byref: return byval a_pointer vs. return *a_pointer

[noop]

Hello,

where is the difference in returning a pointer by-value and returning the dereferenced pointer in a byref-function-result?

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dim shared as integer x

function test_b() byref as integer
    dim as integer ptr px = @x
    return byval px
end function

function test_d() byref as integer
    dim as integer ptr px = @x
    return *px
end function

x = 4

print test_b()
print test_d()
sleep

[Tourist Trap]

I was affraid that you had found a difference. In my opinion the way it works in your example is normal, there shouldn't be any difference or I don't see why at all.

[MichaelW]

It looks to me like test_b should return a pointer, and test_d should trigger a warning.

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#include <stdio.h>
#include <conio.h>

int x = 4;

int *test_b()
{
    int *px = &x;
    return px;    
}

int *test_d()
{
    int *px = &x;
    return *px;   // line 15
}

int main(void)
{
    printf( "%d\n%d\n", test_b(), test_d() );
    _getch();
}

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C:\Users\User\FreeBASIC My\byref>gcc.exe -std=c99 -pedantic -Os -m64 -masm=intel
 testc.c -o testc.exe
testc.c: In function 'test_d':
testc.c:15:5: warning: return makes pointer from integer without a cast
     return *px;
     ^

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4206608
4

[fxm]

No, not warning because this syntax with the BYVAL keyword is allowed by the fb compiler and also documented in manual.
See my previous post:
http://www.freebasic.net/forum/viewtopi ... 15#p210315

[MichaelW]

No, not warning because this syntax with the BYVAL keyword is allowed by the fb compiler and also documented in manual...

What is the reason for allowing it?

Do you agree that test_b should return a pointer? Per the "byref as integer" it's supposed to return a pointer, and per the "return byval px" it's supposed to return a pointer.

[fxm]

test_b() as test_d() returns a reference.
(in fact in internal, always a pointer is passed by value and an internal code dereferences this pointer for the user).
By specifying Return Byval, an address (usually stored in a pointer) can be passed directly as-is, forcing the Byref function result to reference the same memory location which the address pointed to.

[noop]

fxm, I may have misinterpreted your post/posts. So please bear with me here.

test_b() as test_d() returns a reference.
By specifying Return Byval, an address (usually stored in a pointer) can be passed directly as-is, forcing the Byref function result to reference the same memory location which the address pointed to.

I don't understand why the function could point to sth. else. If I return *px then the byref-return would make a @(*px) out of it and then dereference it "upon arrival". Using byval px (I feel like) does exactly the same thing. In what scenario could the returned pointer (internally) point to a different address?


I've also had a look at the linked thread, from which I draw the same conclusion.

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sub FAKEPRINT1(byref E as integer)
    Print E
end sub 

dim as integer e = 1

FAKEPRINT1 (e)
FAKEPRINT1 (*(@e))
FAKEPRINT1 (byval @e)

sleep

Output:

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1
1
1

The only difference I saw was the way the parser handled it in a function call:

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function f(byref i as integer) byref as integer
    print "inside=";@i
    f = i
end function

dim as integer i=123
print "i from outside=";@i
Print i
f(i) = 100  'parsed as equality operator, not assignement
print i
(f(i)) = 200
print i
f(byval @i) = 300    
print i

sleep

Output:

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i from outside=1638080
 123
inside=1638076
 123
inside=1638080
 200
inside=1638080
 300

It's interesting to see a different address in the first function call. It seems like a temporary copy is made from the variable i.

[fxm]

I never wrote that the function could point to something else.