Thread: Character handling help

Oh crap

Thanks for the correction Quzah

I was actually experimenting, and I am not altogether clear on pointers yet, obviously. What I had wanted to do originally was to make the variable 'character' an array like this:

Code:

int main( void )
{
   int character[10];  /* Initializes character  */
 
   printf( "Enter a character: ");    /* Asks for character and inputs it */     
   character = getchar();

I cannot remember what happened when I tried that - something about a cast - anyway, I was mistaken when I thought that

int character[10];

is the same as

int *character;

Now is it correct to say that the array

Code:

int character[10];

would be ok to use, as there is space set aside for it? This would not over write another part or memory, or would it?

Originally Posted by vandalay

Sorry about that. I have to input a character from the keyboard and test it against the functions in the character handling library. I have all of the functions working, except the ones requiring a string (\n, \t, \r, and such). In those, the only thing recognized is \. I've tried everything I can think of, which isnt a whole lot. Thanks for your patience and help.

Err... Let me point out to you that in order to catch \r and \n, you'd have to use two different functions...

As far as I know, you can only get '\r' when you are using the getch() function (which is in the conio.h, and may be none standard...) and when you press enter.

getchar(), on the other hand, will return '\n' when you press enter.

I haven't tried it out myself, but would assume that in order to return '\t', you'd have to press the "tab" button, not type in \t.

[QUOTE=kermit]

Code:

int main( void )
{
   int character[10];  /* Initializes character  */
 
   printf( "Enter a character: ");    /* Asks for character and inputs it */     
   character = getchar(); // Err... you're trying to change the address of the array here...

This won't work either...

Your confusion regarding arrays and pointers may caome from the fact that the character (without the square brackets and number) actually refers to the address at which the array is stored. character[i] however, would then refer to the ith element of the character array.

However, in any case, Quzah is correct in the sense that you don't need arrays for what this program has to do in the first place.


This code is an example showing how you can use a pointer to store the address of an array, as well as how to access a particular element in the array through a pointer.

Code:

int main(void)
{
   int character[10], *pointer;
   pointer = character;
   pointer[0] = 5;
   printf("%d", character[0]);
   return 0;
}