How does malloc work?

Question

C++ / C#

There is this code

#include <malloc.h>
#include <stdio.h>
#include <string.h>

int main()
{
  char **text = (char**)malloc(1000);
  for(int i = 0; i < 1000; i ++)
  {
    text[i] = (char*)malloc(100);
  }	return 0;
}

1 - the size of the array element text[n] is 8 and not 100
2 - the array elements contain garbage

I have one question: why does all of the above work like this?
If the question is incorrect, please suggest an edit.

Answer 1

1. You allocated not an array of characters, but an array of pointers to characters. Accordingly, the size of the pointer is 8 bytes.
2. C does not initialize the memory allocated by malloc.

Answer 2

From myself, I’ll add there are functions for clearing memory that fill the buffer at your address with some value, usually 0

Answer 3

"... void * calloc( size_t number, size_t size );
The calloc function allocates a block of memory for an array of size - num elements, each of which occupies size bytes, and initializes all of its bits to zero."
cppstudio.com/post/846

Answer 4

malloc(size_t size) allocates memory equal to size bytes. The example does not explicitly take this into account, and the following for loop will go through uninitialized memory (1000 bytes / 8 bytes pointer = 128). Your best bet is to use the calloc from the answer above, passing in the required number of elements and their size.