assign dynamic array in a function and send it back to the caller in C

Question

I have been using java for long time but for some reason I need to use C (ANSI C not C++) to write a simple code. I need to pass the pointer from outside to a function, allocate some memory to the pointer and assign some values also before the function return. I have my code like

#include <stdio.h>
#include <stdlib.h>
void test(int *a)
{
 int n=3;
 // I have to call another function t determine the size of the array
 n = estimatesize(); // n >=3
 // I tried fix size n=10 also
 a = (int*)malloc(n*sizeof(int));
 a[0] = 1;
 a[1] = 2;
 a[2] = 3;
}
void main(void)
{
 int *s=NULL;
 test(s);
 printf("%d %d %d", s[0], s[1], s[2]);
}

I don't know why the code crashes. I thought at the beginning it is estimatesize() return wrong number but even I fix n to 10, the error still there. So I cannot pass a pointer to a function for memory allocation? If so, how can I dynamically create memory inside a function and pass it out? I know it may be a safe problem in this way but I just want to know if it is possible and how to do that. Thanks.

Answer 1

There are two solutions to this: Either return the pointer from the function, or pass the argument by reference.

For the first one, you simply don't take any arguments, instead you return the pointer:

int *test(void)
{
 int *a = malloc(...);
 ...
 return a;
}
int main(void)
{
 int *s = test();
 ...
}

For the second one, you need to pass the address of the pointer, in other words a pointer to the pointer, using the address-of operator &:

void test(int **a)
{
 *a = malloc(sizeof(int) * 3);
 for (int i = 0; i < 3; ++i)
 (*a)[i] = i;
}
int main(void)
{
 int *s;
 test(&s);
 ...
}

---

The reason it doesn't work now, is because the pointer (s in main) is passed by copying it. So the function test have a local copy, whose scope is only in the test function. Any changes to a in the test function will be lost once the function returns. And as s is copied for the argument, that means that s in main never actually changes value, it's still NULL after the function call.

• user1285419: Thanks for your reply. I get your idea but I tried your code, it doesn't work. Again, it seems that you are implementing it in C++ but I need ANSI C and only have ANSI C compiler.
• Some programmer dude: @user1285419 No, both solutions are pure C, there is no C++ in it.
• WhozCraig: @user1285419 Arrays not withstanding, remember this mantra: C is a by-value parameter language. If you want to pass a parameter by-address, then passing an address is exactly what you need to do: pass the address of the target variable and declare the parameter to be a formal pointer-to-type. And the "type" makes no difference. Whatever the target variable type is, a pointer to that type is the way the parameter should be declared ( in Joachim's code above, the type is int*). (and +1 on the answer, btw).
• user1285419: I think the syntax you define i in the loop only works in c++. In ANSI C, we need to define all variables before the code. Anyway, by fixing that, I have it run in now. Thanks a lot :)
• Some programmer dude: @user1285419 Declaring variables in a for loop like that has been standardized for a very long time (since the C99 standard). What compiler and what version of it are you using?