Finding the length of a Character Array in C

Provided the char array is null terminated,

char chararray[10] = { 0 };size_t len = strlen(chararray);

If you have an array, then you can find the number of elements in the array by dividing the size of the array in bytes by the size of each element in bytes:

char x[10];int elements_in_x = sizeof(x) / sizeof(x[0]);

For the specific case of char, since sizeof(char) == 1, sizeof(x) will yield the same result.

If you only have a pointer to an array, then there's no way to find the number of elements in the pointed-to array. You have to keep track of that yourself. For example, given:

char x[10];char* pointer_to_x = x;

there is no way to tell from just pointer_to_x that it points to an array of 10 elements. You have to keep track of that information yourself.

There are numerous ways to do that: you can either store the number of elements in a variable or you can encode the contents of the array such that you can get its size somehow by analyzing its contents (this is effectively what null-terminated strings do: they place a '\0' character at the end of the string so that you know when the string ends).

Although the earlier answers are OK, here's my contribution.

//returns the size of a character array using a pointer to the first element of the character arrayint size(char *ptr){    //variable used to access the subsequent array elements.    int offset = 0;    //variable that counts the number of elements in your array    int count = 0;    //While loop that tests whether the end of the array has been reached    while (*(ptr + offset) != '\0')    {        //increment the count variable        ++count;        //advance to the next element of the array        ++offset;    }    //return the size of the array    return count;}

In your main function, you call the size function by passing the address of the first element of your array.

For example:

char myArray[] = {'h', 'e', 'l', 'l', 'o'};printf("The size of my character array is: %d\n", size(&myArray[0]));