java - 为什么添加长变量会导致串联？

Question

Java 在执行加法时如何处理长变量？

错误的版本 1:

Vector speeds = ... //whatever, speeds.size() returns 2
long estimated = 1l;
long time = speeds.size() + estimated; // time = 21; string concatenation??

错误的版本 2:

Vector speeds = ... //whatever, speeds.size() returns 2
long estimated = 1l;
long time = estimated + speeds.size(); // time = 12; string concatenation??

正确的版本:

Vector speeds = ... //whatever, speeds.size() returns 2
long estimated = 1l;
long size = speeds.size();
long time = size + estimated; // time = 3; correct

我不明白，为什么 Java 将它们连接起来。

谁能帮我，为什么要连接两个原始变量？

你好，你好

Answer 1

我猜你实际上是在做类似的事情:

System.out.println("" + size + estimated); 

这个表达式从左到右求值:

"" + size        <--- string concatenation, so if size is 3, will produce "3"
"3" + estimated  <--- string concatenation, so if estimated is 2, will produce "32"

要让它工作，你应该这样做:

System.out.println("" + (size + estimated));

再次从左到右求值:

"" + (expression) <-- string concatenation - need to evaluate expression first
(3 + 2)           <-- 5
Hence:
"" + 5            <-- string concatenation - will produce "5"