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我尝试过各种各样的事情。不幸的是，我没有保留以前的代码记录，所以我只会向您展示我目前拥有的内容：

    string l2n(std::string input) {
    std::string original[26] = {"A","B","C","D","E", "F", "G", "H", "I", "J", "K," "L", "M", "N", "O", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z"};
    //std::string originalLower[26] = {"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"};
    std::string newval[26] = {"1", "2", "3", "4", "5", "6", "7", "8", "9", "10", "11", "12", "13", "14", "15", "16", "17", "18", "19", "20", "21", "22", "23", "24", "25", "26"};
    string neworig = (original[0], original[1], original[2], original[3], original[4], original[5], original[6], original[7], original[8], original[9], original[10], original[11], original[12], original[13], original[14], original[15], original[16], original[17], original[18], original[19], original[20], original[21], original[22], original[23], original[24], original[25]);
    string newnewval = (newval[0], newval[1], newval[2], newval[3], newval[4], newval[5], newval[6], newval[7], newval[8], newval[9], newval[10], newval[11], newval[12], newval[13], newval[14], newval[15], newval[16], newval[17], newval[18], newval[19], newval[20], newval[21], newval[22], newval[23], newval[24], newval[25]);
    size_t posit = input.find(neworig.c_str());
    string hash2345;
    hash2345 = input.replace(posit, std::string(neworig).length(), newnewval.c_str());
    return hash2345;
}

编译得很好，但输出“26HELLO”和我的输入“HELLO”。它实际上只是将 26 附加到给定的任何输入的开头..我猜它与使用 input.replace 使用长度有关，但我不知道还能做什么，谷歌没有找到任何事物。谢谢你的帮助！

4

2 回答 2

1

我认为你误解了什么string::replace。它不会将一个字符串中的字符替换为另一个字符串中的等价字符。它用另一个字符串的全部替换一个字符串中的一个字符范围。

例如，

std::string test = "This is a test.";

// Replace 2 characters, starting at position 5 with "was"
test.replace(5, 2, "was");

std::cout << test;  // Outputs "This was a test."

对于您正在做的事情，您将需要一个循环来对输入中的每个字符执行操作，例如

string hash2345;

// For each character in input
for (int i = 0; i < input.length(); ++i) {

  // Try to find the string equal to the i'th character of input in 
  // the "original" array.
  string* original_val_ptr = std::find(original, original + 26, 
                                       std::string(input[i]));

  // Make sure that it actually found something (this handles the case
  // where input has some other characters, like lowercase or punctuation).
  if (original_val_ptr != NULL) {

    // Find the position of this value in the original array.
    size_t pos = original_val_ptr - original;

    // Add the string in the equivalent position in the "newval" array
    // to the output.
    hash2345 += newval[pos];
  }
}

return hash2345;

现在可以对您的代码进行一些额外的改进，我相信其他答案会建议它们，但鉴于您似乎是一个相对较新的程序员，我选择让这个答案尽可能简单可能并且只关注您显然试图实现的算法的正确性。

此外，这适用于一组字符到另一组字符的任何映射，即使它不是字母到数字。

于 2012-11-22T17:16:11.057 回答
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根据上面的评论，并假设您使用的是 ASCII，那么以下就是您想要的

string l2n(std::string input)
{
    ostringstream buf;
    for (string::const_iterator i = input.begin(); i != input.end() ++i)
    {
        buf << (*i - 'A') + 1;
    }
    return buf.str();
}
于 2012-11-22T17:13:03.777 回答