I am always confused at the way which is used to declare a c string.
so i assume we can do it either by
char *str;
or by doing
char str[10];
so what is the difference here? it seems that if i do it the second way then my length of string is limited to 9, and if i do it in the first way, i can just say str="random blah blah blah" and it can be as long as i want.
can some pros out there give me some enlighten me a lil bit? thx.
It's helpful to keep these tips in mind:
1) Strings are simply one or more char's, which we can elevate to a string, if we add the end of string char after it or them.
'A' is a char. "A\0" is a string. "ab" is two char's. "ab\0" is a two char which has been elevated to a string.
You can declare a string either way. Say you put the name "Stephen", in the first string. Your string would be "Stephen\0". 7 char's + 1 end of string char.
Say now you put the same name into the second array of char's. It would be:
0123456789
Stephen\0JU
Where JU at the end could be anything - junk, NULL's, you name it. The point is that in the second instance, you've wasted 2 bytes of memory. In the first instance, you've wasted zero bytes.
The second instance, with the array of char's is easier for most people to figure out and work with, and the small waste is acceptable - we're not going to protest in the streets over 2 whole wasted bytes.
But if you had a dictionary with hundreds or thousands of words for your game, the waste becomes huge and unacceptable, since in the case of the array, every row would need to be the same length: Axe would have just as much memory as Blacksmith.
Internally, C has no arrays. They are a convenience for us programmers, because people are so good at working things into 1, 2, or 3 dimension arrays. Arrays are immediately transferred by C into pointers with offsets to give their lengths:
Array[10] is immediately changed into *(Array + 10), by the compiler. It's no coincidence that the name of the array, is also the address of the first element of the array (also called the base of the array). In other words it becomes a constant pointer.
Since the compiler does this transformation for us, why not just skip the Array[10] format altogether? Because programming is about idea's, and working out problems, and Array[10] is more helpful than *(Array + 10), is. So I let the compiler do it's job, and I'll do mine.
Last edited by Adak; 01-19-2008 at 08:16 AM.
Another point to keep in mind, speaking of memory storage, is that the declaration str[10] allocates us room for 10 characters (of which we can use 9) while the declaration *str gives us no memory whatsoever.[1] If we use *str, it must either be initialized to a string literal (in which case, we can't change it), or must have allocated memory to it with malloc().
[1] Of course, we get some memory, namely enough memory to put an address in, but not any memory we can use to store characters in.
Array[10], gives us 10 elements we can use, not 9. 0-9 IS 10.
Just a small nit... the end of string character, also called the null terminating character, is an escape sequence defined by the 2-character sequence \0 (back slash, zero), not /0 (forward slash, zero).
char *str = "string";Originally Posted by tabstop
Does create a string literal - however, it is not necessarily located in read-only memory. AFAIK; it's undefined where it gets put and depends on the compiler and architecture.
Thank you, Todd. Made that correction.
I just meant that we must be sure to not use the 10th character, since we need a \0 at the end.
thank you kind people. i can go and do my project now:d
