为什么我得到两个输出？

Question

我一直在为我的生物课写这段代码。这个想法是用户输入DNA和代码输出，相应的mRNA和it.It的氨基酸序列几乎没有任何错误，但是我注意到它有时会给出两个输出而不是一个输出。(例如，"Ile Met“而不是"Met")我也知道我的代码很乱，所以请您也帮忙一下，我应该更改哪些部分和如何更改？

#include <iostream>
#include <cstring>
using namespace std;

int main()
{
    string sdna;
    cout << "Enter codogener strang: ";
    cin >> sdna;
    int n = sdna.size();
    char dna[100];
    char mrna[100];
    strcpy_s(dna, sdna.c_str());
    cout << "DNA: ";
    for (int i = 0; i < n; i++) {
        cout << dna[i];
        if (i % 3 == 2)
            cout << " ";
    }
    for (int i = 0; i < n; i++)
    {
        if (dna[i] == 'A')
            mrna[i] = 'U';
        else if (dna[i] == 'T')
            mrna[i] = 'A';
        else if (dna[i] == 'G')
            mrna[i] = 'C';
        else if (dna[i] == 'C')
            mrna[i] = 'G';
    }
    cout << "\nmRNA: ";
    for (int i = 0; i < n; i++)
    {
        cout << mrna[i];
        if (i % 3 == 2)
            cout << " ";
    }
    cout << "\n";
    for (int k = 0; k < n; k++) {
        if ((mrna[k * 3] == 'A') && (mrna[k * 3 + 1] == 'U') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A'))
            cout << "Ile ";
        if ((mrna[k * 3] == 'A') && (mrna[k * 3 + 1] == 'U') && (mrna[k * 3 + 2] == 'G'))
            cout << "Met ";
        if ((mrna[k * 3] == 'A') && (mrna[k * 3 + 1] == 'C') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A' || 'G'))
            cout << "Thr ";
        if ((mrna[k * 3] == 'A') && (mrna[k * 3 + 1] == 'A') && (mrna[k * 3 + 2] == 'U' || 'C'))
            cout << "Asn ";
        if ((mrna[k * 3] == 'A') && (mrna[k * 3 + 1] == 'A') && (mrna[k * 3 + 2] == 'G' || 'A'))
            cout << "Lys ";
        if ((mrna[k * 3] == 'A') && (mrna[k * 3 + 1] == 'G') && (mrna[k * 3 + 2] == 'U' || 'C'))
            cout << "Ser ";
        if ((mrna[k * 3] == 'A') && (mrna[k * 3 + 1] == 'G') && (mrna[k * 3 + 2] == 'G' || 'A'))
            cout << "Arg ";
        if ((mrna[k * 3] == 'G') && (mrna[k * 3 + 1] == 'U') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A' || 'G'))
            cout << "Val ";
        if ((mrna[k * 3] == 'G') && (mrna[k * 3 + 1] == 'C') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A' || 'G'))
            cout << "Ala ";
        if ((mrna[k * 3] == 'G') && (mrna[k * 3 + 1] == 'A') && (mrna[k * 3 + 2] == 'U' || 'C'))
            cout << "Asp ";
        if ((mrna[k * 3] == 'G') && (mrna[k * 3 + 1] == 'A') && (mrna[k * 3 + 2] == 'A' || 'G'))
            cout << "Glu ";
        if ((mrna[k * 3] == 'G') && (mrna[k * 3 + 1] == 'G') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A' || 'G'))
            cout << "Gly ";
        if ((mrna[k * 3] == 'U') && (mrna[k * 3 + 1] == 'U') && (mrna[k * 3 + 2] == 'U' || 'C'))
            cout << "Phe ";
        if ((mrna[k * 3] == 'U') && (mrna[k * 3 + 1] == 'U') && (mrna[k * 3 + 2] == 'A' || 'G'))
            cout << "Leu ";
        if ((mrna[k * 3] == 'U') && (mrna[k * 3 + 1] == 'C') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A' || 'G'))
            cout << "Ser ";
        if ((mrna[k * 3] == 'U') && (mrna[k * 3 + 1] == 'A') && (mrna[k * 3 + 2] == 'A' || 'G')) {
            cout << "Stopp ";
            return 0;
        }
        if ((mrna[k * 3] == 'U') && (mrna[k * 3 + 1] == 'A') && (mrna[k * 3 + 2] == 'U' || 'C'))
            cout << "Tyr ";
        if ((mrna[k * 3] == 'U') && (mrna[k * 3 + 1] == 'G') && (mrna[k * 3 + 2] == 'A')) {
            cout << "Stopp ";
            return 0;
        if ((mrna[k * 3] == 'U') && (mrna[k * 3 + 1] == 'G') && (mrna[k * 3 + 2] == 'U' || 'C'))
            cout << "Cys ";
        }
        if ((mrna[k * 3] == 'U') && (mrna[k * 3 + 1] == 'G') && (mrna[k * 3 + 2] == 'G'))
            cout << "Trp ";
        if ((mrna[k * 3] == 'C') && (mrna[k * 3 + 1] == 'U') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A' || 'G'))
            cout << "Leu ";
        if ((mrna[k * 3] == 'C') && (mrna[k * 3 + 1] == 'C') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A' || 'G'))
            cout << "Pro ";
        if ((mrna[k * 3] == 'C') && (mrna[k * 3 + 1] == 'A') && (mrna[k * 3 + 2] == 'U' || 'C'))
            cout << "His ";
        if ((mrna[k * 3] == 'C') && (mrna[k * 3 + 1] == 'A') && (mrna[k * 3 + 2] == 'A' || 'G'))
            cout << "Gln ";
        if ((mrna[k * 3] == 'C') && (mrna[k * 3 + 1] == 'G') && (mrna[k * 3 + 2] == 'U' || 'C' || 'A' || 'G'))
            cout << "Arg ";

