在C中保存一个指针地址

Question

I want to save a certain address of a pointer, and use it later.

This is the struct which holds the original pointer:

typedef struct CSV
{
    char *RD;
    ...
}CSV;

This is the called function:

static status_t write_to_buffer(CSV *CSVUtil,...)
{

    // The way i was planning to save address:
    char* temp =  &CSVUtil->RD;

...
    // pointer location ++
    CSVUtil->RD++;
...
    // The way i wanted to restore it:
    &CSVUtil->RD = temp;
}

First, am i doing the address restoring as needed?

I get this error message: expression must be a modifiable lvalue

So i'm guessing i'm not, But what can i do to fix this?

EDIT:

Just to be clear, what i want to do is to copy the address to a certain pointer, change the used address (increment it), and than set back the copied address to the used address.

Cheers.

Answer 1

This assignment is incorrect:

char* temp =  &CSVUtil->RD;

The expression &CSVUtil->RD returns a pointer to a character pointer (ie char** ), but you are assigning it to a character pointer char* , so one level of indirection is missing. There should be a compiler warning in the output telling you about this problem.

The reason the assignment back &CSVUtil->RD = temp does not work is that the result of the "take address & " operator is not assignable. You can obtain an address, but you cannot change it by "assigning" it a new address.

It looks like you need to save and restore the pointer itself, not the location of that pointer. Therefore, you can fix the code by removing the ampersands:

char* temp =  CSVUtil->RD;
...
CSVUtil->RD = temp;

Answer 2

According to your problem

expression must be a modifiable lvalue

That means that your are trying to assign a value to a const or an Un-Changeable type.

Now lets take a look at the assignments you have:

1. char* temp = &CSVUtil->RD;
2. CSVUtil->RD++;
3. &CSVUtil->RD = temp;

A Pointer as you know, holds an address so if i have a:

char *pointer

The pointer variable will hold the address where the actual value i want is, and to access that value i would either use *pointer = XXX or if it's a struct pointer->inside-value = XXX.

What you are trying to do by adding the "&" in front of the pointer is to actually give back the location of "pointer" variable which was declared.

So - instead of getting the address of the value that RD is pointing to, you get the address of RD itself. ALSO and here is the problem - Addresses of variables cannot be changed this way but through malloc as it's being decided in compilation or runtime if it's dynamic (malloc) so when doing &Pointer = XXXX you are actually aren't changing the variable but trying to change it's address so to speak. BUT the "&" operator just returns the address (think of it like a function that returns an int that specifies address) so it cannot be changed.

I would change the assignments to:

1. char *temp = CSVUtil->RD;
2. Is ok;
3. CSVUtil->RD = temp;

That should work.

Hope my explanation was good enough.