c - GCC 关于 const 限定符的警告是否正确？

Question

考虑以下代码，这些代码来自 this question :

const int (*foo(const char *a))[1]
    { return (const int (*)[1]) a; }

当 compiled with GCC 8.2 (and older versions) using -Wcast-qual , GCC 警告:

source>:2:15: warning: cast discards 'const' qualifier from pointer target type [-Wcast-qual]      { return (const int (*)[1]) a; }               ^

Is this warning correct? Clearly the destination type has a const qualifier in it.

It is on the element type rather than on the thing immediately pointed to by the pointer, which is the array type. However, the warning remains even if we use typedef int T[1]; and replace the cast with (const T *). Additionally, per C 2018 6.7.3 10, qualifiers on an array type apply to the element type, not the array type, so the type is the same either way.

Clang does not show this warning.

If we change the cast to (const void *):

const int (*foo(const char *a))[1]
    { return (const void *) a; }

然后警告消失了。如果我们将 -pedantic 添加到编译开关，我们会收到关于 const 的不同警告:

source>:2:15: warning: return discards 'const' qualifier from pointer target type [-Wdiscarded-qualifiers]      { return (const void *) a; }               ^~~~~~~~~~~~~~~~

这看起来像同一个警告，只是它是关于从返回表达式到函数返回类型的隐式转换，而之前的警告是关于强制转换中的显式转换。但是这个只出现在 -pedantic 中。为什么？

Answer 1

这是 GCC bug 81631 . GCC 无法识别指向数组指针的转换保留了 const 限定符，因为应用于数组的限定符实际上应用于数组元素。