Write "int" value to array of bytes

Question

1.00/5 (1 vote)

See more:

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 11-Jul-13 5:42am

vmvictor

Add a Solution

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;

3 solutions

Solution 4

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 );

Posted 12-Jul-13 0:26am

PrafullaVedante

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,

Solution 5

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.

Posted 12-Jul-13 15:28pm

User 59241

Updated 12-Jul-13 20:27pm

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,

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

Pascal-78 · Accepted Answer · 2013-07-11T06:22:00

Solution 1

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

Posted 11-Jul-13 6:22am

Pascal-78

Updated 11-Jul-13 6:26am

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;

vmvictor 11-Jul-13 16:02pm

Ok, thank you...everything is fine now :)