所以你认为你知道运算符在c ++的优先级？

问题描述

我刚刚遇到一个代码，不仅编译，但给出预期的结果（其中x是一个整数值）：

  int y =（int）（0.5 * x * x + + + 0.6 * x + 1.2）; 
  

我花了一段时间才发现会发生什么，我必须说这是一个有趣的运算符问题。没有编译程序，以下操作的结果是什么？为什么？

  int a = 1; 
 int b = 2; 
 int z = a + + + b; 
 int z1 = a +  -  + b; 
 int z2 = a +  -   -  b; 
 int z3 = a  -   -  + b; 
 int z4 = a  -   -   -  b; 
 int z5 = a + -b; 
  

我还有一个问题，标准是给出这样的结果还是编译器特定？ / p>

---

说明：
因为+和 - 运算符之间有空格，所以+ + +不编译为++ +，而是作为一元运算符在右边成员。

  int y =（int）（0.5 * x * x + + + 0.6 * x + 1.2） 
  

实际上是：

  int y =（int）（0.5 * x * x + 0.6 * x + 1.2）; 
 
 
 
 
这是预期结果。
 
 
 
  z = a + + + b = a + +（+ b）= a +（+ b）= a + b = 3; 
 z1 = a +  -  + b = a +  - （+ b）= a +（-b）= a-b = 
 z2 = a +  -  b = a +  - （-b）= a +（+ b）= a + b = 
 z3 = a  -  + b = a  -   - （+ b）= a  - （-b）= a + b = 
 z4 = a  -   -   -  b = a  -   - （-b）= a  - （+ b）= a  -  b = 
 z5 = a + -b = a +（-b）= a-b = -1; 
  
 
 
解决方案
 z = 3 
 
 z1 = -1 
 
 z2 = 3 
 
 z4 = 3 
 
 z5 = -1 
 
 
 
不编译，我会说这是一个简单的数学运算符组合
 
我学会了这个短语（从巴西葡萄牙语翻译）：
 b 
 $ b 
  int z = a + + + b; //'+'+'+'='+'，生成的'+'+'+'= +。 
 int z1 = a +  -  + b; //'+'+' - '=' - '，结果' - '+'+'=  - 。 
 int z2 = a +  -   -  b; //'+'+' - '=' - '，结果' - '+' - '= +。 
 int z3 = a  -   -  + b; //' - '+' - '='+'，结果'+'+'+'= +。 
 int z4 = a  -   -   -  b; //' - '+' - '='+'，结果'+'+' - '=  - 。 
 int z5 = a + -b; // simpledifferent signals = subtract:) 
  
 
I just ran into a piece of code that not only compiles, but gives the expected result (where x is an integer value):
int y = (int)(0.5 * x * x + + + 0.6 * x + 1.2);
It took me a while to figure out what happens and I must say it was an interesting operator problem. Without compiling the program, what are the results of the following operations and why?
int a = 1;
int b = 2;
int z = a + + + b;
int z1 = a + - + b;
int z2 = a + - - b;
int z3 = a - - + b;
int z4 = a - - - b;
int z5 = a +- b;
I still have one question, though: does the standard give such results or is it compiler specific?
Explanation:Because the + and - operators have spaces between them, the "+ + +" sequence is not compiled as "++ +", but as unary operators on the right member. So
int y = (int)(0.5 * x * x + + + 0.6 * x + 1.2);
actually gives:
int y = (int)(0.5 * x * x + 0.6 * x + 1.2);
which was the expected result.
So,
z  = a + + + b = a + + (+b) = a + (+b) = a + b = 3;
z1 = a + - + b = a + - (+b) = a + (-b) = a - b = -1;
z2 = a + - - b = a + - (-b) = a + (+b) = a + b = 3;
z3 = a - - + b = a - - (+b) = a - (-b) = a + b = 3;
z4 = a - - - b = a - - (-b) = a - (+b) = a - b = -1;
z5 = a +- b = a + (-b) = a - b = -1;
 解决方案 
z = 3
z1 = -1
z2 = 3
z4 = 3
z5 = -1  
Without compiling, i'd say it's a simple math operator combining
I've learned it with this phrase (translated from Brazilian portuguese):
So
int z = a + + + b; // '+' + '+' = '+' , the resulting '+' + '+' = + again.
int z1 = a + - + b; // '+' + '-' = '-' , the resulting '-' + '+' = - again.
int z2 = a + - - b; // '+' + '-' = '-' , the resulting '-' + '-' = +.
int z3 = a - - + b; // '-' + '-' = '+' , the resulting '+' + '+' = + again.
int z4 = a - - - b; // '-' + '-' = '+' , the resulting '+' + '-' = -.
int z5 = a +- b; //simple "different signals = subtract" :)
                        
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