making an array-based program work with no elements - c

Question

The exercise it to take an array of strings and, using pointers, concatenate them in a new string. I made it work by hard coding the values at each string array index, but I can't figure out how to make it work if the array has NO elements. I think I need to entirely restructure my code...
#include <stdio.h>
#include <stdlib.h>
int main(int argc, const char * argv[]) {
char *newArray[3];
//original code didn't have newArray as a pointer, and had the following:
//char *firstString = &newArray[0]; This was pointing to the value at the address of newArray[0] which was null at this point in time
//firstString = "Knicks"; This was reassigning the value of firstString to "Knicks" instead of pointing to the value located at the proper index address.
//this code correctly assigns values to each index of the array (which itself is just a pointer that is pointing to these values
newArray[0] = "Knicks";
newArray[1] = "Warriors";
newArray[2] = "Bulls";
//this code correctly assigns pointers that point to the index values of the array, which are all stated above.
char *firstString = newArray[0];
char *secondString = newArray[1];
char *thirdString = newArray[2];
int firstStringCount = 0;
int secondStringCount = 0;
int thirdStringCount = 0;
//count length of first element
for (int i = 0; i < 1000; i++) {
if (firstString[i] == '\0') {
break;
}
else {
firstStringCount++;
}
}
//count length of second element
for (int i = 0; i < 1000; i++) {
if (secondString[i] == '\0') {
break;
}
else {
secondStringCount++;
}
}
//count length of third element
for (int i = 0; i < 1000; i++) {
if (thirdString[i] == '\0') {
break;
}
else {
thirdStringCount++;
}
}
//int len = firstStringCount + secondStringCount + thirdStringCount;
char *concatenatedString = malloc(firstStringCount + secondStringCount + thirdStringCount);
for (int i = 0; i < firstStringCount; i++) {
concatenatedString[i] = firstString[i];
}
for (int i = 0; i < secondStringCount; i++) {
concatenatedString[i+firstStringCount] = secondString[i];
}
for (int i = 0; i < thirdStringCount; i++) {
concatenatedString[i+firstStringCount+secondStringCount] = thirdString[i];
}
//add in null value
concatenatedString[firstStringCount + secondStringCount + thirdStringCount] = '\0';
printf("%s\n", concatenatedString);
printf("%i\n", firstStringCount);
printf("%i\n", secondStringCount);
printf("%i\n", thirdStringCount);
return 0;
}

Answer 1

Your program will work when you have empty strings. This is not the same as having no array elements, because there must still be a \0 terminator in each, such as would be created with
newArray[0] = "";
But you have not allocated enough memory for the destination's string terminator. By using a 1 +
char *concatenatedString = malloc(1 + firstStringCount + secondStringCount +
thirdStringCount);
you now have room for the terminator with
concatenatedString[firstStringCount + secondStringCount + thirdStringCount] = '\0';