* 지금 테스트 코드는 전부 전역변수 > Data 영역에서 메모리 할당
* Alt + 커서 : 세로로 블록 설정
* 대입 연산자
=
* 산술 연산자
+, -, *, /, %
+=, -=, *=, /=, %=
#include <iostream>
using namespace std;
int a, b, result;
int main(void)
{
00DC1000 push ebp
00DC1001 mov ebp,esp
// 대입 연산
a = 10;
00DC1003 mov dword ptr [a (0DC33D0h)],0Ah
b = 3;
00DC100D mov dword ptr [b (0DC33C8h)],3
// + - * / %
result = a + b;
00DC1017 mov eax,dword ptr [a (0DC33D0h)]
00DC101C add eax,dword ptr [b (0DC33C8h)]
00DC1022 mov dword ptr [result (0DC33CCh)],eax
result = a - b;
00DC1027 mov ecx,dword ptr [a (0DC33D0h)]
00DC102D sub ecx,dword ptr [b (0DC33C8h)]
00DC1033 mov dword ptr [result (0DC33CCh)],ecx
result = a * b;
00DC1039 mov edx,dword ptr [a (0DC33D0h)]
00DC103F imul edx,dword ptr [b (0DC33C8h)]
00DC1046 mov dword ptr [result (0DC33CCh)],edx
result = a / b;
00DC104C mov eax,dword ptr [a (0DC33D0h)]
00DC1051 cdq
00DC1052 idiv eax,dword ptr [b (0DC33C8h)]
00DC1058 mov dword ptr [result (0DC33CCh)],eax
result = a % b;
00DC105D mov eax,dword ptr [a (0DC33D0h)]
00DC1062 cdq
00DC1063 idiv eax,dword ptr [b (0DC33C8h)]
00DC1069 mov dword ptr [result (0DC33CCh)],edx
return 0;
00DC106F xor eax,eax
}
00DC1071 pop ebp
00DC1072 ret
* 증감 연산자
++, --
#include <iostream>
using namespace std;
int hp;
int main(void)
{
002E1000 push ebp
002E1001 mov ebp,esp
hp = 100;
002E1003 mov dword ptr [hp (02E33C8h)],64h
// ++, --
hp = hp + 1;
002E100D mov eax,dword ptr [hp (02E33C8h)]
002E1012 add eax,1
002E1015 mov dword ptr [hp (02E33C8h)],eax
hp += 1;
002E101A mov ecx,dword ptr [hp (02E33C8h)]
002E1020 add ecx,1
002E1023 mov dword ptr [hp (02E33C8h)],ecx
hp++;
002E1029 mov edx,dword ptr [hp (02E33C8h)]
002E102F add edx,1
002E1032 mov dword ptr [hp (02E33C8h)],edx
++hp;
002E1038 mov eax,dword ptr [hp (02E33C8h)]
002E103D add eax,1
002E1040 mov dword ptr [hp (02E33C8h)],eax
hp--;
002E1045 mov ecx,dword ptr [hp (02E33C8h)]
002E104B sub ecx,1
002E104E mov dword ptr [hp (02E33C8h)],ecx
--hp;
002E1054 mov edx,dword ptr [hp (02E33C8h)]
002E105A sub edx,1
002E105D mov dword ptr [hp (02E33C8h)],edx
return 0;
002E1063 xor eax,eax
}
002E1065 pop ebp
002E1066 ret
* 시프트 연산자
: 왼쪽 혹은 오른쪽으로 밀어주는 역할
<<, >>
#include <iostream>
using namespace std;
int hp;
int main(void)
{
006E1000 push ebp
006E1001 mov ebp,esp
hp = 0b100;
006E1003 mov dword ptr [hp (06E33C8h)],4
hp >>= 1;
006E100D mov eax,dword ptr [hp (06E33C8h)]
006E1012 sar eax,1
006E1014 mov dword ptr [hp (06E33C8h)],eax
return 0;
006E1019 xor eax,eax
}
006E101B pop ebp
006E101C ret
* 최상단 비트가 부호 비트이기 때문에 아래와 같은 예상치 못한 상황이 발생할 수 있음
[해결] unsigned 처리
* 비교 연산자
<, >, <=, >=, ==, !=
#include <iostream>
using namespace std;
int a, b;
bool result;
int main(void)
{
00EC1000 push ebp
00EC1001 mov ebp,esp
00EC1003 sub esp,18h
a = 100;
00EC1006 mov dword ptr [a (0EC33D0h)],64h
b = 50;
00EC1010 mov dword ptr [b (0EC33C8h)],32h
result = (a < b);
00EC101A mov eax,dword ptr [a (0EC33D0h)]
00EC101F cmp eax,dword ptr [b (0EC33C8h)]
00EC1025 jge main+30h (0EC1030h)
00EC1027 mov dword ptr [ebp-4],1
00EC102E jmp main+37h (0EC1037h)
00EC1030 mov dword ptr [ebp-4],0
00EC1037 mov cl,byte ptr [ebp-4]
00EC103A mov byte ptr [result (0EC33CCh)],cl
result = (a > b);
00EC1040 mov edx,dword ptr [a (0EC33D0h)]
00EC1046 cmp edx,dword ptr [b (0EC33C8h)]
00EC104C jle main+57h (0EC1057h)
00EC104E mov dword ptr [ebp-8],1
00EC1055 jmp main+5Eh (0EC105Eh)
00EC1057 mov dword ptr [ebp-8],0
