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我知道 noreturn 应该用于预期不会将控制权返回给调用者的函数，但我在生成的汇编代码中找不到差异。有谁知道会生成不同代码的示例？

编辑： noreturn 之后不会生成清理代码。

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1 回答 1

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returns_twice禁用一些 gcc 优化。

举个例子，在一个旧的 gcc 版本中，我有：尾调用优化、全局公共子表达式消除、跳转绕过。

returns_twice使用calls_setjmp机器（在源代码树周围）：

          if (flags & ECF_RETURNS_TWICE)
            cfun->calls_setjmp = true;

总成本 ( gcc/tree-tailcall.c):

static bool
suitable_for_tail_call_opt_p (void)
{
  [...]
  /* Any function that calls setjmp might have longjmp called from
     any called function.  ??? We really should represent this
     properly in the CFG so that this needn't be special cased.  */
  if (cfun->calls_setjmp)
    return false;

高考（gcc/gcse.c）：

static int
gcse_main (rtx f ATTRIBUTE_UNUSED)
{
  [...]
  /* We do not construct an accurate cfg in functions which call
     setjmp, so just punt to be safe.  */
  if (cfun->calls_setjmp)
    return 0;

跳转绕过（gcc/gcse.c）：

static int
bypass_jumps (void)
{
  [...]
  /* We do not construct an accurate cfg in functions which call
     setjmp, so just punt to be safe.  */
  if (cfun->calls_setjmp)
    return 0;

该函数既不能是纯函数也不能是 const ( gcc/ipa-pure-const.c)：

/* Check the parameters of a function call to CALL_EXPR to see if
   there are any references in the parameters that are not allowed for
   pure or const functions.  Also check to see if this is either an
   indirect call, a call outside the compilation unit, or has special
   attributes that may also effect the purity.  The CALL_EXPR node for
   the entire call expression.  */

static void
check_call (funct_state local, gimple call)
{
  [...]
      /* When bad things happen to bad functions, they cannot be const
         or pure.  */
      if (setjmp_call_p (callee_t))
        {
          local->pure_const_state = IPA_NEITHER;
          local->looping = false;
        }

该函数不能内联 ( gcc/tree-inline.c)：

/* A callback for walk_gimple_seq to handle statements.  Returns
   non-NULL iff a function can not be inlined.  Also sets the reason
   why. */

static tree
inline_forbidden_p_stmt (gimple_stmt_iterator *gsi, bool *handled_ops_p,
                         struct walk_stmt_info *wip)
{
  [...]
     /* We cannot inline functions that call setjmp.  */
      if (setjmp_call_p (t))
        {
          inline_forbidden_reason
            = G_("function %q+F can never be inlined because it uses setjmp");
          *handled_ops_p = true;
          return t;
        }

它还会影响函数堆栈帧中的寄存器保存区域。

示例（tco）：

函数.c：

int func(void)
{
        return 0;
}

tco.c:

extern int func(void)  /*__attribute__((returns_twice))*/;

int main()
{
        return func();
}

不返回两次：

00000000004003a0 <main>:
  4003a0:       e9 0b 01 00 00          jmpq   4004b0 <func>
  4003a5:       90                      nop
  4003a6:       90                      nop
  4003a7:       90                      nop

返回两次：

00000000004003a0 <main>:
  4003a0:       48 83 ec 08             sub    $0x8,%rsp
  4003a4:       e8 17 01 00 00          callq  4004c0 <func>
  4003a9:       48 83 c4 08             add    $0x8,%rsp
  4003ad:       c3                      retq   
  4003ae:       90                      nop
  4003af:       90                      nop
于 2013-08-30T12:33:06.143 回答