Difference between revisions of "Help File:old flash (div 8)"
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(Created page with '<hr><div align="left"><font face="Arial" color="#010101" class="Arial2"> This is just the money type adjusted to 8 (in case you didn't get it in the previous example)</font></d…') |
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| − | + | == Divide by 8 (old flash) == | |
| − | This is just the money type adjusted to 8 (in case you didn't get it in the previous example) | + | |
| − | alloc(TypeName,256) | + | |
| − | alloc(ByteSize,4) | + | This is just the money type adjusted to 8 (in case you didn't get it in the previous example). |
| − | alloc(ConvertRoutine,1024) | + | |
| − | alloc(ConvertBackRoutine,1024) | + | |
| − | TypeName: | + | alloc(TypeName,256) |
| − | db 'Civ 5 Float',0 | + | alloc(ByteSize,4) |
| − | ByteSize: | + | alloc(ConvertRoutine,1024) |
| − | dd 4 | + | alloc(ConvertBackRoutine,1024) |
| − | //The convert routine should hold a routine that converts the data to an | + | |
| − | nteger (in eax) | + | |
| − | //function declared as: stdcall int ConvertRoutine(unsigned char | + | TypeName: |
| − | *input); | + | db 'Civ 5 Float',0 |
| − | //Note: Keep in mind that this routine can be called by multiple threads | + | |
| − | at the same time. | + | |
| − | ConvertRoutine: | + | ByteSize: |
| − | [32-bit] | + | dd 4 |
| − | push ebp | + | |
| − | mov ebp,esp | + | |
| − | push ecx | + | //The convert routine should hold a routine that converts the data to an nteger (in eax) |
| − | mov ecx,[ebp+8] | + | //function declared as: stdcall int ConvertRoutine(unsigned char *input); |
| − | [/32-bit] | + | |
| − | //at this point ecx contains the address where the bytes are stored | + | |
| − | //save the used registers | + | //Note: Keep in mind that this routine can be called by multiple threads at the same time. |
| − | push edx //fun fact about ce's assembler, because push ebx does not | + | |
| − | exist in 64-bit it becomes the 64-bit push rdx automatically | + | |
| − | push ebx | + | ConvertRoutine: |
| − | //put the bytes into the eax register | + | [32-bit] |
| − | mov eax,[ecx] //second fun fact, addressing with 32-bit registers | + | push ebp |
| − | doesn't work in 64-bit, it becomes a 64-bit automatically (most of the | + | mov ebp,esp |
| − | time) | + | push ecx |
| − | xor edx,edx | + | mov ecx,[ebp+8] |
| − | mov ebx,#8 | + | [/32-bit] |
| − | div ebx //divide eax by 8and put the result in eax (and leftover in edx) | + | |
| − | pop ebx | + | |
| − | pop edx | + | //at this point ecx contains the address where the bytes are stored |
| − | //and now exit the routine | + | //save the used registers |
| − | [64-bit] | + | push edx //fun fact about ce's assembler, because push ebx does not exist in 64-bit it becomes the 64-bit push rdx automatically |
| − | ret | + | push ebx |
| − | [/64-bit] | + | |
| − | [32-bit] | + | |
| − | pop ecx | + | //put the bytes into the eax register |
| − | pop ebp | + | mov eax,[ecx] //second fun fact, addressing with 32-bit registers doesn't work in 64-bit, it becomes a 64-bit automatically (most of the time) |
| − | ret 4 | + | |
| − | [/32-bit] | + | |
| − | //The convert back routine should hold a routine that converts the given | + | xor edx,edx |
| − | integer back to a row of bytes (e.g when the user wats to write a new | + | mov ebx,#8 |
| − | value) | + | div ebx //divide eax by 8and put the result in eax (and leftover in edx) |
| − | //function declared as: stdcall void ConvertBackRoutine(int i, unsigned | + | |
| − | char *output); | + | |
| − | ConvertBackRoutine: | + | pop ebx |
| − | [32-bit] | + | pop edx |
| − | push ebp | + | //and now exit the routine |
| − | mov ebp,esp | + | [64-bit] |
| − | push edx //save the registers | + | ret |
| − | push ecx | + | [/64-bit] |
| − | mov edx,[ebp+0c] | + | [32-bit] |
| − | mov ecx,[ebp+08] | + | pop ecx |
| − | [/32-bit] | + | pop ebp |
| − | //at this point edx contains the address to write the value to | + | ret 4 |
