My question is aboutt dynamic memory allocation in C. I have been asked to dynamically allocate an array of n
longs, and return the pointer to the first element of this array. I have some code to test the output of this but the memory allocation is failing.
long* make_long_array(long n)
{
int i;
int *a;
a = (int*)malloc(sizeof(int)*n);
if (a == NULL) {
printf("ERROR: Out of memory\n");
return 1;
}
for (i = 0; i < n; *(a + i++) = 0);
return *a;
}
Im getting an error on two lines saying
'error: return makes pointer from integer without cast'
this occurs for the lines
return 1;
and
return *a;
I'm not entirely sure how to fix this. I think the error in return 1;
being that I am trying to return an integer when it is looking for a pointer? But I am not sure how to fix it for the return of the pointer. Any help would be much appreciated.
To fix your original version:
long* make_long_array(/* long not the correct type for sizes of objects */ size_t n)
{
// int i; define variables where they're used.
/* int you want to return a */ long *a; // array.
a = /* (int*) no need to cast */ malloc(sizeof(/* int */ you want */ long /*s, remember? *) */ ) * n);
if (a == NULL) {
printf("ERROR: Out of memory\n"); // puts()/fputs() would be sufficient.
return /* 1 */ NULL; // 1 is an integer. Also it is uncommon to return
} // anything other than NULL when a memory allocation
// fails.
for (size_t i = 0; i < n; /* *(a + i++) = 0 that falls into the category obfuscation */ ++i )
/* more readable: */ a[i] = 0;
// return *a; you don't want to return the first long in the memory allocated
return a; // but the address you got from malloc()
}
A Better Way tm to write such allocations is
FOO_TYPE *foo = malloc(NUM_ELEMENTS * sizeof(*foo)); // or
BAR_TYPE *bar = calloc(NUM_ELEMENTS, sizeof(*bar));
By using *foo
and *bar
as the operand of sizeof
you don't have to worry about changing it when the type of foo
or bar
changes.
Your function can be simplified to
#include <stddef.h> // size_t
#include <stdlib.h> // calloc()
long* make_long_array(size_t size) // size_t is guaranteed to be big enough to hold
{ // all sizes of objects in memory and indexes
return calloc(size, sizeof(long)); // into them. calloc() initializes the memory
} // it allocates with zero.
// if you really want an error-message printed:
long* make_long_array(size_t size)
{
long *data = calloc(size, sizeof(long));
if (!data) // calloc() returned NULL
fputs("Out of memory :(\n\n", stderr); // Error messages should go to stderr
return data; // since it is unbuffered*) and
} // might be redirected by the user.
*) so the user gets the message instantly.
Also there is no need to cast the result of *alloc()
since they return a void*
which is implicitly convertible in every other pointer type.
Could be written as a macro so it not only works for long
but for any type:
#include <stddef.h>
#include <stdlib.h>
#define MAKE_ARRAY(TYPE, COUNT) calloc((COUNT), sizeof((TYPE)))
// sample usage:
int main(void)
{
int *foo = MAKE_ARRAY(*foo, 12);
long *bar = MAKE_ARRAY(*bar, 24);
char *qux = MAKE_ARRAY(*qux, 8);
free(qux);
free(bar);
free(foo);
}
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