C- GCC编译器中的字符数组

Question

This may seem trivial but I am a newbie to programming. In case of a string, we are supposed to keep a space reserved for the null character, so a string of size 10 will only give us 9 memory locations to use. I was wondering whether the same happens when we input letters individually into the array.Can anyone please explain?

Answer 1

C arrays aren't aware of their own size, so there's a standardised convention of using the special character '\\0' to mark the end of a string.

So if you want to implement your own puts() you can implement it like:

int puts(const char *s){
   int i=0;
   while(s[i]!='\0'){
      putc(s[i]);
      i++
   }
   return i;

}

It's just a language convention, non mandatory because the computer needs it but because the programmers need it. Otherwise you wouldn't know how to stop reading.

When you use a character (or any other type), you know how it's size beforehand so there's no need for marking an end of string.

Answer 2

In C, a string basically is an array. And the size of that array must always be big enough for the terminating null character. So if I have

char array[10];

I can do

array[0] = 'H';

or

array[5] = 'x';

but I had better stop putting in characters at array[8] so I can do

array[9] = '\0';

at the end.

Bottom line: I can do this:

array[0] = 'C';
array[1] = 'a';
array[2] = 't';
array[3] = '\0';
printf("%s\n", array);

or this:

array[0] = 'S';
array[1] = 'p';
array[2] = 'a';
array[3] = 'g';
array[4] = 'h';
array[5] = 'e';
array[6] = 't';
array[7] = 't';
array[8] = 'i';
array[9] = '\0';
printf("%s\n", array);

but not this:

array[0] = 'H';
array[1] = 'e';
array[2] = 'l';
array[3] = 'i';
array[4] = 'c';
array[5] = 'o';
array[6] = 'p';
array[7] = 't';
array[8] = 'e';
array[9] = 'r';
array[10] = '\0';    /* XXX WRONG */