Does calloc initialize structures arrays?

Question

I have the doubt if calloc initialize to zero all of the elements of a struct array like:

#define MAXDATA    10

struct Est2 {
int dato0;      // Index k
int dato1;      // Index j
int dato2;      // Index i
double dato3;   // Y Coordinate
};

Est2 *myArray = (Est2*) calloc(MAXDATA, sizeof(Est2));

I'm asking this because I don't want the initial data of myArray have garbage or Is there a problem if I don't initialize the array with any value if later in the code I will initialize it anyway for example storing the result of some arithmetical operations? Thanks in advance.

Did you print the structure values to see their values are 0? To answer you question, yes calloc will initialize all elements to 0 — aakash, Apr 10 '14 at 01:09
`malloc` is the one that doesn't initialise the memory first, `calloc` sets it to value zero. — Leigh, Apr 10 '14 at 01:12
No, `calloc` set its to all bytes zero. This may or may not be value zero. — M.M, Apr 10 '14 at 01:23

score 5 · Answer 1 · answered Apr 10 '14 at 01:10

From man calloc:

void *calloc(size_t count, size_t size);

The calloc() function contiguously allocates enough space for count objects that are size bytes of memory each and returns a pointer to the allocated memory. The allocated memory is filled with bytes of value zero.

calloc() guarantees that it's going to be pointing at zeroed data.

score 1 · Answer 2 · answered Apr 10 '14 at 01:09

1

All of the bytes in the structure are set to 0.

This means that the ints have value 0. The double could be a trap , although most common systems use IEEE 754 representation for double, in which the value would be 0.0.

answered Apr 10 '14 at 01:09

M.M

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BTW, don't cast `calloc`. – M.M Apr 10 '14 at 01:09

Does calloc initialize structures arrays?

2 Answers2