java - Java如何知道跳出循环时跳转到哪里？

Question

while (condition) {

    if (condition) {
        statement1;
        statement2;

        break;
    } else {
        statement3;
        statement4;
    }

}

通过在 if 子句中使用 break，我们确保循环停止并退出。

我不明白 break 语句是如何“知道”它在循环中首先退出的，或者它是如何“知道”跳转到哪里的。这是怎么发生的？

Answer 1

I don't understand how the break statement "knows" that it is within a loop for it to exit out of in the first place.

break 语句不知道它在 switch 或循环语句中。编译器验证 break 语句在switch 或循环语句中。如果在循环语句中遇到 not 的 break 语句，它将发出编译时错误。

If no switch, while, do, or for statement in the immediately enclosing method, constructor, or initializer contains the break statement, a compile-time error occurs.

如果编译器能够验证 break 语句是否在 switch 或循环语句中，它将发出 JVM 指令立即突然跳转到第一条语句在最近的封闭循环之后。

因此:

for(int i = 0; i < 10; i++) {
    if(i % 2 == 0) {
         break;
    }
}

会被编译器翻译成:

0:  iconst_0        # push integer 0 onto stack
1:  istore_1        # store top of stack in local 1 as integer                  
                    # i = 0
2:  iload_1         # push integer in local 1 onto stack
3:  bipush 10       # push integer 10 onto stack
5:  if_icmpge 23    # pop and compare top two (as integers), jump if first >= second
                    # if i >= 10, end for
8:  iload_1         # push integer in local 1 onto stack
9:  iconst_2        # push integer 2 onto stack
10: irem            # pop top two and computes first % second and pushes result
                    # i % 2
11: ifne 17         # pop top (as integer) and jump if not zero to 17
                    # if(i % 2 == 0) 
14: goto 23         # this is the break statement
17: iinc 1, 1       # increment local 1 by 1
                    # i++
20: goto 2          # go to top of loop
                    # loop
23: return          # end of loop body