You are not bothering me. It is just hard to help when there isn't much debugging data to help analyze the problems. However, the output you've provided helps. It is also clear from the debugging code you've added and the output you've displayed that you are having problems understanding how pointers work in C. Let me walk you through some of your code and explain what it means.

First, the declarations:

Code:

uint8_t *M = malloc(MAX_M_LEN*sizeof(uint8_t));
uint8_t *C = malloc((2*KEYDIGITS*NN_DIGIT_LEN + 1 + MAX_M_LEN + HMAC_LEN)*sizeof(uint8_t));	
uint8_t *dM = malloc(MAX_M_LEN*sizeof(uint8_t));

create M, C, and dM as pointers to objects of type uint8_t and the calls to malloc() allocate buffers to hold data to be stored into the areas pointed to by those pointers. Then you have:

Code:

printf("C before encrypt %p\n",*C);
printf("M before encrypt %p\n",*M);
printf("dM before encrypt %p\n",*dM);

The %p in these printf format strings are used to convert the value of a pointer (the address to which the pointer points) to a printable value. But the operands you're providing for these conversions are objects of type uint8_t, not pointers. (C is a pointer, *C is the uint8_t object pointed to by that pointer; M is a pointer, *M is the uint8_t object pointed to by M; dM is a pointer, *dM is the uint8_t object pointed to by dM.) If you want to print the contents of the pointers, the above code needs to be changed to:

Code:

printf("C before encrypt %p\n",C);
printf("M before encrypt %p\n",M);
printf("dM before encrypt %p\n",dM);

If you want to print the unsigned byte pointed to be these pointers (both as an unsigned decimal value and as a printable character), the above code needs to be changed to something like:

Code:

printf("*C before encrypt %hhu(%c)\n",*C,*C);
printf("M before encrypt %hhu(%c)\n",*M,*M);
printf("dM before encrypt %hhu(%c)\n",*dM,*dM);

Then your code:

Code:

if (dM == M){printf("Works\n");}
else{printf("Not Works\n");}

is comparing the pointers (not the unsigned 8-bit integers to which they point). Since these pointers were initialized by separate calls to malloc() and not changed since they were initialized, they can't have the same value. They might point to bytes that have the same value, but that would be coded as:

Code:

if (*dM == *M){printf("Works\n");}
else{printf("Not Works\n");}

or as:

Code:

printf("%sWorks\n", (*dM == *M) ? "" : "Not ");

While we're talking about pointers, you may also see an ampersand followed by an object (i.e., &object). This gives you a pointer to the object. So, for example, the following code segment:

Code:

char *ptr = "abcd";
printf("&ptr=%p, ptr=%p, *p=%c\n", &ptr, ptr, *ptr);

will print the address in memory where ptr is allocated (&ptr), the address in memory of the first byte in the string "abcd" (ptr), and the character pointed to by ptr (*ptr) which in this case is 'a'.

I hope this helps.