Why malloc+memset is slower than calloc?

Question

It's known that `calloc` is different than `malloc` in that it initializes the memory allocated. With `calloc`, the memory is set to zero. With `malloc`, the memory is not cleared.So in everyday work, I regard `calloc` as `malloc`+`memset`.Incidentally, for fun, I wrote the following code for a benchmark.The result is confusing.Code 1: #include<stdio.h> #include<stdlib.h> #define BLOCK_SIZE 1024*1024*256 int main() { int i=0; char *buf[10]; while(i<10) { buf[i] = (char*)calloc(1,BLOCK_SIZE); i++; } }Output of Code 1: time ./a.out **real 0m0.287s** user 0m0.095s sys 0m0.192s Code 2: #include<stdio.h> #include<stdlib.h> #include<string.h> #define BLOCK_SIZE 1024*1024*256 int main() { int i=0; char *buf[10]; while(i<10) { buf[i] = (char*)malloc(BLOCK_SIZE); memset(buf[i],'\\0',BLOCK_SIZE); i++; } }Output of Code 2: time ./a.out **real 0m2.693s** user 0m0.973s sys 0m1.721s Replacing `memset` with `bzero(buf[i],BLOCK_SIZE)` in Code 2 produces the same result.**My question is:** Why is `malloc`+`memset` so much slower than `calloc`? How can `calloc` do that?

Answer 1

The process of malloc then memset is a two step operation, and depending on how the memset is implemented, it's most likely going to have to walk all 268,435,456 elements of your array to assign each the value of 0, and that's going to take time. calloc (or as I like to think of it, clear allocation) does this at the same time it is allocating your memory request, and as it knows what the value is going to be (0) it can use a different assembly code to implement the operation.

A good compiler would be able to optimize this; try building it with no debug and optimization turned up, and see if your test results aren't different.