Compiler/TMS320F28379D: memcpy function receives size input as a pointer

Dear Katta,

Thank you for your reply.

Yes, I've checked those posts, but unfortunately they do not answer my question.

I managed to find the memcpy source code in the TI compiler directory:

#include <string.h>

void *memcpy(void * __restrict to, const void * __restrict from, size_t n)
{
     size_t rn;

     for (rn = 0; rn < n; rn++)
         ((unsigned char *)to)[rn] = ((const unsigned char *)from)[rn];
     return to;
}

Let me describe what I'm confused about. The "to" and "from" inputs are memory addresses where the original code is, and where the original code needs to be copied. The "n" input is not a pointer (i.e., a memory addresses), but an unsigned integer that tells the memcpy function how many "memory locations" need to be copied. However, in all your examples you call the memcpy function as follows:

memcpy(&RamfuncsRunStart, &RamfuncsLoadStart, (uint32_t) &RamfuncsLoadSize);

I get why the first two inputs have the "&" operator - to send the actual memory location where the original code is, and where the code needs to be copied. But what I don't get is the third input:

(uint32_t) &RamfuncsLoadSize

This will send the memory location where the RamfuncsLoadSize is placed, but not the actual size. Here is the simple program that does not work with the "&" operator near the third input:

#include <stdio.h>

void *memcpy(int * to, const int * from, unsigned n)
{
    printf("size in memcpy is %u\n", n);
    
    size_t rn;

    for (rn = 0; rn < n; rn++)
        ((unsigned int *)to)[rn] = ((const unsigned int *)from)[rn];
    return to;
}

int main()
{
    int src[2] = { 1, 2 };
    int dst[2];
    int size = sizeof(src) / sizeof(int);
    
    printf("size is %d (0x%08X)\n", size, &size);
    
    // Does work
    memcpy(&dst, &src, size);
    
    // Does not work!
    //memcpy(&dst, &src, (unsigned)&size);
    
    printf("src = {%d,%d}\n", src[0], src[1]);
    printf("dst = {%d,%d}\n", dst[0], dst[1]);

    return 0;
}

I'm trying to understand where does the "&" operator near the third input come from. Is this some compiler-specific stuff, or the way the linker does things? I was unable to find any documentation that describes this.