快捷方式要记住评价在C秩序和precedence运营商 [英] Shortcut to remember the Order of Evaluation and Precedence of Operators in C

问题描述

有没有什么快捷方式或百威记得在C precedence和评估秩序东阳它起着主要的作用和我（或我们大多数人），通常忘记和混乱中结束。请帮我....

我想举一个例子这个......说..

 无效的主要（）
{
    INT A = 1;
    A = A + + + + A;
    的printf（％d个，一）;
} //输出5;无效的主要（）
{
    INT A = 1;
    INT X;
    X = A + + + + A;
    的printf（％D，X）;
} //输出4;
 

也是前pression结果
X = + A + A ++; 结果
给出了不同的结果。

我希望，如果后增量高优先级，那么 A ++ 应先进行评估，请如果IAM错误清除我，请给我解释一下它是如何工作的。

解决方案

首先，形式的前pressions A + + + + A ， + A + A ++ 等，导致的未定义行为;的任何的结果是可能的。从语言标准（ n1256 ）：

6.5防爆pressions 的结果
...结果
2 previous和下一个序列点的对象应具有其存储的值之间
由前pression的评价最多一次修改。 ^72）而且，前值
应仅被理解为确定的值被存储。 ^73）结果
...结果
72）浮点状态标志不是一个对象，并且可以多次设置一个前pression内。搜索结果
73）本款呈现不确定的语句前pressions如

        I = + I + 1;
        A [++] =我;

同时允许

        I = I + 1;
        一个[我] =我;

所以，千万不要那样做。

一个定义良好的前pression像 X = A + + + + b 会被解析为 X =（（A ++）+ （++ b））;在 ++ 运营商的两种形式有更高的precedence比另外，加入具有较高的precedence比分配。在结果的前任pression将是一样的 X = A +（B + 1）。

其次，记得 ++ 和 - 运营商拥有的结果的和一个副作用的，如下所示：

防爆pression结果副作用
---------- ------ -----------
       我++ I I = I + 1
       ++ I I + 1 I = I + 1
       我 - 我I = I  -  1
       --i我 -  1 I =  -  1

重要提示要记住：副作用不具备前pression进行评估后，立即应用;它只有下一个序列点之前被应用。这是可能的 X = A + + + + b 来进行评价如下：

  T1 =一个;
T2 = B + 1;
X = T1 + T2;
B = B + 1;
A = A + 1;
 

在这种情况下，更新 A 和 B 被推迟到加法赋值为<$之后C $ C> X 。

至于precedence而言，这里是从高分到低分顺序一般：

1. Postfix的运营商（都具有相同的precedence，因此运营商的序列进行评估左到右）
   
   
   
   • 数组下标运算符 []
     
   • 函数调用操作（）
     
   • 组件选择运营商 和 - >
     
   • 后缀 ++ 和 -
     
   
   
2. 一元运算符（都具有相同的precedence，因此运营商的序列进行评估左到右）
   
   
   
   • preFIX ++ 和 -
     
   • 的sizeof
     
   • 按位求反运算符〜
     
   • 逻辑否定运算符！
     
   • 单目标志运营商 - 和 +
     
   • 地址运算符和
     
   • 引用操作 *
     
   
   
3. 铸造前pressions （ 类型名称 ）
   
4. 乘法运算符 * ， / ，％
   
5. 添加剂经营者 + 和 -
   
6. 移位运算＆LT;＆LT; 和＆GT;＆GT;
   
7. 关系运算符＆LT; ，＆GT; ，＆LT; = ，＆GT; =
   
8. 相等运算符 == 和！=
   
9. 位与和
   
10. 位异或 ^
    
11. 位或 |
    
12. 逻辑与 &&
    
13. 逻辑或 ||
    
14. 条件运算符：
    
15. 赋值运算符 = ， + = 。 - = ， * = ， / = ，（％）= ，＆LT;＆LT; = ，＆GT;＆GT; = ，＆= ， ^ = ， | =
    
