Struct array gets wrong size with malloc

Question

I'm trying to make a dynamic struct array using malloc.

I have a struct: 
typedef struct{
    char question[200];
    char answer1[30];
    char answer2[30];
    char answer3[30];
    char correctAnswer[2];
}Questions;

And the following code:

int length = 10;
Questions * qArray = malloc(length*sizeof(*qArray));

For some weird reason the sizeof(qArray) returns 4 even if I change length. How do I actually set the correct size of the array?

You seem to be working on a 32 bit platform with 32 bit pointers. Now, read your question again - carefully. — too honest for this site, Dec 03 '15 at 15:57
@Jullix993 that's a correct way to set the size of the array. The way you read it is wrong. — Manos Nikolaidis, Dec 03 '15 at 16:03
@ManosNikolaidis I see. So is there anyway to get the size of the array? I will need it for my for loops. — Jullix993, Dec 03 '15 at 16:05
@JohnnyMopp - that's not the problem. The problem is the questioner is expecting to get the size of the array from the pointer via `sizeof(qArray)` *after* calling `malloc()`. There's no difference between `sizeof(Questions)` and `sizeof(qArray)`. — Andrew Henle, Dec 03 '15 at 16:08
@AndrewHenle Exactly, I need the size, which in my case should be 10. — Jullix993, Dec 03 '15 at 16:09
What is your actual problem? You already allocate 10 entries. You just have to differentiate between the size of the pointer (which you don't need) and the **length** of he array. — too honest for this site, Dec 03 '15 at 16:09
@AndrewHenle, There is no difference between `sizeof(Questions)` and `sizeof(*qArray)`. `sizeof(qArray)` is different. — Haris, Dec 03 '15 at 16:10
@JohnnyMopp - *That was not clear from the question.* True. I had to read it several times myself. The real question is buried after the code blocks in the final sentence. — Andrew Henle, Dec 03 '15 at 16:13
@Jullix993 You already know the size. You set it to 10 to call `malloc()`, so save it somewhere. — Andrew Henle, Dec 03 '15 at 16:15

score 3 · Answer 1 · answered Dec 03 '15 at 15:57

3

The allocated size is in no way associated with the pointer to the first element.

In other words you cannot do that, sizeof(qArray) is statically computable from the type.

You need to track the allocated size at run-time separately yourself.

answered Dec 03 '15 at 15:57

unwind

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Manos Nikolaidis · Accepted Answer · 2015-12-03T16:20:43.140

With these lines you are allocating memory for an array of 10 Questions

int length = 10;
Questions * qArray = malloc(length*sizeof(*qArray));

You could achieve the same result with :

Questions * qArray = malloc(length*sizeof(Questions));

The former syntax makes it easier to change the type in the future and is preferred by a lot of people.

After that :

sizeof(qArray[0]) // size of the first element of the array in bytes (292 probably)
sizeof(*qArray)   // size of the first element of the array in bytes (292 probably)
sizeof(Questions) // size of a struct of type Questions in bytes (292 probably)
sizeof(qArray)    // size of a pointer in your system in bytes (4 or 8 bytes usually)
length            // number of elements in qArray
length*sizeof(*qArray) // total size of the array in bytes

Haris · Answer 3 · 2015-12-03T16:16:31.813

0

You are setting the size of the array correctly only. The problem is the way you are using to get back the size.

For some weird reason the sizeof(qArray) returns 4 even if I change length.

Because qArray is a pointer. So sizeof(qArray) is the size of a pointer, that is size of your address, which is 4 bytes in your case.

Doing sizeof(*qArray) will actually give you the size of the structure Questions, since *qArray is pointing to a structure Questions.

How do I actually set the correct size of the array?

To get the full size of the array, you need to keep length handy everytime. Since you are using pointers. Just like unwind said.

edited Dec 03 '15 at 16:16

answered Dec 03 '15 at 15:56

Haris

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In actual question OP actually uses `sizeof(*qArray)`. – ameyCU Dec 03 '15 at 16:00
@ameyCU *In actual question OP actually uses `sizeof(*qArray)`* No, he doesn't. The real question is at the very bottom: *For some weird reason the `sizeof(qArray)` returns `4` even if I change length. How do I actually set the correct size of the array?* – Andrew Henle Dec 03 '15 at 16:21
@ameyCU, Thats for declaring, he is asking why `sizeof(qArray)` gives `4`. – Haris Dec 03 '15 at 16:22
@Haris Ohh fine then , my bad . But FYI I did not downvote :-) . – ameyCU Dec 03 '15 at 16:28
@ameyCU, Not a problem. – Haris Dec 03 '15 at 16:29

score -1 · Answer 4 · answered Dec 03 '15 at 16:11

-1

You can't get the size of the array you allocated from the pointer. You already know the size since you had to compute it to call malloc():

Questions * qArray = malloc(length*sizeof(*qArray));

If you need to retain the size of the array:

size_t qArraySize = length * sizeof( *qArray );
qArray = malloc( qArraySize );

answered Dec 03 '15 at 16:11

Andrew Henle

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sizeof *qArray should be Questions – Robert Jacobs Dec 03 '15 at 16:11
@RobertJacobs The question *after* the code blocks is "For some weird reason the sizeof(qArray) returns 4 even if I change length. How do I actually set the correct size of the array?" Please read. – Andrew Henle Dec 03 '15 at 16:12
1

@RobertJacobs - *sizeof *qArray should be Questions* And no, it should **not** be. Because if someone were to change the definition of qArray, to something like `struct NewQuestions`, then one would also have to remember to change the size calculation code when there is no direct connection between the two other than the pointer used to hold the address of the allocated block. – Andrew Henle Dec 03 '15 at 16:17

Robert Jacobs · Answer 5 · 2015-12-03T16:10:03.987

-2

depending on what you want.

Number of elements in array

length

Number of bytes in array. This should be the argument to malloc.

length*sizeof(Questions)

edited Dec 03 '15 at 16:10

answered Dec 03 '15 at 16:05

Robert Jacobs

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It still gives me the number 4. Where do the number 4 come from? – Jullix993 Dec 03 '15 at 16:07
1

Number of bytes in array --> `length*sizeof(*qArray)` – Haris Dec 03 '15 at 16:08
@Jullix993, Number `4` is the size of your address. – Haris Dec 03 '15 at 16:08
My Bad. Editing answer. – Robert Jacobs Dec 03 '15 at 16:09
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The calcilation of the allocated size in the question is fine and the recommended way. Telling OP now that it should be done by using the type is just confusing. – too honest for this site Dec 03 '15 at 16:16

Struct array gets wrong size with malloc

5 Answers5