1 unsigned char i;
2 i=-20;
3
上面一段很簡單的代碼，如果輸出會是什么 呢？
今天群里的新學c++同學問我，unsighed char 和char 有啥區(qū)別，上面的輸出會有什么 不同
在這里，我先鄙視一下自己，我直觀的理解為通常的 首位符號位，然后丟下的就是輸入無符號的。好吧，估計各位看官說我太菜，但確實是輸出的結果和我想的不大一樣，如果各位沒明白我說的是什么問題，可以試一下。然后回來看下面的內容。
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In an unsigned type, all the bits represent the value.
If a type is defined for a particular machine to use 8 bits, then the
unsigned version of this type could hold the values 0 through 255.
無符號型中，所有的位都表示數值。如果在某種機器中，定義一種類型使用 8 位表示，那么這種類型的
unsigned 型可以取值 0 到 255。
The C++ standard does not define how signed types are
represented at the bit level. Instead, each compiler is free to decide how it
will represent signed types. These representations can affect the range
of values that a signed type can hold. We are guaranteed that an 8-bit
signed type will hold at least the values from 127 through 127; many
implementations allow values from 128 through 127.
C++ 標準并未定義 signed 類型如何用位來表示，而是由每個編譯器自由決定如何表示
signed 類型。這些表示方式會影響 signed 類型的取值范圍。8 位 signed
類型的取值肯定至少是從 -127 到 127，但也有許多實現允許取值從 -128 到 127。
Under the most common strategy for representing signed
integral types, we can view one of the bits as a sign bit. Whenever the sign bit
is 1, the value is negative; when it is 0, the value is either 0 or a positive
number. An 8-bit integral signed type represented using a sign-bit can
hold values from 128 through 127.
表示 signed 整型類型最常見的策略是用其中一個位作為符號位。符號位為 1，值就為負數；符號位為
0，值就為 0 或正數。一個 signed 整型取值是從 -128 到 127。
Assignment to Integral Types
整型的賦值
The type of an object determines the values that the object can
hold. This fact raises the question of what happens when one tries to assign a
value outside the allowable range to an object of a given type. The answer
depends on whether the type is signed or unsigned.
對象的類型決定對象的取值。這會引起一個疑問：當我們試著把一個超出其取值范圍的值賦給一個指定類型的對象時，結果會怎樣呢？答案取決于這種類型是
signed 還是 unsigned 的。
For unsigned types, the compiler must adjust the out-of-range value so that it will fit.
The compiler does so by taking the remainder of the value modulo the number of
distinct values the unsigned target type can hold. An object that is an
8-bit unsigned char, for example, can hold values from 0 through 255
inclusive. If we assign a value outside this range, the compiler actually
assigns the remainder of the value modulo 256. For example, we might attempt to
assign the value 336 to an 8-bit signed char. If we try to store 336 in
our 8-bit unsigned char, the actual value assigned will be 80, because
80 is equal to 336 modulo 256.
對于 unsigned 類型來說，編譯器必須調整越界值使其滿足要求。編譯器會將該值對
unsigned 類型的可能取值數目求模，然后取所得值。比如 8 位的 unsigned char，其取值范圍從 0 到
255（包括 255）。如果賦給超出這個范圍的值，那么編譯器將會取該值對 256 求模后的值。例如，如果試圖將 336 存儲到 8 位的
unsigned char 中，則實際賦值為 80，因為 80 是 336 對 256 求模后的值。
For the unsigned types, a negative value is always out
of range. An object of unsigned type may never hold a negative value.
Some languages make it illegal to assign a negative value to an
unsigned type, but C++ does not.
對于 unsigned 類型來說，負數總是超出其取值范圍。unsigned
類型的對象可能永遠不會保存負數。有些語言中將負數賦給 unsigned 類型是非法的，但在 C++ 中這是合法的。
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In C++ it is perfectly legal to assign a negative number to an
object with unsigned type. The result is the negative value modulo the
size of the type. So, if we assign 1 to an 8-bit unsigned char, the
resulting value will be 255, which is 1 modulo 256.
C++ 中，把負值賦給 unsigned
對象是完全合法的，其結果是該負數對該類型的取值個數求模后的值。所以，如果把 -1 賦給8位的 unsigned char，那么結果是
255，因為 255 是 -1 對 256 求模后的值。
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When assigning an out-of-range value to a signed type,
it is up to the compiler to decide what value to assign. In practice, many
compilers treat signed types similarly to how they are required to
treat unsigned types. That is, they do the assignment as the remainder
modulo the size of the type. However, we are not guaranteed that the compiler
will do so for the signed types.
當將超過取值范圍的值賦給 signed 類型時，由編譯器決定實際賦的值。在實際操作中，很多的編譯器處理
signed 類型的方式和 unsigned
類型類似。也就是說，賦值時是取該值對該類型取值數目求模后的值。然而我們不能保證編譯器都會這樣處理 signed 類型。
以上摘自 c++ primer,慚愧，還是再細細的從頭品一次這書吧。