memset on float array does not fully zero out array

Question

I have some array,

float *large_float_array;

that I calloc and attempt to memset after a loop

large_float_array = (float*)calloc(NUM, sizeof(float));

// .. do stuff with float array

memset(large_float_array, 0, sizeof(large_float_array);

// .. do other stuff

but it seems large_float_array is not actually zero'd out. Why is this? Is this due to the special properties of floating point numbers? If so, I'm wondering how I could fix this.

PS does calloc actually work here too?

Answer 1

Since large_float_array is a pointer, you cannot use sizeof(large_float_array) to specify the number of bytes to clear: sizeof(large_float_array) is just the size of the pointer, typically 4 or 8 bytes. To clear the whole array, you must write:

memset(large_float_array, 0, NUM * sizeof(float));

or

memset(large_float_array, 0, NUM * sizeof(*large_float_array));

Note that memset sets the memory to all bits zero. The C standard does not guarantee that all bits zero be a representation of 0.0 , But in all current architectures it is the case, so it will work as expected.

The strictly conforming way to clear your array is this:

#include <stdlib.h>

...
for (size_t i = 0; i < NUM; i++) {
    large_float_array[i] = 0;
}

If the target systems represents 0.0 as all bits zero, modern compilers will optimize the above loop as a call to memset as can be verified with Goldbolt's Compiler Explorer .

PS: under the same assumption, calloc() would be a better choice: the block returned is initialized to all bits zero and for large sparse matrices, allocating via calloc() may actually be more cache efficient on many systems.

Answer 2

You can't use memset because it accepts int as value type, but float may be stored in different way, so you should use a regular cycle

for (size_t i = 0; i < NUM; ++i) {
    large_float_array[i] = 0.;
}

Answer 3

The effect of intializing memory area by calloc and using memset function with 0 value is actually the same. In other words, calloc is no more "type-wise" than memset . They both just set memory with all bits zero*.

C11 (N1570) 7.22.3.2/2 The calloc function :

The calloc function allocates space for an array of nmemb objects, each of whose size is size. The space is initialized to all bits zero. ²⁹⁶⁾

Footnote 296 (informative only):

Note that this need not be the same as the representation of floating-point zero or a null pointer constant.

---

*) It's conceivable that calloc could return address of memory location that is already pre-initialized with zeros, thus it may be faster than malloc + memset combo.

Answer 4

Post has multiple issues.

Following does not have balanced () . Unclear what OP's true code may be.

memset(large_float_array, 0, sizeof(large_float_array);

Assuming the above simple used another ) , then code was only zero filling a few bytes with zero as sizeof(large_float_array) is the size of a pointer.

memset(large_float_array, 0, sizeof(large_float_array));

OP likely wanted something like which would zero fill the allocated space with zero bits. This will bring the allocated memory into the same state as calloc() .

memset(large_float_array, 0, NUM *sizeof *large_float_array);

Code should use the following. A good compiler will optimize this trivial loop into fast code.

for (size_t i=0; i<NUM; i++) {
  large_float_array[i] = 0.0f;
}

---

A way to use the fast memcpy() is to copy the float into the first element of the array, then the first element to the 2nd element, then the first 2 elements to elements 3 & 4, then the first 4 elements to elements 4,5,6,7 ...

void *memset_size(void *dest, size_t n, const char *src, size_t element) {
  if (n > 0 && element > 0) {
    memcpy(dest, src, element);  // Copy first element
    n *= element;
    while (n > element) {
      size_t remaining = n - element;
      size_t n_this_time = remaining > element ? element : remaining;
      memcpy((char*)dest + element, dest, n_this_time);
      element += n_this_time;
    }
  }
  return dest;
}

float pi = M_PI;
float a[10000];
memset_size(a, 10000, &pi, sizeof pi);

As with all optimizations, consider clearly written code first and then employ candidates such as the above in the select cases that warrant it.