Malloc a pointer to a pointer to a structure array by reference
The code below compiles, but immediately crashes for reasons obvious to others, but not to me. I can't seem to get it right, can anyone tell me how to fix this.
*array_ref[2] = array[0];
*array_ref[3] = array[1];
It crashes on that part everytime.
typedef struct test {
char *name;
char *last_name;
} person;
int setName(person ** array, person ***array_ref) {
*array = malloc (5 * sizeof(person));
*array_ref= malloc(5 * sizeof(person*));
array[0]->name = strdup("Bob");
array[1]->name = strdup("Joseph");
array[0]->last_name = strdup("Robert");
array[1]->last_name = strdup("Clark");
*array_ref[2] = array[0];
*array_ref[3] = array[1];
return 1;
}
int 开发者_Python百科main()
{
person *array;
person **array_r;
setName(&array,&array_r);
printf("First name is %s %s\n", array[0].name, array[0].last_name);
printf("Second name is %s %s\n", array_r[3]->name, array_r[3]->last_name);
while(1) {}
return 0;
}
Operator []
has higher precedence than unary operator*
. Hence, this:
*array_ref[2] = array[0];
*array_ref[3] = array[1];
actually means:
*(array_ref[2]) = array[0];
*(array_ref[3]) = array[1];
Types are correct here, which is why it compiles. But from your code it's clear that your intent actually was:
(*array_ref)[2] = array[0];
(*array_ref)[3] = array[1];
So just use parentheses.
You allocated space for the array_ref pointers, but not for what they point to.
Try changing the following in setName()
*array_ref[2] = array[0];
*array_ref[3] = array[1];
to
*(*array_ref+2) = array[0];
*(*array_ref+3) = array[1];
This works.
array[1]->name
is your problem. This should be (*array)[1].name
. Notice how the two aren't equivalent. All the similar uses have the same problem, except for [0]
, which accidentally does the right thing.
Remember that array
, the function parameter, isn't your array, it's a pointer to your array.
In functions like this I prefer code like:
int setName(person ** out_array, person ***out_array_ref) {
person* array = malloc(5 * sizeof(person));
person** array_ref = malloc(5 * sizeof(person*));
array[0].name = strdup("Bob");
array[1].name = strdup("Joseph");
array[0].last_name = strdup("Robert");
array[1].last_name = strdup("Clark");
// I'm guessing this was your intent for array_ref, here:
array_ref[2] = &array[0];
array_ref[3] = &array[1];
*out_array = out_array;
*out_array_ref = array_ref;
return 1;
}
Note that this catches both array[1]->name
as noted by Roger Pate, and *array_ref[2] = array[0]
as (almost) noted by Pavel - whose solution (*array_ref)[2] = array[0] assigns from an unallocated person*
array[1]
- both of which are hard to notice with the extra dereference.
Of course, I mostly do this because I use C++, and this increases exception safety ;).
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