On Dec 29, 2010, at 07:30 AM, Olin Lathrop wrote:

> PETER ONION wrote:
>>>> if(diff >=3D 0)
>>>>     byte =3D  0;
>>>> else
>>>>=20
>>>> {
>>>>     byte =3D  8;
>>>>     diff =3D - diff;
>>>> }
>>=20
>> The ";" just terminates the "if true statement".
>=20
> But if a semicolon doesn't end the IF statement, what does end it
> unambiguously?  Or put another way, what if there was a outer IF not show=
n
> above, and ELSE was meant to be part of it, not the IF shown.  How would =
you
> then end the IF shown so that the ELSE would not be part of it?
>=20

Semicolons do not terminate an if statement.  In C, if statements will exec=
ute a single block of code following the if and the else (note: 'else if' s=
tatements just build on the above method).  The single block of code in the=
 above is considered to be the line following the if and the code with the =
{ } following the else.

An un-compiliable example:

    if(i >=3D 0)
        byte =3D 0;   //executes on condition
        byte++;     //always runs
    else            //syntax error, HAL is not pleased
    {
        byte =3D 99;
    }

If this could be run, line 2 is executed if i >=3D 0 is true.  Line 3 is al=
ways executed (note the nesting makes this ambiguous).  Then line 4, would =
be an orphaned else with no preceding if and the code will not compile.

If the programmer intents to execute what most of us see when we look at th=
e above, the block to execute following the if needs to be enclosed:

    if(i >=3D 0)=20
    {
        byte =3D 0;
        byte++;
    }
    else
    {
        byte =3D 99;
    }


--=20
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