Re: for loops without braces?

  • From: "Littlefield, Tyler" <tyler@xxxxxxxxxxxxx>
  • To: programmingblind@xxxxxxxxxxxxx
  • Date: Fri, 19 Nov 2010 07:42:44 -0700

Ur, that you run one, as opposed to many. Now, that one may be an if statement that also takes one or more statements after it, etc etc. 

On 11/19/2010 7:39 AM, Littlefield, Tyler wrote:
It doesn't specify a different statement to be run, it just means that you run, as opposed to many. 
On 11/19/2010 5:46 AM, Alex Hall wrote:

Oh, I never thought of it like that! So, anywhere you have braces, you
are really just running a sort of list of statements inside the
braces, but that list is considered one big statement. Therefore,
leaving off the braces does not change the underlying behavior, it
just specifies a different statement to be run. It makes sense, and I
never knew that. Thanks.

On 11/19/10, Sina Bahram<sbahram@xxxxxxxxx>  wrote:

This is very simple.


For(blah)
If(blah)
{
Do blah
}

Simplifies to

For(blah)
If statement doing blah

You see?
The for loop only repeats one item, ever. It's just that, most of the time, 
that one item is a block noted by an open and closed
brace, but otherwise it can be any other one item, like another for loop, an 
if statement, a single statement, or so on.

Take care,
Sina

-----Original Message-----
From: programmingblind-bounce@xxxxxxxxxxxxx
[mailto:programmingblind-bounce@xxxxxxxxxxxxx] On Behalf Of Alex Hall
Sent: Thursday, November 18, 2010 9:09 PM
To: programmingblind@xxxxxxxxxxxxx
Subject: for loops without braces?

Hi all,
I am desperately trying to understand Prim's algorithm. I have found it in 
pseudocode and C code, but in both examples they use for
loops without braces, so it might be:
for(i=0; i<4; i++)
if(i%2==0){
print(i);
}
Obviously this is understandable, but when you mix it into other code it 
becomes much harder to tell what is going on. First, how
can one do this syntactically and have it be correct? Second, what is the 
rule to figure out where the loop ends if it is not in
braces? The C compiler must have a way...

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Thanks,
Ty

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