Bool 运算符 + + 和——

今天在编写一些 Visual C + + 代码时，我遇到了一些让我感到惊讶的事情。似乎 C + + 支持 bool 的 + + (递增) ，但不支持——(递减)。这只是一个随机的决定，还是有什么原因？

结果显示:

static HMODULE hMod = NULL;
static bool once = false;
if (!once++)
hMod = LoadLibrary("xxx");

这不是:

static HMODULE hMod = NULL;
static bool once = true;
if (once--)
hMod = LoadLibrary("xxx");
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ANSI ISO IEC 14882 2003 (c++03):

5.2.6-2

The operand of postfix -- is decremented analogously to the postfix ++ operator, except that the operand shall not be of type bool. [Note: For prefix increment and decrement, see 5.3.2. ]

And unsurprisingly...

5.3.2-2

The operand of prefix -- is modified by subtracting 1. The operand shall not be of type bool. The requirements on the operand of prefix -- and the properties of its result are otherwise the same as those of prefix ++. [Note: For postfix increment and decrement, see 5.2.6. ]

Also the 5.6.2-1 and 5.3.2-1 mention that ++ for bools shall be true and Annex D-1 says that ++ on bools in deprecated.

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最佳答案

It comes from the history of using integer values as booleans.

If x is an int, but I am using it as a boolean as per if(x)... then incrementing will mean that whatever its truth value before the operation, it will have a truth-value of true after it (barring overflow).

However, it's impossible to predict the result of -- given knowledge only of the truth value of x, as it could result in false (if the integral value is 1) or true (if the integral value is anything else - notably this includes 0 [false] and 2 or more [true]).

So as a short-hand ++ worked, and -- didn't.

++ is allowed on bools for compatibility with this, but its use is deprecated in the standard and it was removed in C++17.

This assumes that I only use x as an boolean, meaning that overflow can't happen until I've done ++ often enough to cause an overflow on it's own. Even with char as the type used and CHAR_BITS something low like 5, that's 32 times before this doesn't work any more (that's still argument enough for it being a bad practice, I'm not defending the practice, just explaining why it works) for a 32-bit int we of course would have to use ++ 2^32 times before this is an issue. With -- though it will only result in false if I started with a value of 1 for true, or started with 0 and used ++ precisely once before.

This is different if we start with a value that is just a few below 0. Indeed, in such a case we might want ++ to result in the false value eventually such as in:

int x = -5;
while(++x)
doSomething(x);

However, this example treats x as an int everywhere except the conditional, so it's equivalent to:

int x = -5;
while(++x != 0)
doSomething(x);

Which is different to only using x as a boolean.

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Due to historical reasons this was supported. But note that ... The use of an operand of type bool with the ++ operator is deprecated see Section 5.3.2 in the C++ Standard(n3092)

5.3.2 Increment and decrement [expr.pre.incr]

  • The operand of prefix ++ is modified by adding 1, or set to true if it is bool (this use is deprecated). The operand shall be a modifiable lvalue. The type of the operand shall be an arithmetic type or a pointer to a completely-defined object type. The result is the updated operand; it is an lvalue, and it is a bit-field if the operand is a bit-field. If x is not of type bool, the expression ++x is equivalent to x+=1 [ Note: see the discussions of addition (5.7) and assignment operators (5.17) for information on conversions. —end note ]
  • The operand of prefix -- is modified by subtracting 1. The operand shall not be of type bool. The requirements on the operand of prefix -- and the properties of its result are otherwise the same as those of prefix ++.
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  • With the old standards (C++98) it is not an error.
  • With the new standards incrementing a boolean is deprecated. (C++11)
  • You can use incrementation on a boolean until C++17.

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