c++ 和 static_cast 的新功能<char> ()</char>

Question

So I'm in college right now and I have this lab that wants me to make a code that inputs an integer and outputs a corresponding letter, I was able to get the code to work but it wants me to use static_cast() for 1-9 " Write a program that prompts the user to input an integer between 0 and 35. The prompt should say Enter an integer between 0 and 35:

If the number is less than or equal to 9, the program should output the number; otherwise, it should output:

A for 10 B for 11 C for 12. . . and Z for 35. (Hint: For numbers >= 10, calculate the ACSII value for the corresponding letter and convert it to a char using the cast operator, static_cast().)"

and here is what I have.

 #include <iostream>

using namespace std;

int main() 
{
 int num;

 cout << "Enter an integer between 0 and 35: ";
 cin >> num;
 cout << endl;

 switch(num)
 {
   case 0:
    cout << "0";
      break;
   case 1:
    cout << "1";
      break;
   case 2:
    cout << "2";
      break;
   case 3:
    cout << "3";
      break;
   case 4:
    cout << "4";
      break; 
   case 5:
    cout << "5";
      break;
   case 6:
    cout << "6";
      break;
   case 7:
    cout << "7";
      break;
   case 8:
    cout << "8";
      break;
   case 9:
    cout << "9";
      break;
   case 10:
    cout << "A";
    break;
   case 11:
    cout << "B";
   case 12:
    cout << "C";
   case 13:
    cout << "D";
   case 14:
    cout << "E";
   case 15:
    cout << "F";
   case 16:
    cout << "G";
   case 17:
    cout << "H";
   case 18:
    cout << "I";
   case 19:
    cout << "J";
   case 20:
    cout << "K";
   case 21:
    cout << "L";
   case 22:
    cout << "M";
   case 23:
    cout << "N";
   case 24:
    cout << "O";
   case 25:
    cout << "P";
   case 26:
    cout << "Q";
   case 27:
    cout << "R";
   case 28:
    cout << "S";
   case 29:
    cout << "T";
   case 30:
    cout << "U";
   case 31:
    cout << "V";
   case 32:
    cout << "W";
   case 33:
    cout << "X";
   case 34:
    cout << "Y";
   case 35:
    cout << "Z";
 }
    // Write your main here
    return 0;
}

Answer 1

Here's a quick improvement to you code. Rather than have 36 case statements, use a table.

const char table[] = "0123456789ABCDEFGHUJKLMNOPQRSTUVWXYZ";

if ((num >= 0) && (num <= 35))
{
    cout << table[num];
}

But I suspect your professor wants you to make use of the fact that chars are integers to and can be added with an offset to get another char. (eg 'A' + 1 == 'B' ). And all character literals like '4' and 'G' have ordinal values that match their values on the ascii char. (eg 'A' is 65 and '0' is 48)

So I think what they are getting at is this:

if ((num >= 0) && (number <= 9))
{
    int ord = '0' + num;
    cout << static_cast<char>(ord) << endl;
}
else if ((num>= 10) && (num <= 35))
{
    int ord = 'A' + num - 10;
    cout << static_cast<char>(ord) << endl;
}

Notice that I'm using character literals in single quotes ( 'A' ) instead of string literals in double quotes like "A" . You can effectively "add" chars together. Math operations applied to strings have a quite different effect.

Pedantic comment on my answer. Technically, the C standard only stipulates that digit chars ( '0' to '9' ) are aligned sequentially in ordinal values. Ancient encodings like EBCIDIC didn't have the letters in sequential order. But your prof called out "ascii", so you're in the clear.

Answer 2

The ideal solution here is to build a time machine and get the ASCII developers to put 9 adjacent to A in order to simplify the logic required. :)

What you have with the big switch is unsatisfactory for the test. I recommend looking at the ASCII table and observing the decimal numbers that represent each character.

For example, 65 is 'A', 66 is 'B'. static_cast<char>(65) will output 'A'. With a little bit of math, you can manipulate the input number to add it to 65. A trick in C/C++ is that you can add to the character value directly, such as 'A' + offset .

Try to get your resulting code as simple as possible!