c - 复合文字的生命周期

Question

6.5.2.5p5说

If the compound literal occurs outside the body of a function, the object has static storage duration; otherwise, it has automatic storage duration associated with the enclosing block.

我将此处的“封闭 block ”解释为“最里面的封闭 block ”是否正确？ (因为如果它不是最里面的，那是哪一个？) Why are gcc and clang behaving as if the lifetime of a literal were its enclosing function?

例子:

long foo(long*);

void call_foo()
{
    {foo(&(long){42});}
    {foo(&(long){42});}
    {foo(&(long){42});}
    {foo(&(long){42});}
}

//for comparison

void call_foo2()
{
    {long x=42;foo(&x);}
    {long x=42;foo(&x);}
    {long x=42;foo(&x);}
    {long x=42;foo(&x);}
}

gcc/clang 在 -O3 生成的代码:

call_foo:
  sub rsp, 40
  mov rdi, rsp
  mov QWORD PTR [rsp], 42
  call foo
  lea rdi, [rsp+8]
  mov QWORD PTR [rsp+8], 42
  call foo
  lea rdi, [rsp+16]
  mov QWORD PTR [rsp+16], 42
  call foo
  lea rdi, [rsp+24]
  mov QWORD PTR [rsp+24], 42
  call foo
  add rsp, 40
  ret
call_foo2:
  sub rsp, 24
  lea rdi, [rsp+8]
  mov QWORD PTR [rsp+8], 42
  call foo
  lea rdi, [rsp+8]
  mov QWORD PTR [rsp+8], 42
  call foo
  lea rdi, [rsp+8]
  mov QWORD PTR [rsp+8], 42
  call foo
  lea rdi, [rsp+8]
  mov QWORD PTR [rsp+8], 42
  call foo
  add rsp, 24
  ret

Answer 1

这似乎没有任何充分的理由。我只想称它为编译器错误。