Javascript减法返回NaN

Question

What is wrong with my code?? Everything except the subtraction works. It just returns NaN. I am new to javascript so i might have written my code poorly.

// Variables
var count = prompt("Choose an arithmetic method: \n1. Addition \n2.     Subtraktion\n3. Multiplikation\n4. Division");
var x = parseInt(prompt("Enter your first number", "0"));
var y = parseInt(prompt("Enter your second number", "0"));
var z = +x + +y;


// Switch function with 4 cases
switch(count) {
case '1':
alert("Answer: " + z);
break;

case '2':
alert("Answer: " + x - y);
break;

case '3':
alert("Answer: " + x * y);
break;

case '4':
alert("Answer: " + x / y);
break;

}

Answer 1

You need to group the operations in parentheses, for instance alert("Answer: " + (x - y)); (and the same for the others). Otherwise JavaScript runs "Answer: " + x first, resulting in a string.

Also, always specify the radix (you want 10) for parseInt: parseInt(input, 10) , otherwise some engines get confused with octal numbers.

Answer 2

Your trouble is here:

alert("Answer: " + x - y);

Due to how the association of operators work, it works as if you had written this:

alert(("Answer: " + x) - y);

You need to write it like this:

alert("Answer: " + (x - y));

Snippet here:

 // Variables var count = prompt("Choose an arithmetic method: \\n1. Addition \\n2. Subtraktion\\n3. Multiplikation\\n4. Division"); var x = parseInt(prompt("Enter your first number", "0")); var y = parseInt(prompt("Enter your second number", "0")); var z = +x + +y; // Switch function with 4 cases switch (count) { case '1': alert("Answer: " + z); break; case '2': alert("Answer: " + (x - y)); break; case '3': alert("Answer: " + x * y); break; case '4': alert("Answer: " + x / y); break; } 

Answer 3

Wrap your subtraction in parentheses: (x - y). You would get an unexpected result from the addition, too, if you put x + y there instead of z.

What's happening is operator precedence. Multiplication and division are higher precedence than addition, so they get done first - before the implicit conversion to string for concatenation.

With subtraction, '+' and '-' are equal in precedence and so get done in order from left to right. So the concat takes place before the math, which leaves you attempting to subtract y from a string (which doesn't work, and so...NaN).

With addition, it would simply concatenate the two numbers onto the string.