GCC x86 64-bit, the statement:

 unsigned int c = a - b ;

generates the following assembly code (my comments):

    mov     eax, DWORD PTR [rbp-4]      ; Load a to EAX
    sub     eax, DWORD PTR [rbp-8]      ; Subtract b from EAX
    mov     DWORD PTR [rbp-12], eax     ; Write result to c

So in that sense your question is not valid - C does not do anything in particular, the processor performs the subtraction intrinsically.

The C standard allows 2's, 1's and sign+magnitude arithmetic, but in practice the world has settled on 2's complement and machines that use other representations are arcane antiques that probably never had C compilers targeted for them in any case.

There are in any case moves to remove the option for anything other than 2's complement in the language: http://www.open-std.org/jtc1/sc22/wg14/www/docs/n2218.htm