How to initialize these pointers? - c

Question

I need to make a list of employees and I can't change these structures, I'm having trouble in how to initialize each of tab[10] to NULL and how to set values
#include <stdio.h>
#include <stdlib.h>
typedef struct employee Employee;
struct employee{
char name[81];
float salary;
};
Employee *tab[10]; /*a table with employee*/
void set(Employee **tab, int i, char *name, float salary){
tab[i]->name = name;
tab[i]->salary = salary;
}
int main(){
Employee *e;
int i = 0;
for(; i < 10; i++) init(i,&e);
return 0;
}
/*a table with an employee, each position must have a name and a salary*/
Employee *tab[10];
void init(int n, Employee **tab);

Answer 1

Everaldo
With commentators helping you, it seems you are getting there. I would like to sum up the suggests given so far and add a couple of my own.
Declaring the Employee array
Declaring the array as a global variable and then passing it as a parameter to functions makes things a little confusing. I usually prefer declaring a local variable and then passing it to the various functions that uses it. Also as suggested by David C. Rankin, to initialize every array element to 0 just requires you to initial the first element in the declaration statement. No FOR loop needed. The compiler will auto initialize the rest of the array elements for you.
main()
{
Employee* tab[10] = { NULL };
. . . .
}
Array memory allocation
As mention by Patrick87, you need to add code to assign memory to every element in the array. An example initialization routine could be coded as follows:
int init(int len, Employee** tab) {
int i = 0;
for (i = 0; i < len; i++)
{
if ( (tab[i] = (Employee*) calloc (1,sizeof(Employee))) == NULL )
return EXIT_FAILURE;
}
return EXIT_SUCCESS;
}
Function usage:
if (init(10, &tab) == EXIT_FAILURE)
{
puts("CALLOC Failed, aborting....");
exit(EXIT_FAILURE);
}
Things to note:
Check to ensure the memory was allocated. On failure return some
type of failure status to alert the caller of the function.
The status codes that are being returned are define in stdlib.h.
They are not necessary but do give a clear indication to the reader
of your code the success and failure paths your code takes.
The FOR loop was moved inside the initialization function. Function
calls are expensive when it comes to processing time. Since the
array size is known, it is faster to perform the loop inside the
function.
Try to always write functions that return a status. This will enable
the caller to perform any error handling if the function's operations
fail.
Set array element values
The following statement is not valid. You cannot directly copy the content from a string pointer to an array of characters. You will need to use statements like strcpy, strncpy, or memcpy to copy the data.
tab[i]->name = name;
There is a method I prefer for copying strings.
sprintf(tab[i]->name, "%.80s", name);
This will copy up to 80 characters from name into tab[i]->name, then insert a null character. The beauty of this statement is that the designation variable does not have to be the same size as the source. If the source variable (in this case name) is shorter, spirntf will simply stop when it encounter a null character and then null terminate the destination string. If the source is longer than 80 characters or if it is missing the null terminator character, sprintf will stop coping at the 80st character position and then auto insert a null character in the 81st character position.
An example SET routine could look like the following:
void set(Employee** tab, int i, char* name, float salary) {
sprintf(tab[i]->name, "%.80s", name);
tab[i]->salary = salary;
}
Usage:
for (i = 0; i < 10; i++)
{
set(&tab, i, "Bob", 35000. + i); // bogus values, demo purposes only
}
Main program logic
Your main program as you currently have outline will need to change. For starters, the declaration of variable “e” should be replace with the declaration of variable “tab” (see Patrick87 comments) . On initializing the array, see my suggestion above. To set values to the array elements see SET function comments above.
Free memory
Every time you allocate memory, you must free it when you are done. Forgetting to free allocated memory will create memory leaks in your program. Note technically, in this demonstration program, the system will free the memory when your code exits, so you do not need to free it. But it is good practice so when you start writing real applications you will not forget to do so.
Here is an example on how this could be done:
for (i = 0; i < 10; i++)
free (tab[i]);

Answer 2

tab is an array of pointers, so to initialize them all to NULL, you can use a for loop, e.g.
for (i = 0; i < n; i++)
tab[i] = 0;
To set a value, allocate some space for an instance of your struct (either on the stack via function parameter or local variable, or else on the heap by with malloc/calloc/realloc) and then set one of the tab[k] to the address of the memory you allocated (using & or just the pointer directly if allocated).