Front and back operators

This is the answer from Thunder: [cpp] # include <stdio. h> int main (void) {int a, B; a = 5; a = a + a ++; printf ("a = % d \ n", ); a = 5; B = a + a ++; printf ("B = % d \ n", B); return 0 ;}the output is: disassembly process: [cpp] PUBLIC _ main EXTRN _ printf: NEAR _ data segment $ SG529 DB 'a = % d', 0aH, 00 H $ SG530 DB 'B = % d', 0aH, 00 H _ data ends _ text segment _ a $=-4 _ B $=-8 _ main PROC NEAR; 3: int main (void) {push ebp mov ebp, esp sub esp, 8; 4: int a, B; 5:; 6: a = 5; mov DWORD PTR _ a $ [ebp], 5; Value assignment, a = 5; 7: a = a + (a ++); mov eax, dword ptr _ a $ [ebp] add eax, dword ptr _ a $ [ebp]; a + = a (a = 10) mov dword ptr _ a $ [ebp], eax mov ecx, dword ptr _ a $ [ebp] add ecx, 1; a ++, a = 11 mov dword ptr _ a $ [ebp], ecx; 8: printf ("a = % d \ n", a); mov edx, dword ptr _ a $ [ebp] push edx push offset flat: $ SG529 call _ printf add esp, 8; 9:; 10: a = 5; mov dword ptr _ a $ [ebp], 5; 11: B = a + (a ++); mov eax, DWORD PTR _ a $ [ebp] add eax, dword ptr _ a $ [ebp]; a + = a (a = 10, B = 10) mov dword ptr _ B $ [ebp], eax mov ecx, dword ptr _ a $ [ebp] add ecx, 1; a ++, a = 11, B unchanged mov DWORD PTR _ a $ [ebp], ecx; 12: printf ("B = % d \ n", B); mov edx, dword ptr _ B $ [ebp] push edx push OFFSET FLAT: $ SG530 call _ printf add esp, 8; 13: return 0; xor eax, eax; 14:} mov esp, ebp pop ebp ret 0 _ main ENDP _ text ends end first, it must be clear that: the Operation Order of the expression is 1 from left to right. for a = a + a ++, the operation procedure is: a = 5 + 5 = 10, and then a automatically adds 2. for B = a + a ++, the computation process is: B = 5 + 5 = 10. Then, a then adds itself and then looks at a slightly complex program: [cpp] # include <stdio. h> int main (void) {int a, B; a = 5; a = a ++; printf ("a = % d \ n ", a); a = 5; B = (++ a) + (a ++) + (++) + (a --); printf ("a = % d \ n", a); printf ("B = % d \ n", B); return 0 ;} the calculation result is: Do not paste the ASM code ------------------------------- first of all, it must be clear that the operation order of the expression is from left to right. The statement a = a ++; the execution process is: 1. the result of the first a ++ is 5, and the result of the second a ++ is 52. in this case, the expression returns a = 5 + 5 = 10, and then a executes two self-added statements B = (++ a) + (a ++) + (a ++) + (++ a) + (a --); the execution process is: 1. the first result of ++ a is 62. the results of both a ++ are 63. the result of the first ++ a is 7, and the result of the last ++ a is 84. the result of a -- is 8. The final expression result is B = 6 + 6 + 6 + 7 + 8 + 8 = 41. Then, the value of a is updated, a has a total of two post-auto-increment operations and one post-auto-subtraction operation, so a = a + 2-1 = 9 (because the pre-operation has been performed during expression operation, at the end of the expression operation, the value of a is 8)