2015年软考程序员考试复习笔试知识点整理（8）
发布时间：2011/4/25 11:52:33 来源：城市学习网 编辑：ziteng
12、C++操作符优先级:
记忆方法：
去掉一个最高的，去掉一个最低的，剩下的是一、二、三、赋值;双目运算符中，顺序为算术、关系和逻辑，移位和逻辑位插入其中。
--摘自《C语言程序设计实用问答》
问题：如何记住运算符的15种优先级和结合性?
解答：C语言中运算符种类比较繁多，优先级有15种，结合性有两种。
如何记忆两种结合性和15种优先级?下面讲述一种记忆方法。
结合性有两种，一种是自左至右，另一种是自右至左，大部分运算符的结合性是自左至右，只有单目运算符、三目运算符的赋值运算符的结合性自右至左。
优先级有15种，记忆方法如下：
记住一个最高的：构造类型的元素或成员以及小括号。
记住一个最低的：逗号运算符。
剩余的是一、二、三、赋值——意思是单目、双目、三目和赋值运算符。
在诸多运算符中，又分为：算术、关系、逻辑。
两种位操作运算符中，移位运算符在算术运算符后边，逻辑位运算符在逻辑运算符的前面。
再细分如下：
算术运算符*，/，%高于+，-。
关系运算符中：>，>=，<和<=高于==，!=。
逻辑运算符中，除了逻辑求反(!)是单目外，逻辑与(&&)高于逻辑或(||)。
逻辑位运算符中，除了逻辑按位求反(~)外，按位与(&)高于按位半加(^)，高于按位或(|)。
Prece dence Operator Description Example Over loadable Associativity
1 :: Scope resolution operator Class::age = 2; no left to right
2 () Function call printf(“Hello world\n”); yes left to right
() Member initalization c_tor(int x, int y) : _x(x), _y(y * 10) {} yes
[] Array access array[4] = 2; yes
-> Member access from a pointer ptr->age = 34; yes
. Member access from an object obj.age = 34; no
++ Post-increment for (int i = 0; i < 10; i++) cout << i; yes
-- Post-decrement for (int i = 10; i > 0; i--) cout << i; yes
dynamic_cast Runtime-checked type conversion Y& y = dynamic_cast<Y&>(x); no
static_cast Unchecked type conversion Y& y = static_cast<Y&>(x); no
reinterpret_cast Reinterpreting type conversion int const* p = reinterpret_cast<int const*>(0x1234); no
const_cast Cast away/Add constness int* q = const_cast<int*>(p); no
typeid Get type information std::type_info const& t = typeid(x); no
3 ! Logical negation if (!done) ... yes right to left
not Alternate spelling for !
~ Bitwise complement flags = ~flags; yes
compl Alternate spelling for ~
++ Pre-increment for (i = 0; i < 10; ++i) cout << i; yes
-- Pre-decrement for (i = 10; i > 0; --i) cout << i; yes
- Unary minus int i = -1; yes
+ Unary plus int i = +1; yes
* Dereference int data = *intPtr; yes
& Address of int *intPtr = &data; yes
sizeof Size (of the type) of the operand in bytes size_t s = sizeof(int); no
new Dynamic memory allocation long* pVar = new long; yes
new [] Dynamic memory allocation of array long* array = new long[20]; yes
delete Deallocating the memory delete pVar; yes
delete [] Deallocating the memory of array delete [] array; yes
(type) Cast to a given type int i = (int)floatNum; yes
4 ->* Member pointer selector ptr->*var = 24; yes left to right
.* Member object selector obj.*var = 24; no
5 * Multiplication int i = 2 * 4; yes left to right
/ Division float f = 10.0 / 3.0; yes
% Modulus int rem = 4 % 3; yes
6 + Addition int i = 2 + 3; yes left to right
- Subtraction int i = 5 - 1; yes
7 << Bitwise shift left int flags = 33 << 1; yes left to right
>> Bitwise shift right int flags = 33 >> 1; yes
8 < Comparison less-than if (i < 42) ... yes left to right
<= Comparison less-than-or-equal-to if (i <= 42) ... yes
> Comparison greater-than if (i > 42) ... yes
>= Comparison greater-than-or-equal-to if (i >= 42) ... yes
9 == Comparison equal-to if (i == 42) ... yes left to right
eq Alternate spelling for ==
!= Comparison not-equal-to if (i != 42) ... yes
not_eq Alternate spelling for !=
10 & Bitwise AND flags = flags & 42; yes left to right
bitand Alternate spelling for &
11 ^ Bitwise exclusive OR (XOR) flags = flags ^ 42; yes left to right
xor Alternate spelling for ^
12 | Bitwise inclusive (normal) OR flags = flags | 42; yes left to right
bitor Alternate spelling for |
13 && Logical AND if (conditionA && conditionB) ... yes left to right
and Alternate spelling for &&
14 || Logical OR if (conditionA || conditionB) ... yes left to right
or Alternate spelling for ||
15 ? : Ternary conditional (if-then-else) int i = a > b ? a : b; no right to left
16 = Assignment operator int a = b; yes right to left
+= Increment and assign a += 3; yes
-= Decrement and assign b -= 4; yes
*= Multiply and assign a *= 5; yes
/= Divide and assign a /= 2; yes
%= Modulo and assign a %= 3; yes
&= Bitwise AND and assign flags &= new_flags; yes
and_eq Alternate spelling for &=
^= Bitwise exclusive or (XOR) and assign flags ^= new_flags; yes
xor_eq Alternate spelling for ^=
|= Bitwise normal OR and assign flags |= new_flags; yes
or_eq Alternate spelling for |=
<<= Bitwise shift left and assign flags <<= 2; yes
>>= Bitwise shift right and assign flags >>= 2; yes
17 throw throw exception throw EClass(“Message”); no
18 , Sequential evaluation operator for (i = 0, j = 0; i < 10; i++, j++) ... yes left to right
