转换为十六进制 - 代码审查 [英] Converting double to hexadecimal - Code review

问题描述

我有以下代码，它接受一个double值并将其转换为十六进制表示，反之亦然。我想知道是否有任何潜在的问题 - 是否我忽略了某些东西。

I have the following code that takes a double value and converts it to a hexadecimal representation, and vice versa. I would like to know whether there are any potential problems with it - whether I have overlooked something.

double hex_to_double2(string &hexString)
{
    unsigned char byte_string[sizeof(double)];
    int number;

    int j = 0;

    for(int i = 0; i < hexString.size() ; i += 2)
    {
        sscanf(&hexString[i], "%02x", &number);
        byte_string[j] = (unsigned char)number;
        ++j;
    }

    double p = (double&)byte_string;
    return p;
}

std::string double_to_hex_string(double d)
{
    unsigned char *buffer = (unsigned char*)&d;
    int bufferSize = sizeof(double);

    char converted[bufferSize * 2];

    int j = 0;
    for(int i = 0 ; i < bufferSize ; ++i)
    {
        sprintf(&converted[j*2], "%02X", buffer[i]);
        ++j;
    }
    string hex_string(converted);
    return hex_string;
}

它似乎工作得很好。但是我有人告诉我， char convert [bufferSize * 2]; 应该是 char convert [bufferSize * 2 + 1];

It seems to work fine. But I have been told by someone that char converted[bufferSize * 2]; should be char converted[bufferSize * 2 + 1];

是这样的吗？

推荐答案

，转换的大小应为 bufferSize * 2 + 1 。您需要在字符串结尾处留空零字符或空字符。

Yes, the size of converted should be bufferSize * 2 + 1. You need room for a zero or null character at the end of the string.

sprintf 正在插入空字符在字符串的结尾，但你没有为它腾出空间;这意味着它会破坏缓冲区之后的内存中的某些值。很难预测这可能有什么副作用;它可能会弄乱一些其他变量的值，它可能会崩溃你的程序，或者它可能只是看起来工作正常。

sprintf is inserting a null character at the end of the string, but you didn't make space for it; this means it's destroying some value in memory that comes just after the buffer. It's hard to predict what side effect this might have; it might mess up the value of some other variable, it might crash your program, or it might just appear to work fine.

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