array dynamic allocation

Applications Programming
1

Hi,
I have the following problem: i must allocate a dynamic array from a subroutine which should return such array to main function. The subroutine has already a return parameter so i thought of pass the array as I/O parameter. I tried the following program but it doesn't work (segmentation fault), i think because i should have a pointer to array, but i'm not sure.

void main(){
int *array;
int **ptr = &array;
sub(ptr);
}

void sub(int **ptr){
....//calculates the array dimension
*ptr = calloc(n,sizeof(int));
for(i=0;i<n;i++)*ptr=i;
}
2

Because you are skipping a step in allocating memory...*ptr has been calloced w/o first allocating memory for ptr which is a dual-level pointer while *ptr is a single-level.

3

Also: main always returns an int. It is never void. If your compiler does not complain, it is probably an ancient one. There are a lot of reasons here are two

  1. You will get random garbage returned to the calling program or shell

  2. Some compilers implement the stack frame for void functions differently than for int functions - for optimization reasons - resulting in a almost impossible to debug crash when main() ends.

4

shamrock could you correct my code please??

5 
void sub(int **ptr) {
    ptr=(int**)calloc(n,sizeof(int*));
    for(i=0;i<n;i++) {
         ptr=(int*)calloc(1,sizeof(int));
        *ptr=i;
    }
}
6

ok but in this way i lose the array when sub returns...

7

If you pass a pointer to the array you won't lose it then.

8

Hi Shamrock,

*ptr has already been initialized with address of local variable 'array' as -

*ptr = &array; in statement int **ptr = &array;

ptr(int**) -> array(int*) -> (uninitialized/invalid reference to integer).

ptr = &array;

ptr (or array) = (int)calloc(n,sizeof(int)); will initialize array to hold address of allocated memory.

9

Correction -

*ptr has already been initialized with address of local variable 'array' as -
*ptr = &array; in statement int **ptr = &array;

ptr has already been initialized with address of local variable 'array' as -
ptr = &array; in statement int **ptr = &array;

10

Sorry Shamrock but I did not understand your answer.Should ptr be a pointer to array??
I tried the following and it seems to work fine:

int main(){
int *array;
int **ptr = &array;
sub(ptr);
}
void sub(int *ptr){
//calculates n...
int *tmp=(int)calloc(n,sizeof(int));
for(i=0;i<n;i++)
tmp[i]=i;
*ptr=tmp;
}

11

littleboyblu,

Can you try this one -

void main(){
int *array;
int **ptr = &array;
sub(ptr);
}

void sub(int **ptr){
....//calculates the array dimension
*ptr = calloc(n,sizeof(int));
for(i=0;i<n;i++)
(*ptr)[i]=i; // subscript operator [] have higher precedence than dereference * operator
}

12

thanks pshaikh