• BCS <ao pathlink.com> Jul 26 2007
• Gilles G. <schaouette free.fr> Jul 26 2007
• BCS <ao pathlink.com> Jul 26 2007
• Bill Baxter <dnewsgroup billbaxter.com> Jul 26 2007
• Manfred Nowak <svv1999 hotmail.com> Jul 26 2007
• Sean Kelly <sean f4.ca> Jul 26 2007
• Bill Baxter <dnewsgroup billbaxter.com> Jul 26 2007
• Robert Fraser <fraserofthenight gmail.com> Jul 26 2007

BCS <ao pathlink.com> writes:

this is just odd

void main()
{
  bool a;
  a ? assert(a) : assert(!a);
}

http://www.digitalmars.com/d/expression.html

Gilles G. <schaouette free.fr> writes:

I tried this code and... it won't produce any error.  I still don't understand
why. Could you please report the (odd) explaination?

--
Gilles 
BCS Wrote:

 this is just odd
 
 void main()
 {
   bool a;
   a ? assert(a) : assert(!a);
 }


 
 http://www.digitalmars.com/d/expression.html
 
 

BCS <ao pathlink.com> writes:

Reply to Gilles G.,

 I tried this code and... it won't produce any error.  I still don't
 understand why. Could you please report the (odd) explaination?
 
 --
 Gilles
 BCS Wrote:
 this is just odd
 
 void main()
 {
 bool a;
 a ? assert(a) : assert(!a);
 }
 http://www.digitalmars.com/d/expression.html
 




in this case is shouldn't produce an error, however it shows that assert 
is in fact an *expression* and not a statement as I would expect

this for instance, also works:

void main()
{
	int i = 5 + (assert(false), 6);
}

while witting this it just occurred to me why assert should be an expression. 
you can stuff it in logical expressions

while(ptr2ptr != null && (assert(*ptr2ptr != null), **ptr2ptr == 5)) {...}

Bill Baxter <dnewsgroup billbaxter.com> writes:

BCS wrote:
 Reply to Gilles G.,
 
 I tried this code and... it won't produce any error.  I still don't
 understand why. Could you please report the (odd) explaination?

 -- 
 Gilles
 BCS Wrote:
 this is just odd

 void main()
 {
 bool a;
 a ? assert(a) : assert(!a);
 }
 http://www.digitalmars.com/d/expression.html




 in this case is shouldn't produce an error, however it shows that assert 
 is in fact an *expression* and not a statement as I would expect


It looks like a function call and smells like a function call, so why 
shouldn't it act like a function call?  Calling a function is an 
expression, so why shouldn't assert() be an expression too?  Seems 
perfectly reasonable to me.  Expressions can also be used in more places 
than statements, as you point out, so if it makes sense as an expression 
then might as well allow it to be used that way.

--bb

Manfred Nowak <svv1999 hotmail.com> writes:

Bill Baxter wrote

 so if it makes sense as an expression 
 then might as well allow it to be used that way


Problem: what happens to an expression which has an `assert' as 
subexpression, if -release is given to the compiler?

-manfred

Sean Kelly <sean f4.ca> writes:

Manfred Nowak wrote:
 Bill Baxter wrote
 
 so if it makes sense as an expression 
 then might as well allow it to be used that way


 Problem: what happens to an expression which has an `assert' as 
 subexpression, if -release is given to the compiler?


I would guess that the assert() calls would be equivalent to an empty 
statement.  Kind of like how:

     if( sometimes() )
         assert( false );
     printf( "hello\n" );

should always print "hello," regardless of whether -release is specified 
or not.


Sean

Bill Baxter <dnewsgroup billbaxter.com> writes:

Manfred Nowak wrote:
 Bill Baxter wrote
 
 so if it makes sense as an expression 
 then might as well allow it to be used that way


 Problem: what happens to an expression which has an `assert' as 
 subexpression, if -release is given to the compiler?
 
 -manfred


Same thing as if you replace "assert(blah blah blah)" with "void", which 
is what it evaluates to as an expression.

--bb

Robert Fraser <fraserofthenight gmail.com> writes:

Of course it won't produce any error. It's a conditional expression, so if X
and Y are expressions, a ? X : Y will evaluate to X if a is true, and Y if a is
false.

In this expression, this means that if a is true, the expression evaluates to
assert(a), which is assert(true), which is always true. If a is false (as in
the case with this code, because boolean values default to false when first
initialized), the conditional will evaluate to assert(!a), which is
assert(!false), which is also obviously true. a ? assert(a) : assert(!a); will
_always_ evaluate to true.

Gilles G. Wrote:

 I tried this code and... it won't produce any error.  I still don't understand
why. Could you please report the (odd) explaination?
 
 --
 Gilles 
 BCS Wrote:
 
 this is just odd
 
 void main()
 {
   bool a;
   a ? assert(a) : assert(!a);
 }


 
 http://www.digitalmars.com/d/expression.html