[C advanced] 17, ++ and - operator analysis

Summary

1) ++ and - participate in the mixed operation result is uncertain, such as r = (i++) + (i++); etc.

• C++ only specifies the relative execution order corresponding to ++ and - (the value and auto-increment)
• ++ and - The assembly instructions corresponding to
• In the mixed operation, the assembly instruction of ++ and - may be interrupted (the value and auto-increment may be interrupted, and other codes are inserted in the middle)

2) The greedy method of the

• The compiler as many characters as possible from left to right, one by one 16176b873dc766
• When the read character cannot form a legal symbol with the already read character

3) The space can be used as the rest of a complete symbol of compiler reads the space, it will immediately process the previously read symbol.

Analysis of ++ and - Operator

1、Demo1

• What is the output of the following code?

  int i = 0;
  int r = 0;
  
  r = (i++) + (i++) + (i++);
  printf("i = %d\n, r = %d\n", i, r);
  
  r = (++i) + (++i) + (++i);
  printf("i = %d\n, r = %d\n", i, r);

• Preliminary analysis of the code:

  第一行输出：i = 3, r = 3;    // 按照从左向右的计算方法，i先取值，再自增，
                              // 则r = 0 + 1 + 2 = 3;
  第二行输出：i = 6, r = 16;   // 同上，r = 4 + 5 + 6 = 15；

The actual output of the above code in the VS2015 compiler is:

i = 3, r = 0;
i = 6, r = 18;

View the disassembly result with the help of the Vs compiler:

Code analysis:

对于r = (i++) + (i++) + (i++);    // 先取了i的值做了相加，赋值给r；然后i自增3次
    在汇编层面，做的操作依次是：
    1）00C542BC mov 将i代表的这个地址中的值放到寄存器eax中，为0
    2）00C542BF add eax的值和i的值累加，累加和仍然为0
    3）00C542C5 mov 将eax里的值放到r变量代表的内存里，
            所以r的值为0
    4）后面做了3次相同的操作：将i变量内存里的值放到ecx寄存器里，然后加1，
       再把ecx里的值放回i变量的内存里；重复2次
            所以i的值为3

对于r = (i++) + (i++) + (i++);    // 先给i自增了3次；然后把i的值加3次给了r
    在汇编层面，做的操作依次是：
    1）009A42F8 到 009A42FE 将i代表的这个地址中的值放到寄存器eax中；
                            然后eax里的值自增1；
                            写回i的内存里，此时i的值为4
                            重复2次后，i的值为6
            所以i的值为6
    2）009A4313 mov 把i里的值移动到eax寄存器中，eax里的值为6
    3）009A4316 add eax里的值加上i的值，即6+6，eax里的值为12
    4）009A4319 add eax里的值加上i的值，即12+6，eax里的值为18
    5）009A4319 mov eax里的值写回r代表的内存里
            所以r的值为18
                            

This piece of code, our analysis and the actual compiler results are very different; then what about other compilers?
bcc compiler and the vc compiler are the same, while gcc compiler are "r = 0 and r = 16"

The output result of the above code in the java compiler:

// test.java
public static void main(String[] args) {
    System.out.println("test.java file name must be equal to the class name test");
    int i = 0;
    int r = 0;
    
    r = (i++) + (i++) + (i++);
    System.out.println("i = " + i + ", r = " + r);

    r = (++i) + (++i) + (++i);
    System.out.println("i = " + i + ", r = " + r);
}

// 编译：javac test.java
// 执行：java test
// 输出：i = 3, r = 3
        i = 6, r = 15   和分析的一致

The above test description: ++ and - participate in the mixed operation result is uncertain

• C++ only specifies the relative execution order of ++ and-corresponding instructions
• ++ and-the corresponding assembly instructions are not necessarily executed continuously
• In mixed operations, the assembly instructions of ++ and - may be interrupted in execution

2、Demo2

• What is the output of the following code?

  int i = 0;
  int j = ++i+++i+++i;
  
  int a = 1;
  int b = 4;
  int c = a+++b;
  
  int* p = &a;
  b = b/*p;   
  
  printf("i = %d\n", i);
  printf("j = %d\n", j);
  printf("a = %d\n", a);
  printf("b = %d\n", b);
  printf("c = %d\n", c);

compiler's greedy method

• The compiler as many characters as possible one by one from left to right
• When the read character cannot form a legal symbol with the character that has been read

Code analysis:

int j = ++i+++i+++i;    
    // 编译器先读到1个'+'，不知道啥意思；
    // 继续读到1个'+'，它觉得这是个前置的'++'
    // 然后读到了i，确定了这是个'++i'表达式
    // 继续读到1个'+'，这可能是一个加法的运算符 '+'
    // 继续读到1个'+'，这时候判断是一个后置的 '++'，后面再读任何数都不对了，读到变量，不合法；读到符号，也不对；
    // 这时候编译器就停止处理了，所以编译器就得到了 “++i++”
    // 然后计算得到了 “1++”，对一个右值1进行自增，自然会编译错误！

int c = a+++b;
    // 编译器依次读了3个字符'a++'，知道这是个后置的++
    // 然后读到了1个'+'，觉得这个可能是个加法运算符
    // 继续读到了b，后面也没有其他字符了，所以得到了 (a++) + b
    // 计算得到了c = 5

b = b/*p;    // error，编译器会将/*识别为注释，因此整个程序会编译失败
// 如果愿意是用b的值除以*p的值，那么就应该用括号或者空格表明
b = b / (*p); 或 b = b / *p; 

This article is summarized from "C Language Advanced Course" by Tang Zuolin from "Ditai Software Academy".
If there are any errors or omissions, please correct me .