Help File:Money type

Money type (divided by 100)

The following custom type script will handle values that need to be divided by 100 to get to the correct value This type is used by some games like civilization 5 where the money and research technology is stored using this floating point type.

Example:

100.35 gold would be stored in memory as 10013

103.89 gold would be stored in memory as 10389

with this script you'd be able to scan for 100 and 103 respectively, and if you wanted to change the value to 700 you'd just change the value to 700 instead of 70000

 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,#100
 div ebx // divide eax by 100 and 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,#100
 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]