为什么−1>大小（int）？
考虑下面的代码：
template<bool> class StaticAssert;
template<> class StaticAssert<true> {};
StaticAssert< (-1 < sizeof(int)) > xyz1; // Compile error
StaticAssert< (-1 > sizeof(int)) > xyz2; // OK
为什么 -1 > sizeof(int) 是吗？
- 是否真的将
-1提升为unsigned(-1)然后unsigned(-1) > sizeof(int)。 -1 > 是真的吗？ sizeof(int)相当于-1 > size_t(4)如果 sizeof(int) 是 4。如果是这样，为什么-1 > > size_t(4) 是假的？
这个C++标准符合吗？
Consider the following code:
template<bool> class StaticAssert;
template<> class StaticAssert<true> {};
StaticAssert< (-1 < sizeof(int)) > xyz1; // Compile error
StaticAssert< (-1 > sizeof(int)) > xyz2; // OK
Why is -1 > sizeof(int) true?
- Is it true that
-1is promoted tounsigned(-1)and thenunsigned(-1) > sizeof(int). - Is it true that
-1 > sizeof(int)is equivalent to-1 > size_t(4)if sizeof(int) is 4. If this is so why-1 > size_t(4)is false?
Is this C++ standard comformant?
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这很简单，也很悲伤。在C/C++中：
1U < 2U（IOW0U是最小的无符号值）sizeof的类型为size_tsize_t是无符号整数类型-1-1-1-1 1U被解释为unsigned(-1)1U，并且unsigned(-1)=- 1U，并且显然- 1U1U，所以-1 < 1U 是正确的。sizeof Something充当（大部分）作为等效类（！！！）。-1sizeof Something结论：这是从 C 继承的设计错误。
规则：
仅使用无符号类型进行模算术、位操作（
&、|、^、<<、>>、~运算符） 、字节操作（unsigned char在 C/C++ 中表示“字节”）和字符（unsigned char在 C/C++ 中表示字符）。不要使用无符号类型进行算术运算。
如果函数需要一个永远不应该为负数的整数值，请采用有符号整数，并可以选择在函数中检查该值是否在范围内。
It's simple and sad. In C/C++:
1U < 2U(IOW0Uis the smallestunsignedvalue)sizeofhas typesize_tsize_tis an unsigned integer type-1 < 1Uis interpreted asunsigned(-1) < 1U, andunsigned(-1)=- 1U, and obviously- 1U < 1U, so-1 < 1Uis true.sizeof somethingacts (mostly) as an equivalent class (!!!).-1 < sizeof somethingThe conclusion: this is a design error inherited from C.
Rule:
Only use unsigned types for modular arithmetic, bits manipulations (
&,|,^,<<,>>,~operators), byte manipulations (unsigned charmeans "byte" in C/C++), and characters (unsigned charmeans character in C/C++).Do not use unsigned types to do arithmetic.
If a function expects an integer value that should never be negative, take a signed integer, and optionally check in the function that the value is in range.
以下是标准（ISO 14882）如何解释 abort -1 > sizeof(int)
关系运算符“>”定义于 5.9 (expr.rel/2)
通常的算术转换定义在 5 (expr/9)
... 该模式称为通常的算术转换，其定义如下：
double, ...
积分促销在 4.5 (conv.prom/1) 中定义
sizeof的结果在5.3.3中定义(expr.sizeof/6)
size_t 在 C 标准（ISO 9899）中定义，它是无符号整数类型。
所以对于
-1 > sizeof(int)，>触发通常的算术转换。通常的算术转换将 -1 转换为 unsigned int，因为 int 无法表示size_t的所有值。-1变成一个非常大的数字，具体取决于平台。所以-1 > sizeof(int)为true。The following is how standard (ISO 14882) explains abort -1 > sizeof(int)
Relational operator `>' is defined in 5.9 (expr.rel/2)
The usual arithmetic conversions is defined in 5 (expr/9)
... The pattern is called the usual arithmetic conversions, which are defined as following:
double, ...
The integral promotions is defined in 4.5 (conv.prom/1)
The result of sizeof is defined in 5.3.3 (expr.sizeof/6)
size_t is defined in C standard (ISO 9899), which is unsigned integer type.
So for
-1 > sizeof(int), the > triggers usual arithmetic conversions. The usual arithmetic conversion converts -1 to unsigned int because int cannot represent all the value ofsize_t.-1becomes a very large number depend on platform. So-1 > sizeof(int)istrue.因为 unsigned 比有符号更强，并且 -1 从
size_t开始转换为无符号值，所以实际上-1 == 0xFFFFFFFF > 4根据 C++ 标准，它应该是这样工作的
Because unsigned is stronger then signed and -1 converted to unsigned value as of
size_t, so actually-1 == 0xFFFFFFFF > 4This is how it should work according to C++ standard
因为 -1 被转换为
size_t并且这是一个无符号数据类型 - 所以(size_t)-1 == 4294967295（在 32 位系统上）它肯定大于 4例如，如果您将
-Wall添加到 gcc 设置中，您会收到一条警告，提示您正在比较有符号和无符号数据类型because -1 gets casted to
size_tand this is an unsigned data type - so(size_t)-1 == 4294967295(on a 32bit system) which is definitely larger than 4if you add
-Wallto the gcc settings for example you get a warning that you are comparing a signed and an unsigned data type