我对Java热点生成的函数序言感到困惑。考虑这个虚拟例子:
public static int getLen(String s) {
return s.length();
}
public static void main(String[] args) {
for(int i = 0; i < 1_000_000; i++) {
getLen("abcd");
}
}
C2生成的getLen
代码如下:
[Entry Point]
[Verified Entry Point]
[Constants]
# {method} {0x00007fbed3fc2318} 'getLen' '(Ljava/lang/String;)I' in 'examples/Main'
# parm0: rsi:rsi = 'java/lang/String'
# [sp+0x20] (sp of caller)
0x00007fbefd11a960: mov %eax,-0x14000(%rsp)
0x00007fbefd11a967: push %rbp
0x00007fbefd11a968: sub $0x10,%rsp ;*synchronization entry
; - examples.Main::getLen@-1 (line 6)
0x00007fbefd11a96c: mov 0xc(%rsi),%r11d ;*getfield value
; - java.lang.String::length@1 (line 623)
; - examples.Main::getLen@1 (line 6)
; implicit exception: dispatches to 0x00007fbefd11a981
0x00007fbefd11a970: mov 0xc(%r12,%r11,8),%eax ;*arraylength
; - java.lang.String::length@4 (line 623)
; - examples.Main::getLen@1 (line 6)
; implicit exception: dispatches to 0x00007fbefd11a991
问题:
mov %eax,-0x14000(%rsp)
的目的是什么?- 样例代码没有同步,那么
sub $0x10,%rsp ;*synchronization
这一行是什么意思?