    }
    return 0;
}

Answer 1

你的窃听器是(mrna[k * 3 + 2] == 'U' || 'C' || 'A')。您可能认为这会检查mrna[k * 3 + 2]是否等于U、C或A。但这不是它的作用..。|| (“或”)操作符结合了三个条件--其中一个条件是mrna[k * 3 + 2] == 'U'，正如您所预期的，但另外两个条件是'C'和'A'。C/C++语言将这些字符常量视为无条件真(因为它们是不等于零的数字)，因此整个表达式是无条件的。

正确的代码将是(mrna[k * 3 + 2] == 'U' || mrna[k * 3 + 2] == 'C' || mrna[k * 3 + 2] == 'A')。

您可以将mrna[k * 3 + 2]保存在一个名称较短的变量中，以使这些表达式更短、更快(然而，现代优化器将自动为您完成此优化，即使您没有显式地进行此优化)。

Answer 2

我对你在mrna[k * 3 + 2] == 'U' || 'C' || 'A'表格中的所有条件表示怀疑。这不检查mrna[k * 3 + 2]与这三个字母中的任何一个，它是无条件的true。

我还建议不要使用char [100]。您已经在阅读一个std::string，它也有operator[]，也有operator==。

您还(可能)在数据结束后建立索引。循环条件是k < n，其中n是mrna的“长度”，但是索引mrna[k * 3]。要么更改条件，要么更改索引方式。

但是，我最建议的是用一个if替换长串的std::map，您可以为每个三元组填充这个字符串。

#include <iostream>
#include <string>
#include <map>

void print_triple(const std::string & s)
{
    for (int i = 0; i < s.size(); i++) {
        std::cout << s[i];
        if (i % 3 == 2)
            std::cout << " ";
    }
}

static std::map<std::string, std::string> mrna_lookup
{
    { "AUU", "Ile " },
    { "AUC", "Ile " },
    { "AUA", "Ile " },
    { "AUG", "Met " },
    // etc
};

int main()
{
    std::string dna;
    std::cout << "Enter codogener strang: ";
    std::cin >> dna;

    std::cout << "DNA: ";
    print_triple(dna);

    std::string mrna;
    for (char c : dna)
    {
        switch (c)
        {
            case 'A': mrna.push_back('U'); break;
            case 'T': mrna.push_back('A'); break;
            case 'G': mrna.push_back('C'); break;
            case 'C': mrna.push_back('G'); break;
        }
    }
    cout << "\nmRNA: ";
    print_triple(mrna);

    cout << "\n";

    std::string_view mrna_view = mrna; // view so .substr doesn't copy
    
    for (int k = 0; k+2 < mrna.size(); k += 3)
    {
        auto it = mrna_lookup.find(mrna_view.substr(k, 3));
        if (it != mrna_lookup.end())
        {
            std::cout << it->second;
            if (it->second == "Stopp ")
            {
                return 0;
            }
        }
    }

    return 0;
}