### Table of Contents

### Chapter 4: More on Strings

Well, we have progressed quite a way in a short time! Let's
back up a little and look at what was done in Chapter 3 on
copying of strings but in a different light. Consider the
following function:

```
char *my_strcpy(char dest[], char source[])
{
int i = 0;
while (source[i] != '\0')
{
dest[i] = source[i];
i++;
}
dest[i] = '\0';
return dest;
}
```

Recall that strings are arrays of characters. Here we have
chosen to use array notation instead of pointer notation to do
the actual copying. The results are the same, i.e. the string
gets copied using this notation just as accurately as it did
before. This raises some interesting points which we will
discuss.

Since parameters are passed by value, in both the passing of
a character pointer or the name of the array as above, what
actually gets passed is the address of the first element of each
array. Thus, the numerical value of the parameter passed is the
same whether we use a character pointer or an array name as a
parameter. This would tend to imply that somehow:

```
source[i] is the same as *(p+i);
```

In fact, this is true, i.e wherever one writes a[i] it can
be replaced with *(a + i) without any problems. In fact, the
compiler will create the same code in either case. Thus we see
that pointer arithmetic is the same thing as array indexing.
Either syntax produces the same result.

This is NOT saying that pointers and arrays are the same
thing, they are not. We are only saying that to identify a given
element of an array we have the choice of two syntaxes, one using
array indexing and the other using pointer arithmetic, which
yield identical results.

Now, looking at this last expression, part of it.. (a + i)
is a simple addition using the + operator and the rules of c
state that such an expression is commutative. That is (a + i)
is identical to (i + a). Thus we could write *(i + a) just as
easily as *(a + i).

But *(i + a) could have come from i[a] ! From all of this
comes the curious truth that if:

```
char a[20];
int i;
writing a[3] = 'x'; is the same as writing
3[a] = 'x';
```

Try it! Set up an array of characters, integers or longs,
etc. and assigned the 3rd or 4th element a value using the
conventional approach and then print out that value to be sure
you have that working. Then reverse the array notation as I have
done above. A good compiler will not balk and the results will
be identical. A curiosity... nothing more!

Now, looking at our function above, when we write:

```
dest[i] = source[i];
```

due to the fact that array indexing and pointer arithmetic yield
identical results, we can write this as:

```
*(dest + i) = *(source + i);
```

But, this takes 2 additions for each value taken on by i.
Additions, generally speaking, take more time than
incrementations (such as those done using the ++ operator as in i++). This may not be true in modern optimizing compilers, but one can never be sure. Thus, the pointer version may be a bit faster than the array version.

Another way to speed up the pointer version would be to
change:

```
while (*source != '\0') to simply while (*source)
```

since the value within the parenthesis will go to zero (FALSE) at
the same time in either case.

At this point you might want to experiment a bit with writing some of your own programs using pointers. Manipulating strings is a good place to experiment. You might want to write your own versions of such standard functions as:

```
strlen();
strcat();
strchr();
```

and any others you might have on your system.

We will come back to strings and their manipulation through pointers in a future chapter. For now, let's move on and discuss structures for a bit.