Malloc or calloc

Question

here is a very small structure used for indexing words of a file. Its members are a string (the word), an array of integers (the lines this word is found at), and an integer representing the index of the first free cell in the lines array.

typedef struct {
    wchar_t * word;
    int * lines;
    int nLine;
} ndex;

ndex * words;

I am trying to allocate (ndex)es nb_words = 128 at a time, and (lines) nb_lines = 8 at a time, using malloc and realloc.

First question, what is the difference between malloc(number * size) and calloc(number, size) when allocating *words and/or *lines? Which should I choose?

Second question, I gdbed this:

Program received signal SIGSEGV, Segmentation fault.
0x0000000000400cb0 in add_line (x=43, line=3) at cx17.3.c:124
124     words[x].lines[words[x].nLine] = line;
(gdb) p words[43].nLine 
$30 = 0

In other words, it consistently fails at

words[43].lines[0] = 3;

Since I allocate words by 128, and lines by 8, there is no reason the indexing worked for the 42 previous words and fail here, except if my allocating was botched, is there?

Third question: here are my allocations, what is wrong with them?

words = malloc(sizeof(ndex *) * nb_words);
short i;
for (i = 0; i < nb_words; i++) {
    words[i].lines = malloc(sizeof(int) * nb_lines);
    words[i].nLine = 0;
}

Should I initialize lines in a for(j) loop? I don't see why leaving it uninitialized would prevent writing to it, so long as it as been allocated.

This C is a very mysterious thing to me, thanks in advance for any hints you can provide.

Best regards.

Your `malloc()` and `calloc()` examples both allocate the same amount of memory. — Carl Norum, Mar 19 '13 at 00:15
For the main difference between malloc and calloc, see here: http://stackoverflow.com/questions/1538420/c-difference-between-malloc-and-calloc — Sebastian, Mar 19 '13 at 00:18

score 4 · Accepted Answer · answered Mar 19 '13 at 00:14

4

This looks suspicious:

sizeof(ndex *)

You probably don't want the size of a pointer - you want the size of a structure. So remove the star.

answered Mar 19 '13 at 00:14

Sebastian

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score 2 · Answer 2 · edited May 23 '17 at 10:32

2

Here:

words = malloc(sizeof(ndex *) * nb_words);

You are allocating space for some number of pointers (i.e., 4 bytes * nb_words). What you really need is to allocate some number of ndex's:

words = malloc(sizeof(ndex) * nb_words);

Also, calloc 0 initializes the returned buffer while malloc does not. See this answer.

edited May 23 '17 at 10:32

Community

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answered Mar 19 '13 at 00:14

perreal

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score 0 · Answer 3 · answered Mar 19 '13 at 00:19

malloc will allocate the requested space only. calloc will allocate the space and initialize to zero.
In your example, the segmentation fault is observed here words[x].lines[words[x].nLine] = line;. There could be 2 possibilities viz., allocation is wrong which I don't feel is the case. The more probable case would be words[x].nLine didn't evaluate to 0. Please print this value and check. I suspect this is some huge number which is forcing your program to access a memory out of your allocated space.
Others have answered this part, so I will skip it.

Malloc or calloc

3 Answers3