指向C中的结构字段的指针

Question

I have:

typedef struct
{
    int id;
    Other_Struct *ptr;
} My_Struct;

Lets say I have the pointer abcd which its type is My_Struct .

How can I get the address of:

abcd->ptr  

?

I mean the address of ptr itself (like pointer to pointer) and not the address which is stored in ptr .

Answer 1

只需使用运算符的&地址，就像这样

Other_Struct **ptrptr = &(abcd->ptr);

Answer 2

这个怎么样：

&(abcd->ptr)

Answer 3

If I understood correctly, in this scenario you have My_Struct *abcd pointing to an address, and what you want is the address of a field inside this structure (it doesn't matter if this field is a pointer or not). The field is abcd->ptr , so its address you want is &abcd->ptr .

You can easily check this by printing the actual pointer values (the difference between the addresses should give you the offset of ptr inside My_Struct ):

struct My_Struct {
    int id;
    void *ptr;
};

main()
{
    struct My_Struct *abcd;
    printf("%p %p\n", abcd, &abcd->ptr);
}

Update: If you want your code to be portable, standards-compliant, and past and future proof, you may want to add casts to void * to the printf() arguments, as per @alk's comments below. For correctness, you can also use a standard entry point prototype ( int main(void) or int main(int argc, char **argv) ) and, of course, include stdio.h to use printf() .

Answer 4

The most unambiguous way to do it is thus:

My_Struct *abcd;
Other_Struct **pptr;
pptr = &(abcd->ptr);

I don't know if the parentheses are really necessary, and I don't care, because it's more readable this way anyway.