Re: embedded questions!!!

Richard Henry wrote: 
"John B" <spamj_baraclough@xxxxxxxxxxxxxxxxxxx> wrote in message
news:43c7b8d3$0$819$4c56ba96@xxxxxxxxxxxxxxxxxxxxxxxxx


The two declarations:

char str1* = "JHONSON"; 

I'll assume that's supposed to be 'char *str1 = "JHONSON";'...


and

    char str2[] = "JHONSON";

both allocate space for a modifiable string. The difference being that the
first one also creates a pointer to that string.

Isn't str2 also a pointer? 

Nope, it's an array.  If it helps to make things clearer, sizeof(str1)
will be different than sizeof(str2).  In particular:

    sizeof(str1) == sizeof(char *)
    and
    sizeof(str2) == strlen(str2)+1

That should help make it clearer that str1 and str2 are different types.
If you were to give them names, str1's type would be "char *" and str2's
type might be "char[8]".  One is a pointer and the other is an array.

That may not make much sense, but think for a moment about how an int
can be automatically converted to a float:

	int i = 1;
	float f = 3.14159;

	printf ("%g", i + f);

What's going on here is that the compiler sees an expression where the
operator (in this case "+") has two arguments which are different types.
It then implicitly converts the integer into a floating point value
when generating the code that will correspond to that expression.

A similar thing happens when arrays are used where a pointer value is
needed.  The compiler thinks "this isn't a pointer, but since it's an
array, I can create a pointer value from it".

Another way they differ, if I understand correctly is that this would
be legal:

	str1 = (char *) "another string";

but this would not be:

	str2 = (char *) "another string";

The reason is this:  str1 is just a pointer, so it can take on any value.
But str2 is the array that contains { 'J', 'H', 'O', 'N', 'S', 'O', 'N', 0 },
and how are you supposed to assign a pointer to that?

  - Logan
.

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