Thread: beginner array question

well, when i started c/c++ programming, the book i read told me that to create an array, you need to put the number of 'elements' in the [...].

for example, lets say i want to make a char-array capable of holding up to 4 letters, i would do:

Code:

char sWord[4];

then i try putting data inside the char-array, for example:

Code:

sWord[0] = 'T';
sWord[1] = 'E';
sWord[2] = 'S';
sWord[3] = 'T';
sWord[4] = '/0';

when i compile and run that program, it works fine most of the time, but if i have another char-array, i get some weird results.. when i set a value to sWord[3] or sWord[4], i get a value on sWord2[0]...

im just wondering, if i wanted an array that could hold 4 letters (no need for a null char at the end), do i need to declare it as

Code:

char sWord[4];

or

char sWord[5];

i was told that you can only use n - 1 elements, so for the first sWord, i would have 0 to 3, and for the second sWord, i would have 0 to 4...

what happens if i try to write to sWord[4]?

im not sure but, if i declare sWord[4], can i set sWord[4] to null? or is that not allowed.. somebody please explain this, ive been doing it this way for so long, and problems are only just starting to appear, and i want to know if im doing it right

thanks

but i dont get 1 thing.. your declaring it as sword[4].. which is 4 elements (0-3) i get that, but.. what is the 5th element (sword[4]) used for?

so if i want to declare a string, i need to include the null terminator in the no. of elements? and is it compulsory? or can i have strings that dont include a null terminator, if im going to treat them as char arrays rather then strings?

btw, thanks for the reply, cleared a few things up, all this time ive been doing it all wrong :S

Originally Posted by X PaYnE X

but i dont get 1 thing.. your declaring it as sword[4].. which is 4 elements (0-3) i get that, but.. what is the 5th element (sword[4]) used for?

There is no sword[4].
When you go like this:

Code:

char sword[4];

You're actually creating 4 slots (including 0):

Code:

sword[0] = '1';
sword[1] = '2';
sword[2] = '3';
sword[3] = '4';

The only reason we always put '\0' is so that when we use the char array as a string, like:

Code:

puts(sword);

The function puts(char *output) will know when to stop, because it stops printing out sword when it hits the '\0' character. But if you're using the char array to just hold characters, and not to do any string manipulation, then it'll be ok as long as you have a variable that holds how many characters are in the char array. Here's some examples:

Code:

char sword[4];
sword[0] = '1';
sword[1] = '2';
sword[2] = '3';
sword[3] = '4';
/* We know it's 4 characters long, but strlen() does not. */
printf("%i\n", strlen(sword));

But if we go like this:

Code:

char sword[5];
sword[0] = '1';
sword[1] = '2';
sword[2] = '3';
sword[3] = '4';
sword[4] = '\0';
/* It knows the length of the string is 4 characters long!*/
printf("%i\n", strlen(sword));

Code:

sWord[0] = 'T';
sWord[1] = 'E';
sWord[2] = 'S';
sWord[3] = 'T';
sWord[4] = '/0';

your problem is '/'0' should be '\0'
but you can leave the last element alone anyway, the compiler will fill it for you.

posted sand_man

Code:

>your problem is '/'0' should be '\0'

i think , you can use

Code:

NULL

i know that but

Code:

char ch[]={'s','s','s',NULL};
printf("%s\n",ch);

compiling >>>>

sss

im assuming the string will have a fixed number of characters so ill just need to use the size of memory.. well i guess that about answers my question, thanks for all your help everyone

enjoy, pease stop spamming, and learn to use the board's formatting tags properly.