GCC：__ATOMIC_ALWAYS_LOCK_FREE与-O3编译，但不使用-O0
示例代码：
int *s;
int foo(void)
{
return 4;
}
int bar(void)
{
return __atomic_always_lock_free(foo(), s);
}
调用：
$ gcc t0.c -O3 -c
<nothing>
$ gcc t0.c -O0 -c
t0.c:10:10: error: non-constant argument 1 to '__atomic_always_lock_free'
有什么想法吗？
相关： https://gcc.gnu.org/onlinedocs/gcc/_005f_005fatomic-内置.html。
Sample code:
int *s;
int foo(void)
{
return 4;
}
int bar(void)
{
return __atomic_always_lock_free(foo(), s);
}
Invocations:
$ gcc t0.c -O3 -c
<nothing>
$ gcc t0.c -O0 -c
t0.c:10:10: error: non-constant argument 1 to '__atomic_always_lock_free'
Any ideas?
Relevant: https://gcc.gnu.org/onlinedocs/gcc/_005f_005fatomic-Builtins.html.
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这似乎并不奇怪。您链接的文档说“
size必须解决到编译时常数”，因此可以预期，通过foo（），您可能会出现错误。但是，典型的是，如果GCC能够在编译时确定表达式的值，那么即使它不符合语言对常数表达式的基本定义，它也会将其视为编译时间常数。这可能被认为是扩展名，并由C17标准在6.6p10中明确允许。优化级别与编译器在编译时试图评估表达式的尝试相关。通过优化，它的作用仅仅是标准所需的基本常数折叠（例如
2*4）。通过优化，您将获得其完整的恒定传播通行证以及功能内部的好处。因此，从本质上讲，在
-O0下，编译器没有注意到foo（）始终返回相同的值，因为您已经禁用了将其允许其进入的优化得出这个结论。使用-O3它可以，因此它将其作为常数接受。This doesn't seem surprising. The documentation you linked says that "
sizemust resolve to a compile-time constant" and so it's to be expected that you might get an error when passingfoo(). However, it's typical that if GCC is able to determine the value of an expression at compile time, then it will treat it as a compile-time constant, even if it doesn't meet the language's basic definition of a constant expression. This may be considered an extension and is explicitly allowed by the C17 standard at 6.6p10.The optimization level is relevant to what the compiler tries in attempting to evaluate an expression at compile time. With optimizations off, it does little more than the basic constant folding that the standard requires (e.g.
2*4). With optimizations on, you get the benefit of its full constant propagation pass, as well as function inlining.So in essence, under
-O0, the compiler doesn't notice thatfoo()always returns the same value, because you've disabled the optimizations that would allow it to reach that conclusion. With-O3it does and so it accepts it as a constant.