The following code crash dmd with no error message:

<--------------------

int function() foo;
typedef foo bar;

<--------------------

I suppose it isn't allowed to typedef function pointers,
but still, an error message would be nice :)

Lars Ivar Igesund

Another one in a similar vein. The following code crash
dmd with no error message:

<----------------------

int function() foo;
int bar(foo fp);

<----------------------

I believe that function pointers as function parameters
are allowed.

Lars Ivar Igesund


Lars Ivar Igesund wrote:

> The following code crash dmd with no error message:
> 
> <--------------------
> 
> int function() foo;
> typedef foo bar;
> 
> <--------------------
> 
> I suppose it isn't allowed to typedef function pointers,
> but still, an error message would be nice :)
> 
> Lars Ivar Igesund

Lars Ivar Igesund wrote:
> Another one in a similar vein. The following code crash
> dmd with no error message:
> 
> <----------------------
> 
> int function() foo;
> int bar(foo fp);
> 
> <----------------------
> 
> I believe that function pointers as function parameters
> are allowed.
But isn't foo a variable not a type?
Should error anyway.
Sam

Lars Ivar Igesund wrote:

> The following code crash dmd with no error message:
> 
> <--------------------
> 
> int function() foo;
> typedef foo bar;
> 
> <--------------------
> 
> I suppose it isn't allowed to typedef function pointers,
> but still, an error message would be nice :)

the expression 'int function() foo;' doesn't declare a type.  It creates a global called foo, which is a pointer to a function that accepts no arguments, and returns an int.

'typedef foo bar;' on the next line is therefore complete nonsense.

What you probably want is

    typedef int function() bar;

Regardless, it shouldn't segfault the compiler. :)

 -- andy

Sam McCall wrote:

> Lars Ivar Igesund wrote:
> 
>> Another one in a similar vein. The following code crash
>> dmd with no error message:
>>
>> <----------------------
>>
>> int function() foo;
>> int bar(foo fp);
>>
>> <----------------------
>>
>> I believe that function pointers as function parameters
>> are allowed.
> 
> But isn't foo a variable not a type?
> Should error anyway.
> Sam


You're right of course. My fault. But it shouldn't segfault.

Lars Ivar Igesund

Andy Friesen wrote:

> Lars Ivar Igesund wrote:
> 
>> The following code crash dmd with no error message:
>>
>> <--------------------
>>
>> int function() foo;
>> typedef foo bar;
>>
>> <--------------------
>>
>> I suppose it isn't allowed to typedef function pointers,
>> but still, an error message would be nice :)
> 
> 
> the expression 'int function() foo;' doesn't declare a type.  It creates a global called foo, which is a pointer to a function that accepts no arguments, and returns an int.
> 
> 'typedef foo bar;' on the next line is therefore complete nonsense.
> 
> What you probably want is
> 
>     typedef int function() bar;
> 
> Regardless, it shouldn't segfault the compiler. :)
> 
>  -- andy

You're right of course. My fault. But it shouldn't segfault anyway.

Lars Ivar Igesund