CODEWITHSUNDEEP
cpointers
guigouz
guigouz

Reputation: 1211

allocating an array inside a c function

I'm trying to allocate and initialize an array inside a function, but I can't seem to fetch the values after returning.

This was my last almost-working attempt

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

int func(int **thing);

int main() {

 int *thing;

 func(&thing);

 printf("%d %d", thing[0], thing[1]);
}

int func(int **thing) {

 *thing = calloc(2, sizeof(int));

 *thing[0] = 1;
 *thing[1] = 2; 

 printf("func - %d %d \n", *thing[0], *thing[1]);
}

but the values printed outside the function are 1 and 0. There are lots of documentation on pointers out there, but I haven't found this specific case covered. Any tips on what I'm doing wrong ?

Upvotes: 3

Views: 2512

Answers (5)

khachik
khachik

Reputation: 28693

After using () for the right precedence, do not forget to change *thing = calloc(2, sizeof(int)); to thing = calloc(2, sizeof(*int)); and to allocate memory for thing elements, since its elements are pointers to integers.

Upvotes: 0

ruslik
ruslik

Reputation: 14870

The [] has higher precedence than derefferensing. Use explicit paranthesis : (*thing)[0] = 1;

Upvotes: 1

Oliver Charlesworth
Oliver Charlesworth

Reputation: 272467

Inside func(), *thing[n] is equivalent to *(thing[n]), i.e. *(*(thing + n)), i.e. array indexing takes precedence over dereferencing. You need to assign using (*thing)[n].

Upvotes: 3

Ben Jackson
Ben Jackson

Reputation: 93700

The precedence of * and [] are such that your assignments mean *(thing[0]). You need to explicitly parenthesize like (*thing)[0].

Upvotes: 5

Greg Hewgill
Greg Hewgill

Reputation: 992877

Rather than passing a pointer-to-pointer, you may find it easier to return the newly allocated array from your function:

int *func();

int main() {

 int *thing;

 thing = func();

 printf("%d %d", thing[0], thing[1]);
}

int *func() {

 int *thing;

 thing = calloc(2, sizeof(int));

 thing[0] = 1;
 thing[1] = 2; 

 printf("func - %d %d \n", thing[0], thing[1]);

 return thing;
}

The reason your code doesn't work is because of this:

*thing[0]

Due to the precedence of operators, you should use:

(*thing)[0]

Upvotes: 6

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