Count characters in an array of character strings? - c

Question

Given an array of character strings such as...
char *example[] = {"s", "ss", "sss"};
How can I write a function to count the total number of chars in the array including the terminating characters, without using the standard library for strlen() etc.
Follows is my attempt
int countChars(char *array[], int len)
{
int total = 0, count = 0;
for (int i = 0; i < len; i++)
{
if (array[i] != NULL)
{
while (*array[i] != '\0') {
count++;
}
count++;
}
total += count;
}
return total;
}
An explanation on how char *array[] actually works for access wold be appreciated. I believe that it is supposed to be an array of pointers to strings.

Answer 1

You have to increment the index to consider each of the character.
Something like this:-
for (int i = 0; i < len; i++)
{
if (array[i] != NULL)
{
int j=0,count=0;
while (array[i][j++] != '\0') {
count++;
}
total += count;
}
}
Also reset the count or add to total at the end of all the calculation.
As an answer to your second question:-
char* array[] is basically denoting an array pointers each pointing
to the string literals with which you initialized it.
So once you use array[i] you should now think that it is nothing
other than a pointer to a string literal.

Answer 2

You need to reinitialize the variable count inside the for loop for each processed string and to increase the expression *array[i] inside the while loop.
Also it is better when the function has the return type size_t (size_t is the type that is returned by the standard C function strlen and by the operator sizeof)
The function can look as it is shown in the demonstrative program.
#include <stdio.h>
size_t countChars( const char *array[], size_t n )
{
size_t count = 0;
while ( n-- )
{
if ( array[n] )
{
size_t i = 0;
do { ++count; } while ( array[n][i++] );
}
}
return count;
}
int main(void)
{
const char * example[] = { "s", "ss", "sss" };
printf( "%zu\n", countChars( example, sizeof( example ) / sizeof( *example ) ) );
return 0;
}
The program output is
9
Each element of this array
char *example[] = {"s", "ss", "sss"};
has type char * and is a pointer to the first character of the corresponding string literal.

Answer 3

Since your array contains string constants you should declare it with const:
const char *example[3];
Without const the compiler will not warn you if you try to assign a character to example[i][j]. For the same reason the formal parameter should also be declared with const.
For a pure function with no side effects it is better to name it so that it reflects the result. Therefor I would use charCount instead of countChars (or maybe totalLength). The focus should be on a noun (namely count or length).
Here is my solution:
#include <stdio.h>
#define LEN(a) (sizeof (a) / sizeof (a)[0])
static int CharCount(const char *strings[], int len)
{
int result, i, j;
result = 0;
for (i = 0; i < len; i++) {
j = 0;
while (strings[i][j] != '\0') {
result++;
j++;
}
}
return result;
}
int main(void)
{
const char *strings[] = { "s", "ss", "sss" };
printf("character count: %d\n", CharCount(strings, LEN(strings)));
}
The length macro LEN is very convenient and is the least error prone way to handle array lengths.

Answer 4

Yes char *array[] = {"aa", "bb", "cc"} is an array of pointers to strings.
array[0] points to "aa"
array[1] points to "bb"
array[2] points to "cc"
You probably want this:
int countChars(char *array[], int len)
{
int count = 0;
for (int arrayindex = 0; arrayindex < len; arrayindex++)
{
const char *stringptr = array[arrayindex];
// stringptr will point successively
// to "s", to "ss" and to "sss"
while (*stringptr++)
count++; // increment count until NUL character encountered
count++; // one more for NUL character
}
return count;
}
int main() {
char *example[] = { "s", "ss", "sss" };
int x = countChars(example, 3); // x contains 9 after the call to countChars
// that is 2 + 3 + 4
}
Instead of hard coding 3 you could use sizeof(example) / sizeof(example[0]).