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Hi,
I have this array :
C++
`BYTE set[6] = { 0x15, 0x12, 0x84, 0x03, 0x00, 0x00 }`
and i need to insert this value : int Value = 900; ....on last 4 bytes. Practically to convert from int to hex and then to write inside the array...
Is this possible ?

I already have "BitConverter::GetBytes" function, but that's not enough :(

Thank you,
Posted
Comments
pasztorpisti 11-Jul-13 12:16pm    
If we assume that int is 32 bit then we can say *(int*)&set[sizeof(set)-sizeof(int)] = Value;. This is ugly but if you want to do this...
Dholakiya Ankit 13-Jul-13 5:30am    
int value = 900;
BYTE* pValue=(BYTE*)&value;

To convert an int to 4 bytes, you do not need to go through hex :


2 solutions,
1) use your int as a byte array :
C++
int value = 900;
BYTE* pValue=(BYTE*)&value;
//pValue[0], pValue[1], pValue[2] & pValue[3] will contain the 4 bytes.


2) use shift on your int in a loop :
C++
int value = 900;
int tmp=value;
for(int i=0;i<4;++i)
{
  initialized_byte_array[i]=(tmp& 0xFF);
  tmp=tmp>>8;
}

The tmp var is used to keep value intact.

with
C++
#define BYTE unsigned char

or
C++
#define BYTE char
 
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v2
Comments
lewax00 11-Jul-13 12:32pm    
I think that covers most of the options, +5.

I'm a little hazy on the details, but could you also use a union for this?
Pascal-78 11-Jul-13 15:21pm    
the union could be :
typedef union { int ivalue; char bvalues[4]; } theUnion;

theUnion v;
v.ivalue; // To Get or Set the int value
v.bvalues[0 to 3]; // To Get or Set the byte values
vmvictor 11-Jul-13 13:45pm    
Ok, i have fix it :)
Great job Pascal-78, was so simple and i lost 2 days reading all over :))
One more lesson learned .
Thank you,
vmvictor 11-Jul-13 14:18pm    
If i want to put this "int value = 900;" into ""cout << 'enter your value' <<endl; - cin >> value"....i need a function for this too ?
Pascal-78 11-Jul-13 15:15pm    
You're right, to let the user enter the value, you just need to have the value variable declared as an int and use the instructions you mentioned :
int value;
cout << "Enter your value" << endl;
cin >> value;
I have two solutions which are based on the copying the contents ... so please check if it suits to your purpose...


Solution 1

You can use stringstream for this purpose.
What you will need to do is ... write it once .... and read it 8 times....

Pseudo code will be like this
C++
int value=900;
stringstream stream;
stream.write( (char*)&value, sizeof( value ) );   ///Writing it once

BYTE bytes[4];
for ( int i = 0; i < 4; i++ )
{
   stream.read( &bytes[i], sizeof( BYTE ) );

} 
Solution 2
Second solution is ..... copy whole memory to the byte array

C++
int value=900;
BYTE bytes[4];
memcpy( bytes, value, 4 );
 
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Comments
vmvictor 12-Jul-13 10:16am    
I already found the solution...posted by Pascal-78, but thank you for your solution.

Now, i have a new problem and i will post it soon :)
vmvictor 12-Jul-13 11:40am    
Hi again :)
I have a new problem now...i discovered that i need 2 values not just one, to make it work.
When i press a key, the program must switch between this two values.
With switch case, or with (GetAsyncKeyState(VK_TAB) & 1) is the best to do it ?
And is this function working in this case ?
Posted by Pascal-78 : int tmp = atkSpd; for(int i=6; i<10; ++i) { byt[i] = (tmp & 0xFF); tmp = tmp >> 8; }
Or i need something else to get the job done ?

Thank you,
I prefer this approach - entirely with pointers and casting. This lets the compiler do all the work and removes the need for byte copying. It will work no matter what the size of set or int.
This will always put the int value at the end of the BYTE array. Reading back is just the inverse.
To be completely robust there should be a check that set is large enough for an int.


C++
typedef unsigned char BYTE;

BYTE set[6] = { 0x15, 0x12, 0x84, 0x03, 0x00, 0x00 };

int inValue = 900;

// Write
* ((int *) (set + (sizeof(set) - sizeof(int)))) = inValue;

// Read
int outValue = * ((int *) (set + (sizeof(set) - sizeof(int))));


In reply to additional question:

C++
BYTE set[16]={0xC7,0x80,0xEA,0x04,0x00,0x00,0xB0,0x04,0x00,0x00,0x8B,0xE5,0x5D,0xC2,0x08,0x00 };

// Write
* ((int *) (set + 5)) = inValue;

// Read
int outValue = * ((int *) (set + 5));


Read up on pointers. Also if you run this code in a debugger you can verify that it works.
 
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v4
Comments
vmvictor 13-Jul-13 1:14am    
And how this function can modify only the 6th,7th,8th,9th bytes from this array : BYTE set[16]={ 0xC7,0x80,0xEA,0x04,0x00,0x00,0xB0,0x04,0x00,0x00,0x8B,0xE5,0x5D,0xC2,0x08,0x00 }; ?
[no name] 13-Jul-13 3:28am    
Have added to the answer.
vmvictor 13-Jul-13 5:45am    
Returns value = 0 ......
[no name] 13-Jul-13 10:04am    
That statement makes no sense. I have done you the courtesy of answering an extra question. Where's your code?
vmvictor 13-Jul-13 13:48pm    
BYTE atkByt[30] = { 0xC7,0x83,0x1A,0x05,0x00,0x00,0x00,0x00,0x00,0x00,0x48,0x8B,0x74,0x24,0x40,
0x48,0x8B,0x6C,0x24,0x48,0x48,0x8B,0x5C,0x24,0x50,0x48,0x83,0xC4,0x58,0xC3 };

int tempA = atkSpd; for(int c=6; c<10; ++c) { atkByt[c] = (tempA & 0xFF); tempA = tempA >> 8; } // Write 8 bytes starting with 6th position from left to right. (solution posted by Pascal-78 earlier). Thank you,

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