Reverse Array of C-Strings - c

Question

I have a few questions regarding array of strings in C.
I have an array char *string. I have a char *string and then I split every 4 characters in a array of strings called sep_str. So for example if char *string = 'The sum';, then char **sep_str is:
0: |_| --> "The "
1: |_| --> "Sum"
My first question is, in an array of strings in C (so array of array of chars), will there be a null terminating character at the end of each sep_str[i], or just at the last position of sep_str? Here is how I copy string into an array of strings:
for (int i = 0; i < str_length; i++) {
sep_str[i/4][i%4] = *ptr;
ptr++;
}
My second question is, how would I reverse the elements of each string in sep_str? Here's how I did it, but I feel like it is stepping out of the array of the substring. (so out of the element of the sep_str):
// Reverse each element in the array
char temp;
for (int i = 0; i < num_strs; i++) {
for (int j = 0, k = 4; j < k; j++, k--) {
temp = sep_str[i][j];
sep_str[i][j] = sep_str[i][k];
sep_str[i][k] = temp;
}
}

Answer 1

The copy of the strings sounds good to me. Since each string has always 4 chars, you can avoid the null terminator \0. Alternatively you need to declare sep_str as a 5x(lenght/4) matrix, to store the \0 char at the end of each string.
To reverse a string you need to iterate from the start to the middle of the string, replacing the i-th char with the length-i-1-th. You need to replace the inner for replacing k=3 to k=2.
You also need to take care of the last string, since the lenght might not be multiple of four.
char temp;
for (int i = 0; i < (num_strs - 1); i++) {
for (int j = 0, k = 3; j < k; j++, k--) {
temp = sep_str[i][j];
sep_str[i][j] = sep_str[i][k];
sep_str[i][k] = temp;
}
}
if (num_strs > 0) {
for (int j = 0, k = strlen(sep_str[i]) - 1; j < k; j++, k--) {
temp = sep_str[i][j];
sep_str[i][j] = sep_str[i][k];
sep_str[i][k] = temp;
}
}

Answer 2

In a C string, there will be only one termination character. But if you need to tokenize the strings, then each string must be null terminated.
But before that -
char *string = "The sum"; // should be const char* string = "The sum";
String literal in the above case resides in read only location and cannot be modified. If you need to modify, then
char string[] = "The sum";

Answer 3

If you don't have the terminating character in your strings then yes, you will be outside the bounds of the array since you are accessing sep_str[i][4], which is not a valid location:
sep_str[0] = 'T'
sep_str[1] = 'h'
sep_str[2] = 'e'
sep_str[3] = ' '
However, I doubt that you want to have the null character at the beginning of your string, so you need k=3 in your for loop, not k=4.

Answer 4

My first question is, in an array of strings in C (so array of array of chars), will there be a null terminating character at the end of each sep_str[i], or just at the last position of sep_str?
Only at the end, but if you want to treat each individual chunk as its own string, you'll need to add the \0 yourself.
My second question is, how would I reverse the elements of each string in sep_str?
You could do it with pointers...
char temp;
// Point to start of string, `str` will decay to first memory position.
char *start = str;
// Point to the end of the string. You will need to `#include <string.h>`
// for `strlen()`. Otherwise, write a `while` loop that goes until `\0` to find
// the last position.
char *end = &str[strlen(str) - 1];
// Do until we hit the middle of the string.
while (start < end) {
// Need a temp char, no parallel assignment in C.
temp = str[start];
// Swap chars.
str[start++] = str[end];
str[end--] = str[temp];
}
Assuming str is your string.