Creating array of pointers to struct inside a struct

Question

Basiclly, I have a struct defined this way:

struct D_Array
{
    int Capacity;
    int Cur_Size;
    struct Student* List;
};

Now, after creating the struct;

struct D_Array* T_Array;
if (T_Array = malloc(sizeof(struct D_Array)) == NULL)
{
    printf("There has been a problem with memory allocation. Exiting the program.");
    exit (21);
}

I have a problem creating the students list inside this struct. I need the list to be an array of pointers, since I should make some kind of ADT program, I've tried to make it something like this:

T_Array->Capacity = 10;
T_Array->Cur_Size = 0;
T_Array->List[10]= (struct Student*) malloc(sizeof (struct Student*));

It doesn't matter how I tried to change it, I'm getting the same error: incompatible types when assigning to type 'struct Student' from type 'struct Student *'|

I've trying to create an array that I will be able to do something like;

T_Array->List[1]=Student_Pointer;

Thanks!

alk · Accepted Answer · 2016-06-04 15:23:09Z

2

Change the type of the array to struct Student** that will be array of pointers :

struct student * --> a pointer to student. struct student ** --> a pointer to a pointer to student (what you need).

struct D_Array
{
    int Capacity;
    int Cur_Size;
    struct Student** List;
}

and the allocation:

T_Array->List = malloc (sizeof (struct Student*) *  lengthofarray);

and then:

T_Array->List[1]=Student_Pointer;

edited Jun 4, 2016 at 15:23

alk

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answered Jun 4, 2016 at 10:40

Guy haimovitz

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13 Comments

n. m. could be an AI Over a year ago

Don't cast the result of malloc. In sizeof (struct student*)x lengthofarray x should be *.

n. m. could be an AI Over a year ago

Don't cast the result of malloc. If you think you need it, you are either using a compiler for a wrong language, or forgetting to #include required headers.

Cherubim Over a year ago

@Dr.Haimovitz void would automatically convert to the the datatype of the variable that it's getting assigned to... Then why take the pain of casting once again when it's done automatically?

alk Over a year ago

"malloc returns void* wich is obviously not student* so you need a cast" not needed, nor recommended in any way in C, C does this implicitly from and to void-pointers. This is different in C++.

alk Over a year ago

If you do not include the prototype for malloc() on 64bit system its return type will default to int, yes. The compiler would detect this and error. If you would have applied an explicit conversion via a cast to void* the compiler just warns (or do log nothing all), it definitely won't error. Still malloc() would return 32 bits only (what the width of an int), and miss the upper 32bits for the void* it should return with or without cast. Conclusion: The cast hid the error, the program will fail. This is not good. Do not cast where it isn't required (for exp. when call accept().

|

Manoel Rebelo Abranches · Accepted Answer · 2016-06-04 10:45:31Z

0

struct Student* List; List is a pointer of Student When you do List[10] what youre doing is basicaly *(List+10) that means get Student from the 11th position

If you want an array of pointers you need to do something like

struct Student **List;
and then
List = (Student **)malloc(10*sizeof(Student **))
List would then have memory allocated for 10 pointers of student wich would be size of int for each in the end

answered Jun 4, 2016 at 10:45

Manoel Rebelo Abranches

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