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remove useless synchronization in exit and quick_exit

calling exit more than once invokes undefined behavior. in some cases
it's desirable to detect undefined behavior and diagnose it via a
predictable crash, but the code here was silently covering up an
uncommon case (exit from more than one thread) and turning a much more
common case (recursive calls to exit) into a permanent hang.
Rich Felker 11 lat temu
rodzic
commit
2e55da9118
2 zmienionych plików z 0 dodań i 11 usunięć
  1. 0 7
      src/exit/exit.c
  2. 0 4
      src/exit/quick_exit.c

+ 0 - 7
src/exit/exit.c

@@ -1,8 +1,6 @@
 #include <stdlib.h>
 #include <stdint.h>
 #include "libc.h"
-#include "atomic.h"
-#include "syscall.h"
 
 static void dummy()
 {
@@ -21,11 +19,6 @@ extern void (*const __fini_array_end)() __attribute__((weak));
 
 _Noreturn void exit(int code)
 {
-	static int lock;
-
-	/* If more than one thread calls exit, hang until _Exit ends it all */
-	while (a_swap(&lock, 1)) __syscall(SYS_pause);
-
 	__funcs_on_exit();
 
 #ifndef SHARED

+ 0 - 4
src/exit/quick_exit.c

@@ -1,6 +1,4 @@
 #include <stdlib.h>
-#include "syscall.h"
-#include "atomic.h"
 #include "libc.h"
 
 static void dummy() { }
@@ -8,8 +6,6 @@ weak_alias(dummy, __funcs_on_quick_exit);
 
 _Noreturn void quick_exit(int code)
 {
-	static int lock;
-	while (a_swap(&lock, 1)) __syscall(SYS_pause);
 	__funcs_on_quick_exit();
 	_Exit(code);
 }