block代码转为c++后,为什么可以获取构造函数的地址

struct __block_imp {
void *isa;              
int flags;
int Reserved;
void *FuncPtr;
}
struct __main_block_imp_0 {
    struct __block_imp impl;  
    struct __main_block_desc_0 *Desc;  

    __main_block_impl_0(void *fp, struct __main_block _desc_0 *desc, int flags = 0) {
        impl.isa = &_NSConcreteStackBlock.  
        impl.Flags = flags;
        impl.FuncPtr = fp;
        Desc = desc
    }
};

struct __main_block_desc_0 {
    unsigned long reserved;
    unsigned long Block_size;
} __main_block_desc_0_DATA = {
    0, 
    sizeof(struct __main_block_impl_0)
};

void (*blk)(void) = (void(*)(void))&__main_block_impl_0 (
(void *)__mainblock_func_0, &__main_block_desc_0_DATA);
这个代码获取了 第二个结构体的构造函数地址,显式调用构造函数会临时产生变量,所以我自己写的demp试的时候,报错Taking the address of a temporary object of type .
可是它为什么就可以获取呢

对于特定的编译器,它的类的方法地址是固定的,可以通过一些运算得到。

Taking the address of a temporary object of type .
这个是让你不要直接返回堆栈上变量的指针,因为出了函数,这个地址是无效的。