C ++中的条件语句问题

Question

Here is the question that I've been trying to solve:

You are given a positive integer, n,:

If 1 ≤ n ≤ 9, then print the English representation of it. That is " one " for 1, " two " for 2, and so on.

Otherwise print " Greater than 9 " (without quotes)

Here is a portion of my suggested answer, but it doesn't work!

int n;

if (1 <= n <= 9) {
    if (n == 1) {
    cout << "one" << endl;
    } else if (n == 2) {
    cout << "two" << endl;
    } else if (n == 3) {
    cout << "three" << endl;
    } else if (n == 4) {
    cout << "four" << endl;
    } else if (n == 5) {
    cout << "five" << endl;
    } else if (n == 6) {
    cout << "six" << endl;
    } else if (n == 7) {
    cout << "seven" << endl;
    } else if (n == 8) {
    cout << "eight" << endl;
    } else if (n == 9) {
    cout << "nine" << endl;
    }
} else {

    cout << "Greater than 9" << endl;
}

What is the issue with the code?

Answer 1

将if (1 <= n <= 9)更改为if (n>= 1 && n<=9)

Answer 2

if (1 <= n <= 9) doesn't do what you think it does. It's evaluated as ((1 <= n) <= 9). <= returns a Boolean, so you're checking if 'true' or 'false' is less than 9.

You want to use the 'and' operator, &&.

Answer 3

if (1 <= n <= 9) { should be if (1 <= n && n <= 9) {

You could also arrive at the solution by using the switch statement:

//Problem states n is a positive integer, so no need to check if n < 1
switch(n)
{
  case 1: cout << "one" << endl; break;
  case 2: cout << "two" << endl; break;
  //etc...
  case 9: cout << "nine" << endl; break;
  default: cout << "Greater than 9" << endl;
}

It does the same thing, but it looks cleaner in my opinion.