I am getting a segmentation fault 11 on my printArr function though I do not know the reason why. My code compiles. There is no segmentation fault if I do not include the printArr function. Any help will be greatly appreciated.

void printArr(int* arr, int arrlen){
  printf("{");
  for(int i = 0; i < arrlen; i++){
    printf("%d,", arr[i]);
  }
  printf("}\n");
}

int countDeLen (int* in, unsigned int inlen){
  int counter = 0;
  if (inlen % 2 == 1){
    return -1;
  }
  if (inlen == 0){
    return 0;
  }
  int i = 0;
  while (i < inlen){
    counter += in[i];
    i += 2;
  }
  return counter;
}

void rle_decode(int* in, unsigned int inlen, int** out, unsigned int* outlen){
  outlen[0] = countDeLen(in,inlen);
  if(outlen[0] == -1){
    fprintf(stderr, "the code must have an even length!!!\n");
    exit(1);
  }
  int* outCode = (int*)malloc(outlen[0] * sizeof(int));
  out = (int**)malloc(outlen[0] * sizeof(int));
  if(outlen[0] == 0){
    return;
  }
  int codeIndex = 0;
  int i = 0;
  while(i < inlen){
    for(int a = 0; a < in[i]; a++){
      outCode[codeIndex] = in[i+1];
      codeIndex++;
    }
    i+=2;
  }
  out[0] = outCode;
  return;
}

int main(int argc, char *argv[])
{
  printf("testing rle_decode()\n");
  int dtest1[] = {2,1,3,4,2,3}; unsigned int dlen1 = 6; int *out1; unsigned int outlen;
  rle_decode(dtest1, dlen1, &out1, &outlen);
  printArr(*out1,outlen);
}

I am getting a segmentation fault 11 on my printArr function though I do not know the reason why. My code compiles. There is no segmentation fault if I do not include the printArr function. Any help will be greatly appreciated.

1 Answers

3
Barmar On

You're not calling printArr() correctly. It should be:

printArr(out1,outlen);

Your code is equivalent to calling

printArr(out1[0], outlen);

But printArr() requires the first argument to be a pointer to an array of int, not a single int.

You should have gotten a warning about the mismatching types.