A very **very** basic question

[mdh]

As I begin to write more little programs without the help of the
exercises, little things pop up that I need to understand more fully.
Thus, below, and although this is not the exact code, the principle of
the question is the same, ( I hope :-) )
#include <stdio.h>
int i = 0;
int main () { return 0; } /* no errors or warnings*/

but

#include <stdio.h>
int i ;
i=0;
int main () { return 0; } /* 2 warnings. */
I think one of the regular contributors has previously alluded to this
issue, but I wish to understand the principle more clearly.

So, ???

1) int i = 0 is allowed because i is declared and initialized as an
ext variable.
2) int i; i = 0 is not allowed because ?

a) even though my intention is to assign '0' to i , this can only
occur within a function?
b) the compiler thinks I am once again declaring 'i', which has
previously been declared, even though my **intent** is to initialize
an external variable.

I assume the same principles would apply if declared i as "static".

What key principle am I missing.

Thank you as usual.

[Richard]

ja*********@ver izon.net writes:

Richard wrote:

>vi******@gmail. com writes:

On Sep 25, 11:41 pm, mdh <m...@comcast.n etwrote:

...

>2) int i; i = 0 is not allowed because ?

a) even though my intention is to assign '0' to i , this can only
occur within a function?

yes.

Wrong. I can also be done as

int i =0; /* this is not in a function */

That's an initialization, not an assignment. This might seem like a

Sigh. The initialisation assigns the value 0 to i. Why be so silly and
petty?

*snip*

Why go through so many complexities to explain trivial things? Stop the
language lawyering please. The above confuses no one.

I know loads of C programmers who would say that and none of them
suddenly think they can re-assign without the declaration. Not one.

[mdh]

On Sep 25, 2:03*pm, jameskuy...@ver izon.net wrote:

mdh wrote:

>

2) int i; i = 0 is not allowed because ?

Because that is the combination of an external declaration and an
expression-statement. Statements may only appear in compound-
statements. A compound statement starts with a '{' and ends with a '}'
and may only occur within or as the body of a function definition.

That's what I was missing. Thanks.

[Richard]

mdh <md**@comcast.n etwrites:

On Sep 25, 2:01Â*pm, Richard Heathfield <[e-mail].i nvalidwrote:

>mdh said:
That's a declaration, and it's fine. It's also a tentative definition,
which is also fine.

i=0;

That's an assignment statement, which counts as code. You can't have code
outside a function.

Thank you Richard.

I would disagree. But I would wouldn't I?

int i=0;

is code.

[Richard Tobin]

In article <gb**********@r egistered.motza rella.org>,
Richard <rg****@gmail.c omwrote:

>i=0;

That's an assignment statement, which counts as code. You can't have code
outside a function.

>int i=0;

is code.

It is in one sense of code. But the distinction reflects the
implementationa l difference between things that naturally happen at
run-time and things that can obviously be done at compile time. (Note
"naturally" and "obviously" - I'm not suggesting that it couldn't be
made to work.)

-- Richard
--
Please remember to mention me / in tapes you leave behind.

[Peter Nilsson]

Richard<rgr...@ gmail.comwrote:

vipps...@gmail. com writes:

mdh wrote:

2) int i; i = 0 is not allowed because ?
>
a) even though my intention is to assign '0' to i ,
this can only occur within a function?

yes.

Wrong.

No, it isn't.

I can also be done as

int i =0; /* this is not in a function */

/* Nor is it an assignment. */

You're welcome to point out any section of the standard
which says it is. Of course you've already told us that
you find the thought of reading it too daunting.

Code can only be inside functions. It's possible to
declare and initialize global variables.

Code can only be inside functions?

Yes, if by code he means it to mean whatever he means it
to mean. Like 'global', the term isn't as obvious as
some people think.

So "int i=3*4;" is not code?

Is #include <stdio.hcode?

--
Peter

[Richard]

Peter Nilsson <ai***@acay.com .auwrites:

Richard<rgr...@ gmail.comwrote:

>vipps...@gmail .com writes:

mdh wrote:
2) int i; i = 0 is not allowed because ?

a) even though my intention is to assign '0' to i ,
this can only occur within a function?

yes.

Wrong.

No, it isn't.

Wrong.

Whichever way you look at it i is set to the value 0. In programming
this is known as assignment. And I dont see any reason to confuse a
noob by saying any different.

Of course I am trying to use natural English and not be too clever for
my own good.

>

>I can also be done as

int i =0; /* this is not in a function */

/* Nor is it an assignment. */

It is not in a function.

