Simple memory allocation for embedded application - c

Question

I need to have dynamic memory allocation without the use of malloc in an embedded application.
As we have enough RAM, I would like to allocate a big array (A[A_MAX]) on startup and have buffer arrays (B[]) within this one big array.
They would always begin at A[0] and be as long as they need (at max A_MAX, but mostly shorter).
Example:
at start up:
A[8] = {0,0,0,0,0,0,0,0};
declaring B[4] at A[0] and filling B with ones
B[4] = {1,1,1,1};
now A should look like this:
A[8] = {1,1,1,1,0,0,0,0}
How can I do this?
I looked at the sourcecode for malloc() and it does return the address of the next free part of the heap.
but when I use
float *B[4] = &A[0];
my compiler returns an error (invalid initializer)

Answer 1

As #neon mentioned I wanted to write an address to a float array.
Writing it like float *B = A[0] worked like a charm.

Answer 2

You may do:
#include <stdalign.h>
#define A_MAX 8
alignas(max_align_t) char _memory[A_MAX] = {0};
then you can just use pointers:
int main() {
{
float *floatdata = (void*)_memory;
// use floatdata
}
{
int *intdata = (void*)_memory;
// use intdata
}
}
I would write a handy function to calculate the maximum size for error checking:
// query the max size and allocate
inline void *dataalloc(size_t elemsize, size_t *n) {
*n = sizeof(_data) / elemsize;
return _data;
}
and do:
size_t size;
float *floatdata = dataalloc(sizeof(*floatdata), &size);
// use as-if floatdata[size]
for (size_t i = 0; i < size; ++i) floatdata[i] = i;

Answer 3

do you mean something like ? I'm using pointers to the array
#include <stdio.h>
#include <stdlib.h>
void allocator(float* src,int SrcSize,float **dst,int size,int pos){
if(size <= SrcSize)
for(int i=0;i<size;i++){
dst[i] = &(src+pos)[i];
}
}
int main()
{
float A[8] = {0,0,0,0,0,0,0,0};
float *B[4];
float *C[4];
allocator(A,sizeof(A)/sizeof(float),B,4,0);
*B[2] = 4;
allocator(A,sizeof(A)/sizeof(float),C,4,4);
*C[0] = 6;
printf("A[2] value is : %f\n" , A[2]); // output is 4.000
printf("A[4] value is : %f\n" , A[4]); // output is 6.000
return 0;
}