Get the variable address of structure member in C language

When writing a program , Made a low-level mistake , The main reason is that the concept of value passing and address passing is not well understood . Now I will explain the difference through a small experimental example :
Value passed : When a function passes a value , Formal parameters will first open up new spaces according to their definition types , And re engrave its parameter values into the new memory space , The opening of new memory space also means the change of its storage address , And the operation of its parameters in the function , It is only for the operation in the new space , Not for functions , The transmitted value makes an impact , That is, the two are completely different spaces .
Address delivery : Usually, pointer variables are used as parameters , Because the pointer variable is the variable that saves the address , So when receiving through pointer variables , It's your address , in other words , In memory space , It points to the same memory block , At this time, the operation will be affected at the same time .

demand : In the program , I want to get the address of a structure member variable , And return the address of its member variable by a pointer , By modifying the pointer value , To affect the value of its structure member variables .

C Language :
Wrong way :

#include<stdio.h>// Input and output 
typedef struct {
    
	char a[10];
}node, * nodes;

void init(node b, char** m) {
    
	printf("%p\n", &b);
	*m = &(b.a[2]);
}
int main() {
    
	node a = {
     "123456" };
	char* n;
	init(a, &n);
	*n = '0';
	printf("n=%c \na[2]=%c\n", *n, a.a[2]);
	printf("%p\n", &a);
	printf("%p\n", &(a.a[2]));
	printf("%p\n", (&(&a)->a[2]));
	printf("%p\n", n);
}

It can be seen from it that :
1. Through the structure object . Get the member variable address and pass the structure pointer -> The way to get the address of the member variable is the same , That is, there is no difference between the two methods .
2. Structure object a Passed to the function as an argument by b receive , But we can see that its memory address is different , This is also related to the mechanism of value transmission .
3.n The address of the member variable returned as a pointer is based on the structure variable after copying b There is ,n and a[2] The address of is different . This leads to the final pair n The value of the pointer is modified , Does not affect its a Structure member variable value .

The right way :

#include<stdio.h>// Input and output 
typedef struct {
    
	char a[10];
}node, * nodes;
void init(node* b, char** m) {
    
	printf("b Memory address :%p\n", b);
	*m = &(b->a[2]);
}
int main() {
    
	node a = {
     "123456" };
	char* n;
	init(&a, &n);
	*n = '0';
	printf("n=%c \na[2]=%c\n", *n, a.a[2]);
	printf("a Memory address :%p\n", &a);
	printf(" &(a.a[2]) Memory address :%p\n", &(a.a[2]));
	printf("(&(&a)->a[2]) Memory address :%p\n", (&(&a)->a[2]));
	printf("n Memory address :%p\n", n);
}

From the results :
We go through n The pointer gets a[2] The address of , And through the n Modification of the address value pointed to by the pointer , Affected its a[2] Value .
1. From the memory address , Through to a Pointer passing of ( Address delivery ), from b receive , Its a,b The address is the same , then n A pointer to obtain a[2] And return .