Casting from unsigned into signed char in C

Question

I am converting an input raw pcm stream into mp3 using lame. The encoding function within that library returns mp3 encoded samples in an array of type unsigned char. This mp3-encoded stream now needs to be placed within an flv container which uses a function that writes encoded samples in char array. My problem is that I am passing the array from lame (of type unsigned char) into the flv library. The following piece of code (only symbolic) illustrates my problem:

/* cast from unsigned char to char. */

#include <stdio.h>
#include <stdlib.h>

void display(char *buff, int len) {

  int i = 0;
  for(i = 0; i < len; i++) {
    printf("buff[%d] = %c\n", i, buff[i]);
  }
}

int main() {

  int len = 10;
  unsigned char* buff = (unsigned char*) malloc(len * sizeof(unsigned char));
  int i = 0;

  for(i = 65; i < (len + 65); i++) {
    buff[i] = (unsigned char) i;
    printf("char = %c", (char) i);
  }

  printf("Displaying array in main.\n");
  for(i = 0; i < len; i++) {
    printf("buff[%d] = %u\n", i, 'buff[i]');
  }

  printf("Displaying array in func.\n");
  display(buff, len);

  return 0;
}

My question(s):
1. Is the implicit type conversion in the code below (as demonstrated by passing of buff into function display safe? Is some weird behaviour likely to occur?
2. Given that I have little option but to stick to the functions as they are present, is there a "safe" way of converting an array of unsigned chars into chars?

Answer 1

The only problem with converting unsigned char * into char * (or vice versa) is that it's supposed to be an error. Fix it with a cast.

display((char *) buff, len);

Note: This cast is unnecessary:

printf("char = %c", (char) i);

This is fine:

printf("char = %c", i);

The %c formatter takes an int arg to begin with, since it is impossible to pass a char to printf() anyway (it will always get converted to int, or in an extremely unlikely case, unsigned int.)