char** to cope with string?

问题:

I got quite confused when I came across char**, is it really necessary to cope with string?
For example:

double strtod(const char *nptr,char **endptr);

If endptr is not NULL, a pointer to the character that stopped the scan is stored at the location pointed to by endptr. ---MSDN
This is quite complicated, why not just copy pointer to the character to endptr? All computations after the call can be achieved by passing the pointer's value to endptr as I think.
Is char** really needed?

回答1:

Imagine there's a type called my_type, and you have a function called foo() that needs a pointer to a my_type so that it can modify it:

void foo(my_type *t);

Now, lets look at how my_type is actually defined:

typedef char* my_type;

So it doesn't matter if the type is already a pointer. If you want to pass a pointer to a variable of that type, you need its address. So the decomposition of:

my_type *t

would be:

char **t

Whenever you want a pointer to a variable of type char *, you need a char **.

回答2:

Arguments in C are passed by value. If you want the caller of a function to see a change made to an argument passed to a function you need to pass the address of the argument to the function. This is why the char** is being passed. The caller has a variable of type char* and passes its address. If it is changed in the function the caller can see that change.

For example:

void set_int(int i) { i = 4; }

/* Caller */
int x = 7;
set_int(x);
/* 'x' is still 7 */

void really_set_int(int* i) { *i = 4; }

/* Caller */
int x = 7;
really_set_int(&x);
/* 'x' is now 4 */

The same behaviour for a variable of type char*. If a function needs to change what the char* is pointing to (not its content) it needs the address of the char*, which is of type char**.