++And -- the essence of operators

• ++And -- operators correspond to two assembly instructions

  • Preposition

    • Variable increment (decrement) 1
    • Take variable values

  • Postposition

    • Take variable values
    • Variable increment (decrement) 1

++And -- operator usage analysis

• A troubling pair of brothers

int i = 0;

(i++) + (i++) + (i++);
(++i) + (++i) + (++i);

What are the values of the two expressions?

Case study: a headache brother

#include <stdio.h>

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

output_1: VC
i = 3
r = 0
i = 6
r = 18

//Output 2: GCC
i = 3
r = 0
i = 6
r = 16

//Analysis 1: [compilation - VC]

r = (i++) + (i++) + (i++);

    mov         eax,dword ptr [i]  
    add         eax,dword ptr [i]  
    add         eax,dword ptr [i]  
    mov         dword ptr [r],eax  ; r = i + i + i ==> r = 0 + 0 + 0 = 0
    
    mov         ecx,dword ptr [i]  
    add         ecx,1  
    mov         dword ptr [i],ecx  ; i = i + 1 = 1
    
    mov         edx,dword ptr [i]  
    add         edx,1  
    mov         dword ptr [i],edx  ; i = i + 1 = 2
     
    mov         eax,dword ptr [i]  
    add         eax,1  
    mov         dword ptr [i],eax  ; i = i + 1 = 3

r = (++i) + (++i) + (++i);

    mov         eax,dword ptr [i]  
    add         eax,1  
    mov         dword ptr [i],eax  ; i = i + 1 = 4
    
    mov         ecx,dword ptr [i]  
    add         ecx,1  
    mov         dword ptr [i],ecx  ; i = i + 1 = 5
    
    mov         edx,dword ptr [i]  
    add         edx,1  
    mov         dword ptr [i],edx  ; i = i + 1 = 6
    
    mov         eax,dword ptr [i]  
    add         eax,dword ptr [i]  
    add         eax,dword ptr [i]  
    mov         dword ptr [r],eax  ; r = i + i + i = 6 + 6 + 6 = 18

Analysis 2: [compilation GCC]

r = (i++) + (i++) + (i++);

int i = 0;
movl $0x0,0x1c(%esp)
int r = 0;
movl $0x0,0x18(%esp)

mov 0x1c(%esp),%eax
add %eax,%eax
add 0x1c(%esp),%eax
mov %eax,0x18(%esp)    ; r = i + i + i ==> r = 0 + 0 + 0 = 0

addl $0x1,0x1c(%esp)   ; i = i + 1 = 1
addl $0x1,0x1c(%esp)   ; i = i + 1 = 2
addl $0x1,0x1c(%esp)   ; i = i + 1 = 3

r = (++i) + (++i) + (++i);

    addl $0x1,0x1c(%esp)   ; i = i + 1 = 4
    addl $0x1,0x1c(%esp)   ; i = i + 1 = 5
    
    mov 0x1c(%esp),%eax    
    add %eax,%eax          ; eax = i + i = 10
    
    addl $0x1,0x1c(%esp)   ; i = i + 1 = 6 
    
    add 0x1c(%esp),%eax
    mov %eax,0x18(%esp)    ; r = i + eax = 16

• C only specifies the corresponding execution order of + + and -- corresponding instructions
• ++And -- corresponding assembly instructions may not be executed continuously
• In hybrid operations, assembly instructions for + + and -- may be interrupted [different compilers have different behaviors]

[C language gray area] + and -- participate in mixed operation results are uncertain

**"++ -- Do not mix with other operators"**

"Wonderful flowers" in interview

How does the compiler explain it?

• ++i+++i+++i

• a+++b

  • a++ +b
  • a + ++b

Greedy method: reading skill of + + - expression

• The compiler handles that each symbol should contain as many characters as possible
• The compiler reads as many characters as possible from left to right
• When the read in character is impossible to form a legal symbol with the read in character

Case study: greedy reading

#include <stdio.h>

int main()
{
    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("a = %d\n", a);
    printf("b = %d\n", b);
    printf("c = %d\n", c);
    
    return 0;
}

Output:
a = 2
b = 4
c = 5

//Analysis:
int j = ++i+++i+++i; ==> error: lvalue required as increment operand
                     ==> ++i++ ==> 1 ++ 
c = a+++b;           ==> a++ + b
b = b/*p;            ==> Commentary begins

• Space can be used as a rest of a complete symbol in C language. The compiler processes the previously read symbol immediately after reading the space
• Reasonable use of space, while increasing readability, clearly tells the compiler our purpose

Summary

• ++And -- operators may behave differently in mixed operations
• The compiler uses greedy method to deal with subexpressions in expressions
• Space can be used as a rest of a complete symbol in C language
• The compiler processes the previously read symbols as soon as it reads the spaces

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