What are the contents of the memory just allocated by `malloc()`?

Question

I was curious about what exactly a pointer holds, after malloc() was used to allocate memory space? The manpage tells me that calloc() initializes the allocated memory space with zero.

The malloc() function allocates size bytes and returns a pointer to the allocated memory. The memory is not initialized. If size is 0, then malloc() returns either NULL, or a unique pointer value that can later be successfully passed to free().

and

The calloc() function allocates memory for an array of nmemb elements of size bytes each and returns a pointer to the allocated memory. The memory is set to zero. If nmemb or size is 0, then calloc() returns either NULL, or a unique pointer value that can later be successfully passed to free().

I created a really short example program in C, to C(haha) for myself:

int main() {
    char *dynamic_chars;
    unsigned amount;
    printf("how much bytes you want to allocate?\n");
    scanf("%d", &amount);

    dynamic_chars = (char*)malloc(amount*sizeof(char));
    printf("allocated:\n%s\n", dynamic_chars);

    free(dynamic_chars);
    return 0;

}

However when executing this code, it just outputs nothing. If I initialize the memory my self for example initializing every single byte with 0xFFFF using a loop, then the program shows me exactly what I expect. The memory space actually exists, since I wont get an error claiming that I am trying to access uninitialized variables or so.

Since memory space is usually not deleted but marked as rewritable I wonder if by executing my program, shouldn't I be able to see random previously used Bytes of memory? But I wont see anything, so I am really confused about how exactly malloc() works.

EDIT1

Another thing about malloc() or maybe memory usage in general, that is interesting about my program: If I use calloc(), to allocate memory, I can trace the actual memory usage of my program, by e.g. monitoring it. For example, if I tell my program, to allocate 1.000.000.000 Bytes of memory per calloc() I will see the following in my System monitor:

As you can probably imagine, when using malloc(), I wont see nothing. I understand, that just by allocating memory, I am not really using it at that time, but I am still confused about why my operating system (unix derivate) won't recognize it as being used. Since malloc() just like calloc() returns a physical address to a memory location I don't get, how this memory area seems to be not actually reserved by the OS. Elsewise I could see it in the System Monitor right? If I should rather post this as a new question, please let me know. But I think since the question is still about how malloc() works it fits in here.

Answer 1

No, malloc() returns uninitialized memory, the contents of which is indeterminate. So, attempt to use the value invokes undefined behavior.

Quoting C11, annex §J.2, Undefined behavior

The value of the object allocated by the malloc function is used

In this case, %s expects a null-terminated char array. However, the content of dynamic_chars is indeterminate, so there may very well be no null-terminator, at all, which will cause the out-of-bound memory access, which in turn invokes the UB.

Quoting C11, chapter §7.22.3.5, The malloc function (emphasis mine):

The malloc function allocates space for an object whose size is specified by size and whose value is indeterminate.

That said, please see this discussion on why not to cast the return value of malloc() and family in C..