将二进制文件转换为十六进制表示法 [英] Convert binary file to hex notation

问题描述

我想为我在参数中输入的二进制文件获取此十六进制表示法:

I would like to obtain this hex notation for a binary I enter in parameter:

我获得的输出以及我想要的东西

The output I obtain and what I want:

这是我编写的代码，我没有好的十六进制数字(对于5A之后的部分)，我做错了什么?如何正确地将我读取的字节转换为十六进制? 谢谢.

This is the code I written, I don't have the good hex number (for the part after 5A) , what I am doing wrong ? How to convert properly the byte I read to hex ? Thanks.

int main(int argc, char *argv[])
{

    std::string parameter = "The\\Path\\To\My\exe.exe";
    ifstream::pos_type size;
    char * memblock;

    ifstream file(parametre, ios::in | ios::binary | ios::ate);
    if (file.is_open())
    {
        size = file.tellg();
        memblock = new char[size];
        file.seekg(0, ios::beg);
        file.read(memblock, size);
        file.close();

        cout << "the complete file content is in memory" << endl;
        string str = string(memblock, size);
        string hexContent = "";
        int maxColumn = 0;

        std::stringstream ss;
        int column = 0;
        for (int i = 0; i < size; ++i) 
        {       
            ss << std::hex << (int)str[i];
            if (column == 8)
            {
                ss << '\n';
                column = 0;
            }
            column++;

        }

        std::string mystr = ss.str();
        cout << mystr;
    }
    return 0;
}

推荐答案

看起来像char在您的系统上签名，并且您是符号扩展名的受害者.例如0x90是负数，因此当将其转换为int时，必须进行负运算，结果为0xffffff90.

Looks like char is signed on your system and you are the victim of sign extension. For example 0x90 is a negative, so when it's converted into an int, that negativity has to be carried through, resulting in 0xffffff90.

将文件读取到unsigned char，如果可用，则从<cstdint>读取uint8_t，而不是char的数组.

Read the file into unsigned char, or uint8_t from <cstdint> if it is available, instead of an array of char.

char * memblock;

成为

uint8_t * memblock;

然后

memblock = new char[size];  

成为

memblock = new uint8_t[size];  

，以后不要将其转换为string.

and don't convert it into a string later.

string str = string(memblock, size);

是没有意义的，您可以很容易地从memblock中读取内容，并撤消我们之前建立的无符号性.只需从memblock

is pointless, you could just as easily have read from memblock, and undoes the unsigned-ness we've established earlier. Just read out of memblock

别忘了

delete[] memblock;

完成后.导致

使用std::vector.它会自行清理.

std::vector<uint8_t> memblock(size);
file.seekg(0, ios::beg);
file.read(reinterpret_cast<char*>(memblock.data()), size); 
//or file.read(reinterpret_cast<char*>(&memblock[0]), size); if no or data method 

这篇关于将二进制文件转换为十六进制表示法的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！