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;
}