00EC105E mov al,byte ptr [ebp-8]
00EC1061 mov byte ptr [result (0EC33CCh)],al
result = (a <= b);
00EC1066 mov ecx,dword ptr [a (0EC33D0h)]
00EC106C cmp ecx,dword ptr [b (0EC33C8h)]
00EC1072 jg main+7Dh (0EC107Dh)
00EC1074 mov dword ptr [ebp-0Ch],1
00EC107B jmp main+84h (0EC1084h)
00EC107D mov dword ptr [ebp-0Ch],0
00EC1084 mov dl,byte ptr [ebp-0Ch]
00EC1087 mov byte ptr [result (0EC33CCh)],dl
result = (a >= b);
00EC108D mov eax,dword ptr [a (0EC33D0h)]
00EC1092 cmp eax,dword ptr [b (0EC33C8h)]
00EC1098 jl main+0A3h (0EC10A3h)
00EC109A mov dword ptr [ebp-10h],1
00EC10A1 jmp main+0AAh (0EC10AAh)
00EC10A3 mov dword ptr [ebp-10h],0
00EC10AA mov cl,byte ptr [ebp-10h]
00EC10AD mov byte ptr [result (0EC33CCh)],cl
result = (a == b);
00EC10B3 mov edx,dword ptr [a (0EC33D0h)]
00EC10B9 cmp edx,dword ptr [b (0EC33C8h)]
00EC10BF jne main+0CAh (0EC10CAh)
00EC10C1 mov dword ptr [ebp-14h],1
00EC10C8 jmp main+0D1h (0EC10D1h)
00EC10CA mov dword ptr [ebp-14h],0
00EC10D1 mov al,byte ptr [ebp-14h]
00EC10D4 mov byte ptr [result (0EC33CCh)],al
result = (a != b);
00EC10D9 mov ecx,dword ptr [a (0EC33D0h)]
00EC10DF cmp ecx,dword ptr [b (0EC33C8h)]
00EC10E5 je main+0F0h (0EC10F0h)
00EC10E7 mov dword ptr [ebp-18h],1
00EC10EE jmp main+0F7h (0EC10F7h)
00EC10F0 mov dword ptr [ebp-18h],0
00EC10F7 mov dl,byte ptr [ebp-18h]
00EC10FA mov byte ptr [result (0EC33CCh)],dl
return 0;
00EC1100 xor eax,eax
}
00EC1102 mov esp,ebp
00EC1104 pop ebp
00EC1105 ret
* 논리 연산자
!, &&, ||
#include <iostream>
using namespace std;
bool a, b, result;
int main(void)
{
00161000 push ebp
00161001 mov ebp,esp
00161003 sub esp,0Ch
a = true;
00161006 mov byte ptr [a (01633CAh)],1
b = false;
0016100D mov byte ptr [b (01633C8h)],0
result = !a;
00161014 movzx eax,byte ptr [a (01633CAh)]
0016101B test eax,eax
0016101D jne main+28h (0161028h)
0016101F mov dword ptr [ebp-4],1
00161026 jmp main+2Fh (016102Fh)
00161028 mov dword ptr [ebp-4],0
0016102F mov cl,byte ptr [ebp-4]
00161032 mov byte ptr [result (01633C9h)],cl
result = (a || b);
00161038 movzx edx,byte ptr [a (01633CAh)]
0016103F test edx,edx
00161041 jne main+57h (0161057h)
00161043 movzx eax,byte ptr [b (01633C8h)]
0016104A test eax,eax
0016104C jne main+57h (0161057h)
0016104E mov dword ptr [ebp-8],0
00161055 jmp main+5Eh (016105Eh)
00161057 mov dword ptr [ebp-8],1
0016105E mov cl,byte ptr [ebp-8]
00161061 mov byte ptr [result (01633C9h)],cl
result = (a && b);
00161067 movzx edx,byte ptr [a (01633CAh)]
0016106E test edx,edx
00161070 je main+86h (0161086h)
00161072 movzx eax,byte ptr [b (01633C8h)]
00161079 test eax,eax
0016107B je main+86h (0161086h)
0016107D mov dword ptr [ebp-0Ch],1
00161084 jmp main+8Dh (016108Dh)
00161086 mov dword ptr [ebp-0Ch],0
0016108D mov cl,byte ptr [ebp-0Ch]
00161090 mov byte ptr [result (01633C9h)],cl
return 0;
00161096 xor eax,eax
}
00161098 mov esp,ebp
0016109A pop ebp
0016109B ret
movsx
상위 비트는 값의 부호 비트(최상단 비트)로 채워진다
movzx
상위 비트는 항상 0으로 채워진다
test
비트 단위 AND 연산 후 플래그 설정
ㄴ je는 ZF = 1일 때, 점프
ㄴ jne는 ZF = 0일 때, 점프
00161014 movzx eax,byte ptr [a (01633CAh)]
▶ 00000000 00000000 00000000 00000001
0016101B test eax,eax
▶ 값 1, ZF는 0
0016101D jne main+28h (0161028h)
▶ ZF가 0임
▶ 00161028 mov dword ptr [ebp-4],0
* 해당 글은 게임 프로그래머 입문 올인원 강의를 보고 정리한 글입니다.
[게임 프로그래머 입문 올인원] C++ & 자료구조/알고리즘 & STL & 게임 수학 & Windows API & 게임 서버
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