Thread: Just a simple warning...

I have this warning on a bit of code I can't seem to correct. The code compiles just find but I'd rather not have any warnings. Can anyone correct this?

198 [Warning] assignment makes pointer from integer without a cast

Code:

struct ot_entry {
     char  ot_mnemonic[OP_LENGTH + 1];
     int   ot_format;  
     char  ot_opcode;
};
struct  ot_entry  *optbl[OT_SIZE];
int     ot_count;

int main ( void )
{
        ...

       optbl[ot_count] = malloc ( sizeof ( struct ot_entry ) );    /* line getting the warning */

        ...

       return 0;
}

Put this at the top of your file.

Code:

#include <stdlib.h>

<stdlib.h> contains the prototype for malloc(). The default return type for a function without a prototype is an int.

I don't understand what your trying to do. you already have
Code:

struct ot_entry *optbl[OT_SIZE];


that is an array of OT_SIZE of ptrs to struct ot_entry. so you already have OT_SIZE records to manipulate.

why do you try to allocate more space - which is really another pointer to the struct - by using the same array you've aready declared
Code:

optbl[ot_count] = malloc ( sizeof ( struct ot_entry ) );

and changing its size?

post your code, or explain what your trying to do please.

[edit] if you were just implicitly declaring malloc then....*shrug*[/edit]

In the first line he's declaring an array of pointers. In the second, he's presumabely in a loop that is cycling through each of those pointers and setting them equal to the address returned by malloc.

Beaten, but may still be of use. caroundw5h, the code is valid.

Code:

struct  ot_entry  *optbl[OT_SIZE];

This declares an array of pointers to struct ot_entry. These pointers do not point to allocated memory and an attempt to dereference one will result in very bad things.

Code:

optbl[ot_count] = malloc ( sizeof ( struct ot_entry ) );

This allocates enough memory for a struct ot_entry and assigns its address to a pointer variable for later use.

Arrays can be extremely confusing. Thumper333's code is the array equivalent to:

Code:

int* p; /* Declare a pointer. Does not point to valid memory at this point. */
p = malloc(sizeof(int)); /* Allocate enough memory for an int and assign its address to p. */

Thanks anonytmouse, for some odd reason i felt like leaving out the header. That worked though...