java - String类如何重写+运算符？

Question

为什么在 Java 中，当 String 是一个类时，您可以使用 + 运算符添加字符串？在String.java代码中我没有找到这个运算符的任何实现。这个概念违反了面向对象吗？

Answer 1

让我们看一下 Java 中的以下简单表达式

int x=15;
String temp="x = "+x;

编译器在内部将 "x = "+x; 转换为 StringBuilder 并使用 .append(int) 将整数“添加”到字符串中。

<强> 5.1.11. String Conversion

Any type may be converted to type String by string conversion.

A value x of primitive type T is first converted to a reference value as if by giving it as an argument to an appropriate class instance creation expression (§15.9):

• If T is boolean, then use new Boolean(x).
• If T is char, then use new Character(x).
• If T is byte, short, or int, then use new Integer(x).
• If T is long, then use new Long(x).
• If T is float, then use new Float(x).
• If T is double, then use new Double(x).

This reference value is then converted to type String by string conversion.

Now only reference values need to be considered:

• If the reference is null, it is converted to the string "null" (four ASCII characters n, u, l, l).
• Otherwise, the conversion is performed as if by an invocation of the toString method of the referenced object with no arguments; but if the result of invoking the toString method is null, then the string "null" is used instead.

^{The toString method is defined by the primordial class Object (§4.3.2). Many classes override it, notably Boolean, Character, Integer, Long, Float, Double, and String.}

See §5.4 for details of the string conversion context.

<强> 15.18.1.

Optimization of String Concatenation : An implementation may choose to perform conversion and concatenation in one step to avoid creating and then discarding an intermediate String object. To increase the performance of repeated string concatenation, a Java compiler may use the StringBuffer class or a similar technique to reduce the number of intermediate String objects that are created by evaluation of an expression.

For primitive types, an implementation may also optimize away the creation of a wrapper object by converting directly from a primitive type to a string.

优化版本实际上不会首先执行完全包装的字符串转换。

这是编译器使用的优化版本的一个很好的说明，尽管没有转换原语，您可以在其中看到编译器在后台将内容更改为 StringBuilder:

http://caprazzi.net/posts/java-bytecode-string-concatenation-and-stringbuilder/

<小时/>

这段java代码:

public static void main(String[] args) {
    String cip = "cip";
    String ciop = "ciop";
    String plus = cip + ciop;
    String build = new StringBuilder(cip).append(ciop).toString();
}

生成这个 - 查看两种串联样式如何产生完全相同的字节码:

 L0
    LINENUMBER 23 L0
    LDC "cip"
    ASTORE 1
   L1
    LINENUMBER 24 L1
    LDC "ciop"
    ASTORE 2

   // cip + ciop

   L2
    LINENUMBER 25 L2

    NEW java/lang/StringBuilder
    DUP
    ALOAD 1
    INVOKESTATIC java/lang/String.valueOf(Ljava/lang/Object;)Ljava/lang/String;
    INVOKESPECIAL java/lang/StringBuilder.<init>(Ljava/lang/String;)V
    ALOAD 2
    INVOKEVIRTUAL java/lang/StringBuilder.append(Ljava/lang/String;)Ljava/lang/StringBuilder;
    INVOKEVIRTUAL java/lang/StringBuilder.toString()Ljava/lang/String;

    ASTORE 3

    // new StringBuilder(cip).append(ciop).toString()

   L3
    LINENUMBER 26 L3

    NEW java/lang/StringBuilder
    DUP
    ALOAD 1
    INVOKESPECIAL java/lang/StringBuilder.<init>(Ljava/lang/String;)V
    ALOAD 2
    INVOKEVIRTUAL java/lang/StringBuilder.append(Ljava/lang/String;)Ljava/lang/StringBuilder;
    INVOKEVIRTUAL java/lang/StringBuilder.toString()Ljava/lang/String;

    ASTORE 4
   L4
    LINENUMBER 27 L4
    RETURN

查看上面的示例以及如何生成基于给定示例中的源代码的字节码，您将能够注意到编译器已在内部转换了以下语句

cip+ciop; 

进入

new StringBuilder(cip).append(ciop).toString();

换句话说，字符串连接中的运算符+实际上是更详细的StringBuilder习惯用法的简写。