Why can't I use a pointer in an argument?

[mikie]

Here is an example of a function...

// Function prototype
int foobar(unsigned int a, int *b);

// Function
int foobar(unsigned int a, int &b)
{
    // Acquire some data... then
    b = Data[a];
}

I get the following error:
73, syntax error; found '&' expecting ')'.

This is the normal method of passing in a pointer to an integer to a function and then setting the contents of the integer in the function. Why doesn't CVI let me do this? Is there another way?

[RobertoBozzolo]

Well, in function definition you correctly used the indirect operator (*) while in the function itself you used the address operator instead (&): could it be a transcription error?

[Graziano]

Hi!
  Roberto, as usual, is right.  More, the function prototype line and the function definition should have the same syntax (same name of variables, and same type!)....try to just change & into * ,  (take care on how Data is defined.....).

Graziano

[Former NI Employee MS]

Hi Mikie,

I believe that we all have done errors when using pointers for the first time.

Try this code:

// Function prototype
int foobar(unsigned int a, int *b);

// Function
int foobar(unsigned int a, int *b)
{
    // Acquire some data... then
    *b = Data[a];  // If you want to copy the content of the element a of the tab into b.

      b = &Data[a]; .// If you want that b points to the element a of the table
}

Here is a link to the use of pointers   ______ Bad point for me, it is in French (gonna find an other one)

Regards,

Message Edité par Mathieu Steiner le 11-16-2006 08:10 AM

[mikie]

Well, the function I outlined has been working for me for 20 years.

If I change to the following:

// Function prototype
int foobar(unsigned int a, int *b);

// Function
int foobar(unsigned int a, int *b)
{
    // Acquire some data... then
    b = Data[a];
}

Then then following error occurs on the line trying to set b: Operands of = have illegal types 'pointer to in' and 'int'.

[jr_2005]

You would then need:

    *b = Data[a];

JR

[mikie]

It seems that I can't edit my post, so I'll continue.

It's obvious that in my last example, I should get an error. Both sides are of different type.

Now, I'll admit that 20 years ago I was confused by pointers when I first used them, and I've stayed away from them ever since when possible. But that certainly didn't help me learn any. My first example does work, and it is a correct use. It just doesn't work under CVI.

Now if I use *b = Data[a];

If b already a pointer to b, then isn't *b a pointer to a pointer to b? Or it it different on the left side of an assignment. Is it a de-reference?

Mike

[jr_2005]

You are confusing & and *. & gives you a pointer to, or the address of, a variable. * gives you the contents of whatever is being pointed to, otherwise known as a de-reference.

JR

edit: Of course when you actually call your function, you would then use the & operator, as in: foobar (x, &y); - maybe this was causing some confusion?

Message Edited by jr_2005 on 11-16-2006 02:32 PM

[Manuel H]

I´m also not so familiar with those pointer, but just to sum it up and to see if I´m right:

// Function prototype
int foobar(unsigned int a, int *b);

// Function
int foobar(unsigned int a, int *b)
{
    // Acquire some data... then
    *b = Data[a];
}

//Function call version 1
int a, b;
foobar (a,&b);

//Function call version 2
int a, *b;
foobar (a,b);

is that correct this way?

[jr_2005]

Yes, but in your version 2 code you need to initialise the b pointer - just saying it is a pointer is OK for the compiler, but not enough for the program to work properly. You need to get it to point to something eg:

    //Function call version 2
    int a, *b, z;
    b = &z;
    foobar (a, b);

Then the function would effectively put its output (Data[a]) into the variable z.

JR

Message Edited by jr_2005 on 11-16-2006 03:14 PM