写“int”值到字节数组 [英] Write "int" value to array of bytes

问题描述

我有这个数组：

Hi,
I have this array :

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

我需要插入这个值：int Value = 900; ....最后4个字节。实际上从int转换为十六进制然后写入数组内...

这可能吗？



我已经有了BitConverter :: GetBytes功能，但这还不够:(



谢谢，

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,

推荐答案

转换一个int到4个字节，你不需要通过十六进制：





2解决方案，

1）使用你的int作为字节数组：

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 :

int value = 900;
BYTE* pValue=(BYTE*)&value;
//pValue[0], pValue[1], pValue[2] & pValue[3] will contain the 4 bytes.

2）在循环中对int使用shift：

2) use shift on your int in a loop :

int value = 900;
int tmp=value;
for(int i=0;i<4;++i)
{
  initialized_byte_array[i]=(tmp& 0xFF);
  tmp=tmp>>8;
}

tmp var用于保持价值不变。



with

The tmp var is used to keep value intact.

with

#define BYTE unsigned char

或

or

#define BYTE char

我有两个基于复制内容的解决方案...所以请检查它是否适合您的目的...





解决方案1 ​​



您可以将stringstream用于此目的。

你需要做的是......写一次....并阅读8次....



伪代码将是这样的

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

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 ) );

} 

解决方案2

第二个解决方案是.....复制整个内存到字节数组

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

int value=900;
BYTE bytes[4];
memcpy( bytes, value, 4 );

我更喜欢这种方法 - 完全使用指针和转换。这使编译器可以完成所有工作，并且无需进行字节复制。无论set或int的大小如何，它都可以工作。

这将始终将int值放在BYTE数组的末尾。回读就是反过来了。

为了完全健壮，应该检查一下这个集合是否足够大。

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.

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))));

回复其他问题：

In reply to additional question:

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));

阅读指针。此外，如果您在调试器中运行此代码，则可以验证它是否有效。

Read up on pointers. Also if you run this code in a debugger you can verify that it works.

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