>
You're welcome to point out any section of the standard
which says it is. Of course you've already told us that
you find the thought of reading it too daunting.

And I wont be if it turns me into a robot totally unable to use common
sense when describing something so simple as this.

>

Code can only be inside functions. It's possible to
declare and initialize global variables.

Code can only be inside functions?

Yes, if by code he means it to mean whatever he means it
to mean. Like 'global', the term isn't as obvious as
some people think.

It is if you dont try to be overly clever and confuse people.

>

>So "int i=3*4;" is not code?

Is #include <stdio.hcode?

Why do you ask?

One more:

So "int i=3*4;" is not code?

[jameskuyper]

Richard wrote:

ja*********@ver izon.net writes:

Richard wrote:

vi******@gmail. com writes:

On Sep 25, 11:41 pm, mdh <m...@comcast.n etwrote:

...

2) int i; i = 0 is not allowed because ?

a) even though my intention is to assign '0' to i , this can only
occur within a function?

yes.

Wrong. I can also be done as

int i =0; /* this is not in a function */

That's an initialization, not an assignment. This might seem like a

Sigh. The initialisation assigns the value 0 to i. Why be so silly and
petty?

*snip*

The part you snipped gave the reason why pointing out that difference
was not petty, but was in fact directly relevant to why yours was not
an appropriate answer to the OP's question. If he hadn't known that
"int i=0;" was a way to do this, why would he have included "int i=0;"
in his question? He was clearly asking a question that he did not
consider "int i=0;" to be an answer to.

Assuming that he was paying more attention to the "petty" difference
between initialization and assignment than you were makes his question
reasonable. Assuming that he wasn't making that distinction makes his
question idiotic. From what I've seen, mdh has a lot to learn, but
he's no idiot. He's actually understanding and using some of the
advice he's getting; a welcome change from several of our other recent
visitors.

[August Karlstrom]

Richard Heathfield wrote:

mdh said:

[...]

>i=0;

That's an assignment statement, which counts as code. You can't have code
outside a function.

Wouldn't it be more accurate to say "you can't have *statements* outside
a function?" To me, "code" could be anything.
August

[Richard]

ja*********@ver izon.net writes:

Richard wrote:

>ja*********@ver izon.net writes:

Richard wrote:
vi******@gmail. com writes:

On Sep 25, 11:41 pm, mdh <m...@comcast.n etwrote:
...
2) int i; i = 0 is not allowed because ?

a) even though my intention is to assign '0' to i , this can only
occur within a function?

yes.

Wrong. I can also be done as

int i =0; /* this is not in a function */

That's an initialization, not an assignment. This might seem like a

Sigh. The initialisation assigns the value 0 to i. Why be so silly and
petty?

*snip*

The part you snipped gave the reason why pointing out that difference
was not petty, but was in fact directly relevant to why yours was not
an appropriate answer to the OP's question. If he hadn't known that

It was plenty appropriate.

"int i=0;" was a way to do this, why would he have included "int i=0;"
in his question? He was clearly asking a question that he did not
consider "int i=0;" to be an answer to.

You have lost me. The question was very straightforward and very easily
answered. And is a very common question from C beginners.

>
Assuming that he was paying more attention to the "petty" difference
between initialization and assignment than you were makes his question

It is a petty difference when teaching someone new.

int i=0;

You really want to tell me that in plain english 0 is not assigned to
the global variable i at program initialisation?

Really?

You can wallow in your standard all you like, but this is starting to
get ridiculous and certain posters here are making C almost
impossible to understand to anyone without a degree in the standard vocab.

reasonable. Assuming that he wasn't making that distinction makes his
question idiotic. From what I've seen, mdh has a lot to learn, but
he's no idiot. He's actually understanding and using some of the
advice he's getting; a welcome change from several of our other recent
visitors.

I have no idea what that last little soap box was about.

The question was easily answered. Outside of a declaration you can not
assign a value to a global variable unless you are in a function.

You can fanny around and flap all you like about the use of the word
assign here but its done me and thousands of others fine with zero
negative impact or sudden desire to reassign outside of the declaration
or a function.

And if you try to tell me that

int i=3*4;

is NOT C code then please don your kevlar helmet ....

[Richard]

August Karlstrom <fu********@gma il.comwrites:

Richard Heathfield wrote:

>mdh said:

[...]

>>i=0;

That's an assignment statement, which counts as code. You can't have
code outside a function.

Wouldn't it be more accurate to say "you can't have *statements*
outside a function?" To me, "code" could be anything.
August

and "int i=0;" is not a statement now?