整数值给出意想不到的结果

Question

While working on a project of mine, I came across a problem.

When I run the following program:

public class test {
    public static void main(String[] args) {
        String x = "1234";
        System.out.println("Int value: " + Integer.valueOf(x.charAt(0)));
        System.out.println("String value: " + x.charAt(0));
    }
}

it gives me the result:

Int value: 49
String value: 1

which left me confused because x = "1234" , so when it runs the forth line the result should be the same as:

System.out.println("Int value: " + Integer.valueOf(1));

which should in theory should give me: Int value: 1

So what caused the result to be 49 instead?

Answer 1

There's a difference between

System.out.println("Int value: " + Integer.valueOf(1));

and

System.out.println("Int value: " + Integer.valueOf(x.charAt(0)));

In the second line what's really happening is this -

System.out.println("Int value: " + Integer.valueOf('1'));

Notice that '1' is being passed as a Character not as an Integer. As a result the Integer.valueOf return the ASCII value of character '1', which is 49.

Refer to the ASCII table to see for yourself.

Answer 2

this is because following return UTF-16 value, rather than value itself

        System.out.println("Int value: " + Integer.valueOf(x.charAt(0)));

try using int x = Integer.parseInt("1234"); instead

Answer 3

49 is the decimal value that corresponds to the ASCII character '1'. Take a look at this chart, and look at the "Dec" (decimal value) column that corresponds to Chr (character value) '1'.

A link to an ASCII table that shows you how a given character maps to a given decimal

Answer 4

You used the function valueOf(String s) , not valueOf(int i) . Documentation says, that it "returns an Integer object holding the value of the specified String.". It's mean, that function returns ASCII code of sign "1", which is 49.

Answer 5

This is because you were printing the ASCII value of 1.

Try Character.getNumericValue(char)

Try this (with minimal change from your implementation):

package general;

class CharToInt
{
public static void main(String[] args) {
      String x = "1234";

      System.out.println("Int value: " + Character.getNumericValue(x.charAt(0)));
      System.out.println("String value: " + x.charAt(0));
    }
}

Answer 6

你需要做的是将字符串“1”传递给方法“Integer.valueOf（）”然后它将它解析为整数1.为此，你可以这样做

String x = "1234"; System.out.println("Int value: " + Integer.valueOf(s.substring(0,1));

Answer 7

one thing is how you see the argument passing to the function, but the second thing is what is the type of this argument.

x.charAt(0) returns '1' 

but must remember that the type is character .

So, in fact it returns the binary data like: 00110001 which one can read like 0x31(hex) what is 49(dec) what is the UTF-16 representation of character '1' .

Now... the function:

Integer.valueOf( )

It expectes the number as argument, so it takes those binary data 00110001 and reads it like a normal number... to it is value of 49