allocating and initialising a pointer inside a function

[PeterP24]

Hello,

I am passing a pointer to a function (through call by reference) in order to allocate memory for it with malloc and to put some values into the newly allocated memory cells, i.e. to initialize it.
I came with the following code:


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

void fct_b(int **b)
{
int i;
*b = malloc(10*sizeof(int));
for(i = 0; i < 10; i++)
{
(*b)[i]=i;
// printf("%d ", (*b)[i]);
}

}

void fct_c(int ***c)
{
int i, j;
**c = malloc(10*sizeof(int*));
for(i = 0; i < 10; i++)
{
(*c)[i] = malloc(10*sizeof(int));
for(j = 0; j < 10; j++)
{
(*c)[i][j]=10*i+j;
//printf("%d", (*c)[i][j]);
}
}
}

int main(void)
{
int *b, **c;

fct_b(&b);
fct_c(&c);

int i, j;
// level b
/*for(i = 0; i < 10; i++)
{
printf("%d ", b[i]);
}*/
// level c
for(i = 0; i < 10; i++)
{
for(j = 0; j < 10; j++)
printf("%d ", c[i][j]);
}

return 0;
}

I manage to figure out how to do it for a pointer to pointer type (function fct_b). However I was surprised to see that the initialisation works only if I use:

(*b)[i]=i
because

*b[i]=i
doesn't work.

This would be one issue. Another one would be that I tried to expand the experience learned at fct_b to a pointer to a pointer to a pointer case - like the one implemented in function fct_c. I can check that inside the fct.c the pointer is allocated and initialised however it is not working.

I have two questions:
(a) in case of fct_b why does the initialisation works with (*b)[i] = i instead of *b[i] = i ?
(b) in case of fct_c - why does not work?

[ofnuts]

Wouldn't it be simpler to have :


int * fct_b()
{
int i;
int *b = malloc(10*sizeof(int));
for(i = 0; i < 10; i++)
{
b[i]=i;
// printf("%d ", b[i]);
}
return b;
}
And use it as


b=fct_b()
"[]" has a higher precedence than "*", so "*b[i]" takes the i'th element of b and dereferences it.

[PeterP24]

@ofnuts
I know it is easier to return the pointer instead of passing it by reference in order to be allocated and initialised - this I already tried with respect to the function fct_b - I don't know if it works for function fct_c. However I am interested in the answers to my original questions. It is not like I have to solve some homework or pressing tasks with these answers - it is something I gave some thought and came to a dead end and I would like to find out how to do it in that particular way.

[ofnuts]

@ofnuts
I know it is easier to return the pointer instead of passing it by reference in order to be allocated and initialised - this I already tried with respect to the function fct_b - I don't know if it works for function fct_c. However I am interested in the answers to my original questions. It is not like I have to solve some homework or pressing tasks with these answers - it is something I gave some thought and came to a dead end and I would like to find out how to do it in that particular way.
In fct_c it should be:


*c = malloc(10*sizeof(int*));

[Arndt]

*b[i] means the same as *(b[i]), not (*b)[i].

[Bachstelze]

Your problem with fct_c is that you have not initialised c in main, but you dereference it in fct_c.

[PeterP24]

Ok, I think I got the answer to the first question. It has to be (*b)[i] instead of *b[i] because, as @ofnuts says, I am allocating memory for *b then by using *b[i] I am actually dereferencing b first and not *b. However I did not allocated memory for b with malloc - I allocated for *b. Is it right?

@Bachstelze - I am trying to allocate memory and initialise the pointer c inside the function fct_c.

For function fct_c the following version works (thanks again @ofnuts):

void fct_c(int ***c)
{
int i, j;
*c = malloc(10*sizeof(int*));
for(i = 0; i < 10; i++)
{
(*c)[i] = malloc(10*sizeof(int));
for(j = 0; j < 10; j++)
{
(*c)[i][j]=10*i+j;
//printf("%d ", (*c)[i][j]);
}
}
}