Thread: A newby question on the sizeof operator

let´s say you got this function:

void DoSomething( char* mystring)
{
........\\;
whatever;
}

and you pass an array to it like this:

char teststring[] = "this is some text";
DoSomething( teststring );

how in GODS NAME! can you get the number of elements of the array INSIDE the function?? (without using strlen() ofcourse, or supose the array is of integers)
i´be tried the usual:
int mylen = sizeof mystring / sizeof mystring[0];
but it wont work, my guess is that im getting the size of THE POINTER to the array NOT the array itself.
i also tried changing the function to take an array NOT a pointer (but of course its the same thing, since only the address of the array gets passed. (right?)

is there a way to solve this? (not including passing the size to the function as an argument of course)
thanks.

There is no way in C/C++ that I know of. I use the first element of the array to hold the number of elements (like in turbo pascal with strings : string[0] holds length). If you are using char, then it will limit you to 255 elements.

Code:

#include <string.h>
#include <stdlib.h>
#include <iostream.h>

void doSomething(char *cptr)
{
   int numElements = cptr[0];
   for (int i = 1; i <= numElements; i++)
   {
      cout << "Element " << i << " = " << cptr[i] << endl;
   }
}

int main()
{
   char teststring[] = "this is some text";
   int len = strlen(teststring);
   char * temp = new char[len + 1];
   temp[0] = len;
   strcpy(&temp[1], teststring);
   doSomething(temp);
   delete[] temp;
   return 0;
}

This works fine in Borland C++ 5.02. You can also try using list, vector, etc.