How to get the length of an array from a pointer? [duplicate]

Question

I'm having trouble finding the length of an pointer array. Let's say I have:

char array[40] = "hello"; // size 40
int length =  sizeof array / sizeof array[0]; // no problem returns 40

//How do I get the length of the array with only a pointer to the first element in that array?

 char* pchar = array;

//if

std::strlen(pchar); // this returns the length of 5... i want length 40

//if

int count = 0;
while(true)
{
  if(*(pchar + count) == '\0') // returns 5...
    break;
   count++;
}

How do I get it to return length 40 just from a pointer to the first element in the array?
I found that I can do this.

int count = 0;
    while(true)
    {
      if(*(pchar + count) == '\0' && *(pchar + count + 1) != '\0') 
             break;

       count++;
    }

This returns 39, this is good but I feel like this can be buggy in some situations.

Answer 1

You can't, I'm afraid. You need to pass the length of the array to anyone who needs it. Or you can use a std::array or std::vector or similar, which keep track of the length themselves.

Answer 2

C++ has proper string type:

std::string

which you may find helpful here. Even if you're passing it to function that accepts const char*, it has .c_str() method that allows you to pass it to function that accept a pointer. If the other function needs to modify the string, you can use &str[0] which is valid for many implementations of C++, and is required to work for C++11. Just make sure you resize() them to the correct size.

Some of the other containers in C++ are:

std::array (C++11) Array of constant size. Better than plain old C array, as it has size() method. std::vector Dynamic array (Java ArrayList equivalent)

As for your question - there is no way to find size of a pointed array. How could you even do that? It's just a stupid pointer.