How to read unlimited characters in C

Question

How to read unlimited characters into a char* variable without specifying the size?

For example, say I want to read the address of an employee that may

Answer 1

Just had to answer Ex7.1, pg 330 of Beginning C, by Ivor Horton, 3rd edition. Took a couple of weeks to work out. Allows input of floating numbers without specifying in advance how many numbers the user will enter. Stores the numbers in a dynamic array, and then prints out the numbers, and the average value. Using Code::Blocks with Ubuntu 11.04. Hope it helps.

/*realloc_for_averaging_value_of_floats_fri14Sept2012_16:30  */

#include 
#include 
#define TRUE 1

int main(int argc, char ** argv[])
{
    float input = 0;
    int count=0, n = 0;
    float *numbers = NULL;
    float *more_numbers;
    float sum = 0.0;

    while (TRUE)
    {
        do
        {
            printf("Enter an floating point value (0 to end): ");
            scanf("%f", &input);
            count++;
            more_numbers = (float*) realloc(numbers, count * sizeof(float));
            if ( more_numbers != NULL )
            {
                numbers = more_numbers;
                numbers[count - 1] = input;
            }
            else
            {
                free(numbers);
                puts("Error (re)allocating memory");
                exit(TRUE);
            }
        } while ( input != 0 );

        printf("Numbers entered: ");
        while( n < count )
        {
            printf("%f ", numbers[n]);  /* n is always less than count.*/
            n++;
        }
        /*need n++ otherwise loops forever*/
        n = 0;
        while( n < count )
        {
            sum += numbers[n];      /*Add numbers together*/
            n++;
        }
        /* Divide sum / count = average.*/
        printf("\n Average of floats = %f \n", sum / (count - 1));
    }
    return 0;
}

/* Success Fri Sept 14 13:29 . That was hard work.*/
/* Always looks simple when working.*/
/* Next step is to use a function to work out the average.*/
/*Anonymous on July 04, 2012*/
/* http://www.careercup.com/question?id=14193663 */