how to know dynamic size of struct

Dear all,

Suppose i have a program that is consisted of a struct. this struct will be used as an array.

Code:

---

typedef struct
{
        char name[30];
        int salary;
} person;

---

How do i define an flexible array for this struc ? i don't know how many arrays which i need yet.

and how can i calculate the sum of all person salary ?

I wrote the simple code, but i think not solve my problem

Code:

---

#include <stdio.h>

typedef struct
{
        char name[30];
        int salary;
} person;


int sumStructs(person*, int);

main(void)
{
        person a[10], *pa;                  /* I still put non flexible array*/
        pa = a;
        int i = 0;
        for(i=0; i< 10; i++)
        {
                a[i].i = i+1;
        }
       
        int d = sizeof(a)/sizeof(person);
       
        int c = sumStructs(pa, d);

        printf("sum i adalah %d \n",c);
}

int sumStructs(person*pa, int d)
{

        int sum = 0, j= 0;
        while(j<d)
        {
                sum += (*pa).i;
                pa++;
                j++;
        }
        return sum;
}

---

std::vector.....

Seriously though, just keep track of how big your array needs to be, and over-allocate a buffer. And use realloc() where necessary.

Code:

---

person *a;

a = malloc(sizeof(person) * number_of_elements);

---

Then you'd access the array like this:

Code:

---

a[i].name

---

or

Code:

---

a[i].salary

---

Any which way you do this, you will need to track the number of elements you have allocated.

--
Mats

Yeah, sorry. I was gonna mention that. But I saw that

Code:

---

int sumStructs(person*pa, int d)

---

already has provisions for passing the number of elements.

Yes, and my point was mostly towards when using dynamic allocation.

--
Mats

Quote:

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Originally Posted by nonoob

Yeah, sorry. I was gonna mention that. But I saw that

Code:

---

int sumStructs(person*pa, int d)

---

already has provisions for passing the number of elements.

---

Hi Nonoob,

You mean, i don't need to define variable a[10] instead of just put *a ?
I follow your idea, but i got the 0 result for sumStructs function. I modify my code like this:

Code:

---

#include <stdio.h>

typedef struct
{
        char name[30];
        int salary;
} person;


int sumStructs(person*, int);

main(void)
{
        person *pa;     
        pa = malloc(sizeof(student) * 100);           
       
        int i = 0;
        for(i=0; i< 10; i++)
        {
                pa[i].i = i+1;                                  //put the salary value for each person
        }
       
        int d = sizeof(pa)/sizeof(person);
       
        int c = sumStructs(pa, d);

        printf("sum i is %d \n",c);
}

int sumStructs(person*pa, int d)
{

        int sum = 0, j= 0;
        while(j<d)
        {
                sum += (*pa).i;
                pa++;
                j++;
        }
        return sum;
}

---

and result is :

Code:

---

$ ./a.exe
sum i is 0

---

How do i get the number of element without making the array variable ?

No...............

sizeof(pa) == 4 (or more)
sizeof(person) == 8 or whatever

So your parameter will evaluate to 0 (since .029 is not an integer value--I remember getting into a big argument about this fact once before).

Quote:

---

Originally Posted by master5001

No...............

sizeof(pa) == 4 (or more)
sizeof(person) == 8 or whatever

So your parameter will evaluate to 0 (since .029 is not an integer value--I remember getting into a big argument about this fact once before).

---

You are right, however if i still define
struct a[10];
sizeof(a) == 360

I think my first question still was not answered, because the memory already prepare for 10 variables a

Quote:

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Originally Posted by unikgila

How do i get the number of element without making the array variable ?

---

You don't. You did the malloc -- did you forget between now and then how many elements you malloced? It's your responsibility to remember, not the compiler's.

This is why incrementing by a constant factor is a lovely thing :)

Quote:

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Originally Posted by master5001

This is why incrementing by a constant factor is a lovely thing :)

---

Sorry for not understand what you explain :( . If possible, i hope you can give me the sample code or repair the code that i have wrote above. Thanks

I could. But to be perfectly frank, I am not going to.

Here is why I am being so harsh. To be honest this question gets asked with great regularity.

Code:

---

int main(void)
{
        person *pa;
        int number_I_must_remember_because_its_the_size_of_my_array = 100;     
        pa = malloc(sizeof(student) * number_I_must_remember_because_its_the_size_of_my_array);
           
       
        int i = 0;
        for(i=0; i< number_I_must_remember_because_its_the_size_of_my_array; i++)
        {
                pa[i].i = i+1;                                  //put the salary value for each person
        }
       
        int c = sumStructs(pa, number_I_must_remember_because_its_the_size_of_my_array);

        printf("sum i is %d \n",c);
}

---

Wow. Congratulations unikgila, it took a mere 9 posts for you to make both me and tabstop reach our breaking point. Not too shabby. It takes some people hundreds.

Quote:

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Originally Posted by master5001

Wow. Congratulations unikgila, it took a mere 9 posts for you to make both me and tabstop reach our breaking point. Not too shabby. It takes some people hundreds.

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thanks for both of you :)