1.That is, if you have the value 29.7 and you cast it to a char, is the resulting value 30 or 29?
换言之，假设一个值是29.7，我们把它造型成一个char，那么结果值到底是30还是29呢？
2.When F is an algebraically closed field char F≠2, A is similar over F to an orthogonal matrix if and only if ( 1 ) the invariant factors of A aer self-reciprocal (2) the elementary divi sors of A of the form (x± 1 ) 2k appear even times.
(ii)当F为一个代数闭域且char(F)≠2时,F相似于一个正交矩阵的充要条件为(1)A的不变因子是自反的。 (2)A的形为(X±1)~(2k)的初等因子出现偶数次。
3.Compound assignments do not require casts for char, byte, or short, even though they are performing promotions that have the same results as the direct arithmetic operations.
对于char，byte或者short，混合赋值并不需要造型。即使它们执行转型操作，也会获得与直接算术运算相同的结果。
4.In char, byte, and short you can see the effect of promotion with the arithmetic operators.
在char，byte和short中，我们可看到算术运算符的“转型”效果。
5.What does the cast to char do?
我们的第一个问题是，造型会对char作什么样的处理呢？
6.The switch statement is a clean way to implement multi-way selection (i.e., selecting from among a number of different execution paths), but it requires a selector that evaluates to an integral value such as int or char.
switch语句是实现多路选择的一种易行方式(比如从一系列执行路径中挑选一个)。但它要求使用一个选择因子，并且必须是int或char那样的整数值。
7.(3) If Char F = p > 7, L is a Lie algebra (resp. restricted Lie algebra), L is solvable if and only if the dimension of every (finite-dimensional) irreducible module (resp. restricted module) of L is a power of p.
(3)若char F=P>7,L为李代数(限制李代数),则L为可解当且仅当L的任一不可约模(限制模)的维数为p的幂.
8.Notice in the above code the maximum possible hexadecimal values for char, byte, and short.
注意在上述代码中，最大的十六进制值只会在char，byte以及short身上出现。
9.If you shift a char, byte, or short, it will be promoted to int before the shift takes place, and the result will be an int.
若对char，byte或者short进行移位处理，那么在移位进行之前，它们会自动转换成一个int。
10.The double result is turned back into a char with a cast.
利用一个“造型”，double结果会转换回char。
