字节交换与数组? [英] Byte Swap with an array?

问题描述

首先，请原谅我极其业余的编码知识.我在一家公司实习，并被指派在C ++中创建一个代码，该代码交换字节以获取正确的校验和值.

First of all, forgive my extremely amateur coding knowledge. I am intern at a company and have been assigned to create a code in C++ that swaps bytes in order to get the correct checksum value.

我正在阅读类似于以下内容的列表:S315FFF200207F7FFFFF42A000000000001B000000647CS315FFF2003041A00000FF7FFFFF0000001B00000064EDS315FFF2004042480000FF7FFFFF0000001E000000464F

I am reading a list that resembles something like: S315FFF200207F7FFFFF42A000000000001B000000647C S315FFF2003041A00000FF7FFFFF0000001B00000064ED S315FFF2004042480000FF7FFFFF0000001E000000464F

我已经使程序将该字符串转换为十六进制，然后将其转换为int，以便可以正确读取它.我没有读每行的前12个字符或后2个字符.

I have made the program convert this string to hex and then int so that it can be read correctly. I am not reading the first 12 chars or last 2 chars of each line.

我的问题是如何使转换后的int进行字节交换(小尾数到大尾数)，以便计算机可以读取?再次抱歉，这是一个糟糕的解释.

My question is how do I make the converted int do a byte swap (little endian to big endian) so that it is readable to the computer? Again I'm sorry if this is a terrible explanation.

我基本上需要占用每个字节(4个字母)并将其翻转.即:64C7翻转到C764等，等等.我该怎么做并将其放入新的阵列中?每行现在是一个字符串...

I need to essentially take each byte (4 letters) and flip them. i.e: 64C7 flipped to C764, etc etc etc. How would I do this and put it into a new array? Each line is a string right now...

到目前为止，这是我代码的一部分...

This is part of my code as of now...

    int j = 12;

                    for (i = 0; i < hexLength2 - 5; i++){

                        string convert1 = ODL.substr(j, 4);

                        short input_int = stoi(convert1);
                        short lowBit = 0x00FF & input_int;
                        short hiBit = 0xFF00 & input_int;

                        short byteSwap = (lowBit << 8) | (hiBit >> 8);

我认为我可能需要以某种方式将我的STOI转换为简称.

I think I may need to convert my STOI to a short in some way..

使用下面的答案代码，我得到以下信息...十六进制:8D->以141(十进制)的形式存储到内存(myMem =无符号短整数)->字节交换时:-29440这是怎么了?

Using the answer code below I get the following... HEX: 8D --> stored to memory (myMem = unsigned short) as 141 (decimal) -->when byte swapped: -29440 Whats wrong here??

            for (i = 0; i < hexLength2 - 5; i++){

                        string convert1 = ODL.substr(j, 2);
                        stringstream str1;
                        str1 << convert1;
                        str1 >> hex >> myMem[k];

                        short input_int = myMem[k];                     //byte swap 
                        short lowBit = 0x00FF & input_int;              
                        short hiBit = 0xFF00 & input_int;               
                        short byteSwap = (lowBit << 8) | (hiBit >> 8);  
                        cout << input_int << endl << "BYTE SWAP: " <<byteSwap <<"Byte Swap End" << endl;    
                        k++;
                        j += 2;

推荐答案

您也可以始终按位进行操作.(假设16位字)例如，如果您要字节交换一个int:

You can always do it bitwise too. (Assuming 16-bit word) For example, if you're byte swapping an int:

short input_int = 123; // each of the ints that you have

short input_lower_half = 0x00FF & input_int;
short input_upper_half = 0xFF00 & input_int;

// size of short is 16-bits, so shift the bits halfway in each direction that they were originally
 short byte_swapped_int = (input_lower_half << 8) | (input_upper_half >> 8)

我确切地尝试使用您的代码

My exact attempt at using your code

unsigned short myMem[20];
int k = 0;
string ODL = "S315FFF2000000008DC7000036B400003030303030319A";

int j = 12;
for(int i = 0; i < (ODL.length()-12)/4; i++) { // not exactly sure what your loop condition was

    string convert1 = ODL.substr(j, 4);
    cout << "substring is: " << convert1 << endl;
    stringstream str1;
    str1 << convert1;
    str1 >> hex >> myMem[k];

    short input_int = myMem[k];                     //byte swap
    unsigned short lowBit = 0x00FF & input_int; // changed this to unsigned
    unsigned short hiBit = 0xFF00 & input_int; // changed this to unsigned
    short byteSwap = (lowBit << 8) | (hiBit >> 8);
    cout << hex << input_int << " BYTE SWAPed as: " << byteSwap <<", Byte Swap End" << endl;
    k++;
    j += 4;
}

只重要的是将loBit和hiBit更改为无符号，因为它们是我们正在使用的临时值.

it only matters to change the loBit and hiBit to be unsigned since those are the temporary values we're using.

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