您有两个实现一些代码我在使用Java流修改字符串:
final List<String> strings = new ArrayList<>();
strings.add("abc");
strings.add("bca");
strings.add("xyz");
Method 1:
final List<String> modifiedStrings = strings.stream().map(a -> {
String abc;
abc = a.replace("a", "*");
return abc;
}).collect(Collectors.toList());
Method 2:
final List<String> modifiedStrings2 = strings.stream().map(a ->
a.replace("a", "*")).collect(Collectors.toList());
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我想知道方法一和方法二之间是否存在任何性能差异.
Thi*_*ilo 15
不会.这将编译为基本相同的字节码.第二个版本只是一种更简洁的编写方式.
示例代码:
void foo(List<String> strings) {
strings.stream()
.map(
a -> {
String abc;
abc = a.replace("a", "*");
return abc;
})
.collect(Collectors.toList());
}
void bar(List<String> strings) {
strings.stream().map(a -> a.replace("a", "*")).collect(Collectors.toList());
}
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对应的lambda字节码:
private static java.lang.String lambda$bar$1(java.lang.String);
descriptor: (Ljava/lang/String;)Ljava/lang/String;
flags: ACC_PRIVATE, ACC_STATIC, ACC_SYNTHETIC
Code:
stack=3, locals=1, args_size=1
0: aload_0
1: ldc #8 // String a
3: ldc #9 // String *
5: invokevirtual #10 // Method java/lang/String.replace:(Ljava/lang/CharSequence;Ljava/lang/CharSequence;)Ljava/lang/String;
8: areturn
LineNumberTable:
line 18: 0
private static java.lang.String lambda$foo$0(java.lang.String);
descriptor: (Ljava/lang/String;)Ljava/lang/String;
flags: ACC_PRIVATE, ACC_STATIC, ACC_SYNTHETIC
Code:
stack=3, locals=2, args_size=1
0: aload_0
1: ldc #8 // String a
3: ldc #9 // String *
5: invokevirtual #10 // Method java/lang/String.replace:(Ljava/lang/CharSequence;Ljava/lang/CharSequence;)Ljava/lang/String;
8: astore_1
9: aload_1
10: areturn
LineNumberTable:
line 11: 0
line 12: 9
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因此,字节码是不是很相同-有一个多余的astore_1,并aload_1在foo情况; 但这可能无关紧要,因为它可以被优化掉.
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