第一次在这里发布,所以如果我做错了,请告诉我...对于整个编程工作来说都是新的。

因此,在我的课程中,我们正在执行zip到条形码转换器程序。我的程序运行良好,现在我正试图减少冗余或重复的编码。例如,通过我的功能将邮政编码实际编译为条形码,我有5个switch语句,它们都运行相同的确切情况。唯一的区别是开关的条件(数字A,B,C,D,E;邮政编码的数字):

// Short Zip Code Converter Function
int zip_to_bar(int digitA, int  digitB, int digitC, int  digitD, int digitE)

{
    int sum;
    string bar_Code = "!", check_Code;
    switch(digitA)
    {
        case 0: bar_Code += "!!..."; break;
        case 1: bar_Code += "...!!"; break;
        case 2: bar_Code += "..!.!"; break;
        case 3: bar_Code += "..!!."; break;
        case 4: bar_Code += ".!..!"; break;
        case 5: bar_Code += ".!.!."; break;
        case 6: bar_Code += ".!!.."; break;
        case 7: bar_Code += "!...!"; break;
        case 8: bar_Code += "!..!."; break;
        case 9: bar_Code += "!.!.."; break;
    }
    switch(digitB)
    {
        case 0: bar_Code += "!!..."; break;
        case 1: bar_Code += "...!!"; break;
        case 2: bar_Code += "..!.!"; break;
        case 3: bar_Code += "..!!."; break;
        case 4: bar_Code += ".!..!"; break;
        case 5: bar_Code += ".!.!."; break;
        case 6: bar_Code += ".!!.."; break;
        case 7: bar_Code += "!...!"; break;
        case 8: bar_Code += "!..!."; break;
        case 9: bar_Code += "!.!.."; break;
    }
    switch(digitC)
    {
        case 0: bar_Code += "!!..."; break;
        case 1: bar_Code += "...!!"; break;
        case 2: bar_Code += "..!.!"; break;
        case 3: bar_Code += "..!!."; break;
        case 4: bar_Code += ".!..!"; break;
        case 5: bar_Code += ".!.!."; break;
        case 6: bar_Code += ".!!.."; break;
        case 7: bar_Code += "!...!"; break;
        case 8: bar_Code += "!..!."; break;
        case 9: bar_Code += "!.!.."; break;
    }
    switch(digitD)
    {
        case 0: bar_Code += "!!..."; break;
        case 1: bar_Code += "...!!"; break;
        case 2: bar_Code += "..!.!"; break;
        case 3: bar_Code += "..!!."; break;
        case 4: bar_Code += ".!..!"; break;
        case 5: bar_Code += ".!.!."; break;
        case 6: bar_Code += ".!!.."; break;
        case 7: bar_Code += "!...!"; break;
        case 8: bar_Code += "!..!."; break;
        case 9: bar_Code += "!.!.."; break;
    }
    switch(digitE)
    {
        case 0: bar_Code += "!!..."; break;
        case 1: bar_Code += "...!!"; break;
        case 2: bar_Code += "..!.!"; break;
        case 3: bar_Code += "..!!."; break;
        case 4: bar_Code += ".!..!"; break;
        case 5: bar_Code += ".!.!."; break;
        case 6: bar_Code += ".!!.."; break;
        case 7: bar_Code += "!...!"; break;
        case 8: bar_Code += "!..!."; break;
        case 9: bar_Code += "!.!.."; break;
    }
    // Sum of Zip_Digits
    sum = (digitA + digitB + digitC + digitD + digitE);
    // Calculation of check_Digit_Code
    check_Code = check_Digit_Code(sum);
    // Assignment of check_Bar_Code to check_Digit_Code
    cout << bar_Code + check_Code << endl;
    return sum;
}


我想以某种方式将其最小化到只有一个switch语句的位置,因为它们都是相同的,并且它将循环运行每个数字。我要执行此操作的原因是我们必须运行短(#####),标准(#####-####)和高级(#####-#### + ##)格式的邮政编码。因此,您可以想象该程序看起来像冗长而重复的样子。

我本以为for语句循环会很好地工作,但我坚持设置条件的方式/方式(当函数中有5个变量时应初始化什么?)。显然,我所得到的并不能满足我的期望,但这是我到目前为止所拥有的(有关如何解决此问题的任何建议)?

    // Reformatted Version of Short Zip Code Converter Function
    int zip_to_bar_srt(int digitA, int digitB, int digitC, int  digitD, int digitE)
{

    for(int digitA; digitA < 5; digitA++)
    {
        switch(digitA)
        {
            int sum;
            string barCode = "!", check_bar;

            case 0: barCode += "!!..."; break;
            case 1: barCode += "...!!"; break;
            case 2: barCode += "..!.!"; break;
            case 3: barCode += "..!!."; break;
            case 4: barCode += ".!..!"; break;
            case 5: barCode += ".!.!."; break;
            case 6: barCode += ".!!.."; break;
            case 7: barCode += "!...!"; break;
            case 8: barCode += "!..!."; break;
            case 9: barCode += "!.!.."; break;

                sum += digitA;
                check_bar = check(sum);
                cout<<barCode+check_bar<<endl;
                return sum;
        }
    }
}

最佳答案

编写一个可以让您获得条形码一部分的数字的功能:

string get_barcode_part(int digit)
{
    switch(digitA)
    {
        case 0: return "!!...";;
        case 1: return "...!!";;
        case 2: return "..!.!";;
        case 3: return "..!!.";;
        case 4: return ".!..!";;
        case 5: return ".!.!.";;
        case 6: return ".!!..";;
        case 7: return "!...!";;
        case 8: return "!..!.";;
        case 9: return "!.!..";;
        default: return "Invalid"; // or whatever else you want
    }
}


然后,您可以简单地为每个数字调用它:

// Short Zip Code Converter Function
int zip_to_bar(int digitA, int  digitB, int digitC, int  digitD, int digitE)
{
    int sum;
    string bar_Code = "!", check_Code;

    bar_Code += get_barcode_part(digitA);
    bar_Code += get_barcode_part(digitB);
    bar_Code += get_barcode_part(digitC);
    bar_Code += get_barcode_part(digitD);
    bar_Code += get_barcode_part(digitE);

    // Sum of Zip_Digits
    sum = (digitA + digitB + digitC + digitD + digitE);
    // Calculation of check_Digit_Code
    check_Code = check_Digit_Code(sum);
    // Assignment of check_Bar_Code to check_Digit_Code
    cout << bar_Code + check_Code << endl;
    return sum;
}


这是功能(heh)的基本功能:通过将通用功能放在公共位置来重构和减少复制的代码。

编辑:此外,您可能应该使用数字数组,如下所示:

// Short Zip Code Converter Function
int zip_to_bar(int *digits, int numDigits)
{
    int sum = 0;
    string bar_Code = "!", check_Code;

    for (int i = 0; i < numDigits; i++)
    {
        bar_Code += get_barcode_part(digits[i]);
        // Sum of Zip_Digits
        sum += digits[i];
    }

    // Calculation of check_Digit_Code
    check_Code = check_Digit_Code(sum);
    // Assignment of check_Bar_Code to check_Digit_Code
    cout << bar_Code + check_Code << endl;
    return sum;
}

07-24 09:45
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