C++ 通常的算术转换不转换

Question

First, I'd like to point out that I did read some of the other answers with generally similar titles. However, those either refer to what I assume is an older version of the standard, or deal with promotions , not conversions.

I've been crawling through "C++ Crash Course", where the author states the following in the chapter exploring built-in operators:

If none of the floating-point promotion rules apply, you then check whether either argument is signed. If so, both operands become signed. Finally, as with the promotion rules for floating-point types, the size of the largest operand is used to promote the other operand: ...

If I read the standard correctly, this is not true, because, as per cppreference.com,

If both operands are signed or both are unsigned, the operand with lesser conversion rank is converted to the operand with the greater integer conversion rank

Otherwise, if the unsigned operand's conversion rank is greater or equal to the conversion rank of the signed operand, the signed operand is converted to the unsigned operand's type.

Otherwise, if the signed operand's type can represent all values of the unsigned operand, the unsigned operand is converted to the signed operand's type

Otherwise, both operands are converted to the unsigned counterpart of the signed operand's type.

What confuses me even more is the fact that the following code:

printf("Size of int is %d bytes\n", sizeof(int));
printf("Size of short is %d bytes\n", sizeof(short));
printf("Size of long is %d bytes\n", sizeof(long));
printf("Size of long long is %d bytes\n", sizeof(long long));
unsigned int x = 4000000000;
signed int y = -1;
signed long z = x + y;
printf("x + y = %ld\n", z);

produces the following output:

Size of int is 4 bytes
Size of short is 2 bytes
Size of long is 8 bytes
Size of long long is 8 bytes
x + y = 3999999999

As I understand the standard, y should have been converted to unsigned int , leading to an incorrect result. The result is correct, which leads me to assume that no conversion happens in this case. Why? I would be grateful for any clafirication on this matter. Thank you.

(I would also appreciate someone telling me that I won't ever need this kind of arcane knowledge in real life, even if it's not true - just for the peace of mind.)

Answer 1

A conversion of a negative signed value to an unsigned leads into the binary complement value which corresponds to it. An addition of this unsigned value will lead into an overflow which then results just exactly in the same value as adding the negative value would do.

By the way: That's the trick how the processor does a subtraction (or am I wrong in these modern times?).

Answer 2

If I read the standard correctly, this is not true

You're correct. CCC is wrong.

As I understand the standard, y should have been converted to unsigned int,

It did.

The result is correct, which leads me to assume that no conversion happens in this case.

You assume wrong.

Since -1 is not representable unsigned number, it simply converts to the representable number that is representable and is congruent with - 1 modulo the number of representable values. When you add this to x, the result is larger than any representable value, so again result is the representable value that is congruent with the modulo. Because of the magic of how modular arithmetic works, you get the correct result.

Answer 3

You have an addition of an unsigned int and a signed int, so same conversion rank for both operands. So the second rule will apply (emphasize mine):

[Otherwise,] if the unsigned operand's conversion rank is greater or equal to the conversion rank of the signed operand, the signed operand is converted to the unsigned operand's type.

• -1 is converted to an unsigned type by adding to it the smallest power of two greater that the highest unsigned int (in fact its representation in 2's complement)
• that number is added to 4000000000 and the result is the modulo of the sum and that power of two (in fact discarding its higher bit)
• and you just get the expected result.

The standard mandates conversion to the unsigned type because unsigned overflow is defined by the standard, while unsigned one is not.

Answer 4

From the output of the first print statement from your code snippet it is evident that since the size of int is 4 bytes, the unsigned int values can range from 0 to 4,294,967,295. (2^32-1)

The declaration below is hence perfectly valid:

unsigned int x = 4000000000;

Remember here that declaring x as an unsigned integer gives itself twice the (+ve) range of a signed counterpart, but it falls in exactly the same range and hence conversions between them are possible.

signed int y = -1;
signed long z = x + y;
printf("x + y = %ld\n", z);

Hence for the rest of the code above, it doesn't matter here whether the operands of y and z are signed (+ve/-ve) or unsigned (+ve) because you are dealing with an unsigned integer in the first place, for which the signed integers will be converted to their corresponding unsigned integer versions by adding UINTMAX+1 to the signed version. Here UINTMAX stands for the largest unsigned value which is added with the smallest power of 2 (2^0=1). The end result would obviously be larger than values we can store, hence it is taken as modulo to collect the correct unsigned int.

However, if you were to convert x to signed instead:

signed int x = 4000000000;
int y = -1;
long z = x + y;
printf("x + y = %ld\n", z);

you would get what you might have expected:

x + y = -294967297

Answer 5

I would also appreciate someone telling me that I won't ever need this kind of arcane knowledge in real life

There is one implicit promotion not addressed above that is a common cause of bugs:

unsigned short x = 0xffff;
unsigned short y = 0x0001;
if ((x + y) == 0)

The result of this is false because arithmetic operations are implicitly promoted to at least int.