关于看似兼容的函数指针赋值的编译器警告(const vs no-const)

Question

我遇到了一些我认为应该有效的有趣的事情。首先:

编译器/版本

$ gcc --version
gcc (Debian 4.7.2-5) 4.7.2

编译器选项和警告消息。

$ gcc -c warn.c -o warn.o
warn.c:11:5: warning: initialization from incompatible pointer type [enabled by default]
warn.c:11:5: warning: (near initialization for ‘foo.exec’) [enabled by default]

我想知道为什么“Foo()”与“exec”不兼容。 (添加评论以希望完全清晰)

typedef struct Thing
{
    void (*exec)(char *abc);
} Thing;

// ME: I don't mess with this.. I make const, K?
void Foo(const char *abc)
{
    (void) abc;
}

// GCC: LOL, nope! probably u messed up.
Thing foo = { Foo };

Answer 1

const char * 与 char * 不同。

来自 C 标准 6.7.5.3(我强调):

15 For two function types to be compatible, both shall specify compatible return types. Moreover, the parameter type lists, if both are present, shall agree in the number of parameters and in use of the ellipsis terminator; corresponding parameters shall have compatible types. [...]

来自 C 标准 6.7.5.1(我强调):

2 For two pointer types to be compatible, both shall be identically qualified and both shall be pointers to compatible types.

注意:const 是一个限定符