java 字符串相加 StringBuffer java 字符串相加原理

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mob64ca14010a69 2023-08-04 16:32:52

文章标签 Java String Android 字符串性能 java 文章分类 Java 后端开发

1、背景

话说jdk1.5以前字符串拼接如果使用“+”号时会消耗性能和内存，因为每次使用“+”号编译的时候都会转换出多个String对象（String对象是不可变的），所以都推荐使用StringBuilder。而jdk1.5之后，编译器对“+”号做了改进，会自动转换成StringBuilder再构造String对象，使得字符串拼接变得更方便，然而就代表着程序员可以随便使用"+"号而不考虑性能问题么？答案是：做梦吧。

2、“+”号照妖镜

既然我们所见的“+”号并不是“+”号，那我们通过反编译apk文件来看看“+”号的本质是什么？

首先先看简单的字符串拼接：

String result = "1111"+"2222"+"3333"+"4444";

编译后变成如下smali代码
    const-string v0, "1111222233334444"
    .local v0, "result":Ljava/lang/String;

这种直接拼接字符串常量的用法，编译器已经帮你合成好了，"1111"+"2222"+"3333"+"4444"跟"1111222233334444"编译后是一模一样的。

如果他们中出了个叛徒呢？把"2222"改成数字 2222

String result = "1111"+2222+"3333"+"4444";

编译后如下：
    const-string v0, "1111222233334444"

    .local v0, "result":Ljava/lang/String;

发现结果也是一样，说明编译器会把数字2222一起编译到同一个字符串中

StringBuilder正在热身.... ....

如果我们不在一行拼接完字符串，而是分几次拼接呢？

String result = "1111";
result = result+"2222";
result = result+"3333";
result = result+"4444";

编译后：
    .line 26
    const-string v0, "1111"

    .line 27
    .local v0, "result":Ljava/lang/String;
    new-instance v1, Ljava/lang/StringBuilder;

    invoke-direct {v1}, Ljava/lang/StringBuilder;-><init>()V

    invoke-virtual {v1, v0}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    const-string v2, "2222"

    invoke-virtual {v1, v2}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    invoke-virtual {v1}, Ljava/lang/StringBuilder;->toString()Ljava/lang/String;

    move-result-object v0

    .line 28
    new-instance v1, Ljava/lang/StringBuilder;

    invoke-direct {v1}, Ljava/lang/StringBuilder;-><init>()V

    invoke-virtual {v1, v0}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    const-string v2, "3333"

    invoke-virtual {v1, v2}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    invoke-virtual {v1}, Ljava/lang/StringBuilder;->toString()Ljava/lang/String;

    move-result-object v0
.line 29
    new-instance v1, Ljava/lang/StringBuilder;

    invoke-direct {v1}, Ljava/lang/StringBuilder;-><init>()V

    invoke-virtual {v1, v0}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    const-string v2, "4444"

    invoke-virtual {v1, v2}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    invoke-virtual {v1}, Ljava/lang/StringBuilder;->toString()Ljava/lang/String;

    move-result-object v0

额，“+”号现在是转换成了StringBuilder了，可是每个“+”号都做了 new StringBuilder、append、toString，三倍的操作，这种情况下会导致性能和内存倍增

如果我们把字符串变量在一行拼接起来呢？

String str2 = "2222";
String str3 = "3333";
String str4 = "4444";
String result = "1111"+str2+str3+str4;

编译后：

.line 48
    const-string v0, "2222"

    .line 49
    .local v0, "str2":Ljava/lang/String;
    const-string v1, "3333"

    .line 50
    .local v1, "str3":Ljava/lang/String;
    const-string v2, "4444"

    .line 51
    .local v2, "str4":Ljava/lang/String;
    new-instance v3, Ljava/lang/StringBuilder;

    invoke-direct {v3}, Ljava/lang/StringBuilder;-><init>()V

    const-string v4, "1111"

    invoke-virtual {v3, v4}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    invoke-virtual {v3, v0}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    invoke-virtual {v3, v1}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    invoke-virtual {v3, v2}, Ljava/lang/StringBuilder;->append(Ljava/lang/String;)Ljava/lang/StringBuilder;

    invoke-virtual {v3}, Ljava/lang/StringBuilder;->toString()Ljava/lang/String;

    move-result-object v3

    .line 52
    .local v3, "result":Ljava/lang/String;

这种情况下编译后，只new 一个 StringBuilder，append 4次，toString一次