16. 顺序（逗号）运算符，
    

所以，前pressions像 * X ++ 被解析为 *（X ++），因为后缀 ++ 具有较高的precedence比一元 * 。同样， sizeof的X + 1 被解析为（sizeof的X）+ 1 ，因为的sizeof 具有较高的precedence比加法。一名前pression像点++ - ＆GT; X 被解析为（P ++） - GT; X ;这两个后缀 ++ 和 - ＆GT; 运营商具有相同的precedence，所以他们从左边解析以对。

这是关于短快捷键搞定;有疑问时，请使用括号。

is there any shortcut or bestway to remember the precedence and order of evaluation in C beacause it plays the main role and i or (most of us) usually forgets and end up with the mess. Please Help me....

i want to give an example for this... say..

void main()  
{  
    int a=1;  
    a = a++ + ++a;  
    printf("%d",a);  
}//prints 5; 

void main()  
{  
    int a=1;  
    int x;  
    x = a++ + ++a;  
    printf("%d",x);  
}//prints 4;  

also the expression
x = ++a + a++;
gives a different result

i hope that if post increment have high priority then a++ should be evaluated first please clear me if iam wrong and please explain me how it works..

解决方案

First of all, expressions of the form a++ + ++a, ++a + a++, etc., result in undefined behavior; any result is possible. From the language standard (n1256):

6.5 Expressions
...
2 Between the previous and next sequence point an object shall have its stored value modified at most once by the evaluation of an expression.⁷²⁾ Furthermore, the prior value shall be read only to determine the value to be stored.⁷³⁾
...
72) A floating-point status flag is not an object and can be set more than once within an expression.

73) This paragraph renders undefined statement expressions such as

        i = ++i + 1;
        a[i++] = i;

while allowing

        i = i + 1;
        a[i] = i;

So, don't do that.

A well-defined expression like x = a++ + ++b will be parsed as x = ((a++) + (++b)); both forms of the ++ operator have higher precedence than addition, and addition has higher precedence than assignment. The result of the expression will be the same as x = a + (b + 1).

Secondly, remember that the ++ and -- operators have a result and a side effect, as shown below:

Expression          Result         Side effect
----------          ------         -----------
       i++               i            i = i + 1
       ++i           i + 1            i = i + 1
       i--               i            i = i - 1
       --i           i - 1            i = i - 1

Important note to remember: the side effect doesn't have to be applied immediately after the expression is evaluated; it only has to be applied before the next sequence point. It's possible for x = a++ + ++b to be evaluated as follows:

t1 = a;
t2 = b + 1;
x = t1 + t2;
b = b + 1;
a = a + 1;  

In this case, the updates to a and b are deferred until after the addition and assignment to x.

As far as precedence is concerned, here is the general order from high to low:

1. Postfix operators (all have the same precedence, so sequences of operators will be evaluated left-to-right)
   • array subscript operator []
   • function call operator ()
   • component selection operators . and ->
   • postfix ++ and --
2. Unary operators (all have the same precedence, so sequences of operators will be evaluated left-to-right)
   • prefix ++ and --
   • sizeof
   • bitwise negation operator ~
   • logical negation operator !
   • unary sign operators - and +
   • address-of operator &
   • dereference operator *
3. Cast expressions ( type name )
4. Multiplicative operators *, /, %
5. Additive operators + and -
6. Shift operators << and >>
7. Relational operators <, >, <=, >=
8. Equality operators == and !=
9. Bitwise AND &
10. Bitwise XOR ^
11. Bitwise OR |
12. Logical AND &&
13. Logical OR ||
14. Conditional operator ?:
15. Assignment operators =, +=. -=, *=, /=, %=, <<=, >>=, &=, ^=, |=
16. Sequential (comma) operator ,

So, expressions like *x++ are parsed as *(x++), since the postfix ++ has higher precedence than the unary *. Similarly, sizeof x + 1 is parsed as (sizeof x) + 1, since sizeof has higher precedence than addition. An expression like p++->x is parsed as (p++)->x; both postfix ++ and -> operators have the same precedence, so they're parsed from left to right.

This is about as short as shortcuts get; when in doubt, use parentheses.

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