为什么比较空函数指针会导致“无效的操作数到二进制==”

Question

Can somebody please explain to me why the below code got the "invalid operands to binary ==" error?

typedef int (*func_t)(int);
#define NO_FUNC ((func_t) 0)
struct {
    const char *name;
    func_t func;
} table[] = { {"func1", NO_FUNC} };

if (table[0] == NO_FUNC) { // invalid operands to binary ==

}

Answer 1

并且您应该在结构中引用正确的成员：

if (table[0].func == NO_FUNC)

Answer 2

table[0] is of an unnamed struct type, and NO_FUNC has type int (*)(int) . These two types can't be compared.

Instead, you could use:

if (table[0].func == NO_FUNC)