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As per Why malloc+memset is slower than calloc?

malloc+memset is slower than calloc under certain conditions.

Why wasn't calloc written in such a way that it can take an extra value argument ( like memset) to override default assignment by zero? What would have been the effect of that if it were done?

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user13107
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    From reading the answer in the question you linked, you'd probably lose the speed advantage if calloc was to accept a value argument. – tangrs May 14 '13 at 06:52
  • malloc and calloc are two differrent things.They both are getting used according to requirement. Most of the time, as soon as you allocate memory, you're going to fill in its contents yourself (eg by assigning values to all the elements in an array, or all the fields in a struct). In that case, you don't need the memory to be zeroed before you use it, so you use malloc. If you have a specific need for the memory to be zeroed, you can use calloc. if you are not happy with zero go for (malloc + memset) else try to feel happy with calloc @user13107 thanks – Dayal rai May 14 '13 at 07:05
  • How does that answer the O.P's question? – tangrs May 14 '13 at 07:06
  • @tangrs I am just trying to explain it was designed in such a way that is total requirement dependenet.there is no extra argument in calloc because memset is already defined.Why to give overhead to a single function when things are possible in simple manner. – Dayal rai May 14 '13 at 07:12
  • @Dayalrai Fair enough, that makes more sense. – tangrs May 14 '13 at 07:15
  • If you read the accepted answer for the question you're refering to you'll get all the answers you need. calloc typically gets memory from a pool of zero initialized memory by design. – onemasse May 14 '13 at 07:16
  • http://stackoverflow.com/faq#dontask – xaxxon May 14 '13 at 07:23
  • @xaxxon exactly which rule am I flouting? – user13107 May 14 '13 at 07:54
  • Chatty, open-ended questions diminish the usefulness of our site and push other questions off the front page. – xaxxon May 14 '13 at 08:18

2 Answers2

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These calloc or memset initializations operate on a byte level, so even memset with a value different from 0 is not that usefull. At least I don't remember having it used with different values. Mostly you allocate memory for a base type that is wider than char.

The other aspect is that calloc is initialization and not assignment. Platforms may have builtins that provide a fast initialization of all bytes to 0, you wouldn't capture this when passing an argument to initialize.

But probably the most important aspect is that this is history of C. These interfaces originate from the very beginning and are impossible to change.

Jens Gustedt
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That's the way it's designed.

Mainly whenever you allocate memory, you are immediately going to assign values to it. Forcing calloc to initialize a particular is going to make unnecessary overhead. For example, you want to allocate an array of 1 million int's and assign values from 1 to 1 million.

If calloc() were to take an argument like:

int *ptr;
int initValue = 3;

ptr = calloc(NumItems, sizeof(int), initValue);

then calloc() is forced to initialize initValue to all the 1 million elements needlessly.

P.P
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