在C++中,我所知道的创建多行字符串的最佳方式是创建相邻的字符串,并让编译器在编译时将它们连接起来,就像这样:
string s = "This is a very long string ...\n"
" and it keeps on going...";
在Java中,我所知道的唯一方法是使用字符串拼接:
String s = "This is a very long string ...\n" +
" and it keeps on going...";
问题是,这是否在运行时生成单个字符串,还是Java在编译时实际上也进行了拼接?之所以会出现这个问题,是因为以下行为:
String s1 = "abc";
String s2 = "abc";
System.out.println(s1 == s2); // this prints true, because the compiler
// generates only one "abc" object
String s3 = "a";
s3 += "bc";
System.out.println(s1 == s3); // false
static final
修饰符将您的String
设置为常量,然后s1 == s3
将为真。编辑:哦,您甚至不需要这样做... - Bubletans1==s3
是true
! - Sid Zhang