| − | //and ecx contains the value | + | [/32-bit] |
| − | push eax | + | |
| − | push edx | + | |
| − | push ecx | + | //The convert back routine should hold a routine that converts the given integer back to a row of bytes (e.g when the user wats to write a new value) |
| − | mov eax,ecx //eax gets the given value | + | //function declared as: stdcall void ConvertBackRoutine(int i, unsigned char *output); |
| − | xor edx,edx //clear edx | + | ConvertBackRoutine: |
| − | mov ecx,#8 | + | [32-bit] |
| − | mul ecx //multiply eax and put the results into edx:eax (edx is ignored | + | push ebp |
| − | for this routine) | + | mov ebp,esp |
| − | pop ecx | + | push edx //save the registers |
| − | pop edx | + | push ecx |
| − | mov [edx],eax | + | mov edx,[ebp+0c] |
| − | pop eax | + | mov ecx,[ebp+08] |
| − | [64-bit] | + | [/32-bit] |
| − | //everything is back to what it was, so exit | + | |
| − | ret | + | |
| − | [/64-bit] | + | //at this point edx contains the address to write the value to |
| − | [32-bit] | + | //and ecx contains the value |
| − | //cleanup first | + | push eax |
| − | pop ecx | + | push edx |
| − | pop edx | + | push ecx |
| − | pop ebp | + | |
| − | ret 8 | + | |
| − | [ | + | mov eax,ecx //eax gets the given value |
| + | xor edx,edx //clear edx | ||
| + | mov ecx,#8 | ||
| + | mul ecx //multiply eax and put the results into edx:eax (edx is ignored for this routine) | ||
| + | | ||
| + | | ||
| + | pop ecx | ||
| + | pop edx | ||
| + | mov [edx],eax | ||
| + | pop eax | ||
| + | | ||
| + | | ||
| + | [64-bit] | ||
| + | //everything is back to what it was, so exit | ||
| + | ret | ||
| + | [/64-bit] | ||
| + | | ||
| + | | ||
| + | [32-bit] | ||
| + | //cleanup first | ||
| + | pop ecx | ||
| + | pop edx | ||
| + | pop ebp | ||
| + | ret 8 | ||
| + | [/32-bit] | ||
| + | |||
| + | |||
| + | == Links == | ||
| + | * [[Cheat Engine:Help File|Help File]] | ||
| + | |||
| + | * [[Help_File:Money_type|Back]] | ||
| + | |||
| + | * [[Help_File:Big_endian_2_byte|Next]] | ||
Revision as of 07:57, 10 March 2017
Divide by 8 (old flash)
This is just the money type adjusted to 8 (in case you didn't get it in the previous example).
alloc(TypeName,256) alloc(ByteSize,4) alloc(ConvertRoutine,1024) alloc(ConvertBackRoutine,1024) TypeName: db 'Civ 5 Float',0 ByteSize: dd 4 //The convert routine should hold a routine that converts the data to an nteger (in eax) //function declared as: stdcall int ConvertRoutine(unsigned char *input); //Note: Keep in mind that this routine can be called by multiple threads at the same time. ConvertRoutine: [32-bit] push ebp mov ebp,esp push ecx mov ecx,[ebp+8] [/32-bit] //at this point ecx contains the address where the bytes are stored //save the used registers push edx //fun fact about ce's assembler, because push ebx does not exist in 64-bit it becomes the 64-bit push rdx automatically push ebx //put the bytes into the eax register mov eax,[ecx] //second fun fact, addressing with 32-bit registers doesn't work in 64-bit, it becomes a 64-bit automatically (most of the time) xor edx,edx mov ebx,#8 div ebx //divide eax by 8and put the result in eax (and leftover in edx) pop ebx pop edx //and now exit the routine [64-bit] ret [/64-bit] [32-bit] pop ecx pop ebp ret 4 [/32-bit] //The convert back routine should hold a routine that converts the given integer back to a row of bytes (e.g when the user wats to write a new value) //function declared as: stdcall void ConvertBackRoutine(int i, unsigned char *output); ConvertBackRoutine: [32-bit] push ebp mov ebp,esp push edx //save the registers push ecx mov edx,[ebp+0c] mov ecx,[ebp+08] [/32-bit] //at this point edx contains the address to write the value to //and ecx contains the value push eax push edx push ecx mov eax,ecx //eax gets the given value xor edx,edx //clear edx mov ecx,#8 mul ecx //multiply eax and put the results into edx:eax (edx is ignored for this routine) pop ecx pop edx mov [edx],eax pop eax [64-bit] //everything is back to what it was, so exit ret [/64-bit] [32-bit] //cleanup first pop ecx pop edx pop ebp ret 8 [/32-bit]
