C pointer assignment inside a function

Question

I am trying to assign a pointer to an array element inside the function get_by_id.

The function get_by_id finds the element correctly and assigns to element to lp, however when it comes back to the caller (set_attr), it reverts to NULL.

I'm not understanding why it's reverting to NULL in the caller, let alone how to fix it.

#include <stdio.h>
#include <stdlib.h>


typedef struct
{
    int id;
    char name[128];
    int attr;
} Lookup;


static int num_sources = 0;

Lookup ltable[128];


void add(int id, const char *name)
{
    sprintf(ltable[num_sources].name, name);
    ltable[num_sources].id = id;
    ltable[num_sources].attr = 999;  // example default attr 999
    num_sources++; // increment the number of sources
}


int get_by_id(int id, Lookup *lp)
{
    int status = 1;
    for (int i = 0; i < 128; i++)
    {
        lp = &ltable[i];
        if (id == lp->id)
        {
            status = 0;
            break;
        }
    }
    return status;
}


int set_attr(const int id, const int attr)
{
    Lookup *lp = NULL;
    int status = get_by_id(id, lp);
    if (status == 0)
    {
        lp->attr = attr;
    }
    return status;
}

int main(void) {
    add(88, "main"); // example id 88
    set_attr(88, 9); // example attr 9
    return EXIT_SUCCESS;
}

" how to fix it." Do you expect lp in set_attr() to remain unchanged if a matching id not found? — chux
– chux, Commented Mar 2, 2018 at 17:28
I don't, I expect status to remain == 1 and check that from the caller. — GarethD
– GarethD, Commented Mar 2, 2018 at 19:05

Paul Ogilvie · Accepted Answer · 2018-03-03 09:08:45Z

To change the value of lp you must use a double pointer:

int get_by_id(int id, Lookup **lp)
{
    int status = 1;
    for (int i = 0; i < 128; i++)
    {
        *lp = &ltable[i];
        if (id == (*lp)->id)
        {
            status = 0;
            break;
        }
    }
    return status;
}

and call like:

    int status = get_by_id(id, &lp);

Because C is "pass by value", a copy of lp is made and passed to the function. Any change to that copy cannot be seen by the caller. To allow the called function to change the variable of the caller, you pass the address of the caller's variable, so the called function can change it. In your case, you pass the address of lp written as &lp.

The following version may be peferred:

#define T_SIZE (sizeof(ltable)/sizeof(ltable[0]))

Lookup *get_by_id(int id)
{
    for (int i = 0; i < T_SIZE; i++)
        if (id == ltable[i].id)
            return &ltable[i];

    return 0;
}

Note that it returns a pointer to the table entry and zero if not found.

hafeez · Accepted Answer · 2018-03-02 17:51:10Z

0

You need to just pass the Lookup parameter by reference rather by value. This is because you can only modify the parameter only if it is passed by reference.

int get_by_id(int id, Lookup *lp)
{
    int status = 1;
    for (int i = 0; i < 128; i++)
    {
        *lp = ltable[i];
        if (id == lp->id)
        {
            status = 0;
            printf("lp=%X\n", lp);
            break;
        }
    }
    return status;
}

int set_attr(const int id, const int attr)
{
    Lookup lp = {0};
    int status = get_by_id(id, &lp);
    if (status == 0)
    {
        printf("lp = %X\n", &lp);
        lp.attr = attr;
    }
    return status;
}

This should work.

answered Mar 2, 2018 at 17:51

hafeez

1691 silver badge6 bronze badges

Collectives™ on Stack Overflow

C pointer assignment inside a function

2 Answers 2

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related