[c] Finding the length of a Character Array in C

What is a way in C that someone could find the length of a Character array?

I will happily accept psuedo-code, but am not averse to someone writing it out if they'd like to :)

There is also a compact form for that, if you do not want to rely on strlen. Assuming that the character array you are considering is "msg":

  unsigned int len=0;
  while(*(msg+len) ) len++;

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 array
int 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]));

You can use this function:

int arraySize(char array[])
{
    int cont = 0;
    for (int i = 0; array[i] != 0; i++)
            cont++;
    return cont;
}

using sizeof()

char h[] = "hello";
printf("%d\n",sizeof(h)-1); //Output = 5

using string.h

#include <string.h>

char h[] = "hello";
printf("%d\n",strlen(h)); //Output = 5

using function (strlen implementation)

int strsize(const char* str);
int main(){
    char h[] = "hello";
    printf("%d\n",strsize(h)); //Output = 5
    return 0;
}
int strsize(const char* str){
    return (*str) ? strsize(++str) + 1 : 0;
}

By saying "Character array" you mean a string? Like "hello" or "hahaha this is a string of characters"..

Anyway, use strlen(). Read a bit about it, there's plenty of info about it, like here.

You can use strlen

strlen(urarray);

You can code it yourself so you understand how it works

size_t my_strlen(const char *str)
{
  size_t i;

  for (i = 0; str[i]; i++);
  return i;
}

if you want the size of the array then you use sizeof

char urarray[255];
printf("%zu", sizeof(urarray));

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).

If you want the length of the character array use sizeof(array)/sizeof(array[0]), if you want the length of the string use strlen(array).