如何从 int 和 char 连接字符串？

Question

For example I want to concatanate char a ='A' and int b = 5 into string = "A5" .

String string = a + b; doesn't work.

Answer 1

You may want to use StringBuilder , where you can append any type of primitives :

char a ='A';
int b = 5;

StringBuilder sb = new StringBuilder();

sb.append(a);
sb.append(b);

String result = sb.toString();

Answer 2

The easiest way for me is to precede the String with an empty String.

 String str = "" + 'a'  + 10;

The conversion goes from left to right so you start out with a String.

If you do it this way,

String str = 'a' + 10 + ""; 

you will get a String value of "107" since the numeric addition is done before the conversion to a String.

Answer 3

One way to convert most primitive values to String is to utilize the overloaded method in the String class valueOf() :

public static void main(String[] args) 
{
    char a = 'A';
    int b = 5;

    String str = String.valueOf(a) + b; //Can do either of these two lines, will work the same
    String str2 = a + String.valueOf(b);

    System.out.println(str);
    System.out.println(str2);
}

You only need to convert one of the values into String , because appending them afterward will automatically convert the other into a String .

This method will work on both char , and int in this scenario, but will also work on long , double , float and boolean as well. This is identical to calling Integer.toString(int i) or Character.toString(i) etc... but it is convenient to be able to use the same overloaded method for each case instead of requiring to call methods from different classes.

Answer 4

To all the answers already given I provide one that explain why "it doesn´t work" as you expect. A string is nothing but an immutable collection of char . A char itself is nothing but a (signed) integer and thus can implicitely converted to such. Thus when you write char + int an integer-operation occurs, in your case A + 5 which results in 70 , because ASCII-code for A is 65.

Last but not least String s = 70 surely does not compile, because a number can´t be converted to a string.

So you have to tell the compiler that you do not want an integer-operation, but a string-concatenation, which is by turning one of the operands into a string already:

String s = 'A'.toString() + 5

or

String s = 'A' + 5.toString()