Re: How to avoid code duplication in static if branches?
04.03.2012 3:42, Andrej Mitrovic пишет:
[...code...]
I want to avoid writing check() twice. I only have to statically
check a field of a member if it's of a certain type (Foo).
One solution would be to use a boolean:
void test(T)(T t)
{
bool isTrue = true;
static if (is(T == Foo))
isTrue = t.isTrue;
if (isTrue)
check();
}
But that kind of defeats the purpose of static if (avoiding runtime
overhead). Does anyone have a trick up their sleeve for these types of
situations? :)
Alias maybe?
void test(T)( T t ) {
enum TRUE = true;
static if( is(T == Foo) ) {
alias t.isTrue isTrue;
} else {
alias TRUE isTrue;
}
if( isTrue ) {
check();
}
}
This will still insert a redundant check for one of instantiations, but
compiler should be able to deal with 'if(true)' checks.
Re: How to avoid code duplication in static if branches?
Andrej Mitrovic andrej.mitrov...@gmail.com wrote in message
news:mailman.364.1330825349.24984.digitalmars-d-le...@puremagic.com...
import std.stdio;
void check() { writeln(check); }
struct Foo { bool isTrue = true; }
struct Bar { }
void test(T)(T t)
{
static if (is(T == Foo))
{
if (t.isTrue)
check();
}
else
{
check();
}
}
void main()
{
Foo foo;
Bar bar;
test(foo);
test(bar);
}
I want to avoid writing check() twice. I only have to statically
check a field of a member if it's of a certain type (Foo).
One solution would be to use a boolean:
void test(T)(T t)
{
bool isTrue = true;
static if (is(T == Foo))
isTrue = t.isTrue;
if (isTrue)
check();
}
But that kind of defeats the purpose of static if (avoiding runtime
overhead). Does anyone have a trick up their sleeve for these types of
situations? :)
Have you checked the generated code? When the static if check fails, it
should be reduced to:
void test(T)(T t)
{
bool isTrue = true;
if (isTrue)
check();
}
And the compiler should be able to tell that isTrue is always true.
Otherwise,
void test(T)(T t)
{
enum doCheck = is(T == Foo);
bool isTrue = true;
static if (is(T == Foo))
auto isTrue = t.isTrue;
if (!doCheck || isTrue)
check();
}
The compiler takes care of it because doCheck is known at compile time.
Re: How to avoid code duplication in static if branches?
You're right it should be able do that dead-code elimination thing.
Slipped my mind. :)
On 3/4/12, Daniel Murphy yebbl...@nospamgmail.com wrote:
Andrej Mitrovic andrej.mitrov...@gmail.com wrote in message
news:mailman.364.1330825349.24984.digitalmars-d-le...@puremagic.com...
import std.stdio;
void check() { writeln(check); }
struct Foo { bool isTrue = true; }
struct Bar { }
void test(T)(T t)
{
static if (is(T == Foo))
{
if (t.isTrue)
check();
}
else
{
check();
}
}
void main()
{
Foo foo;
Bar bar;
test(foo);
test(bar);
}
I want to avoid writing check() twice. I only have to statically
check a field of a member if it's of a certain type (Foo).
One solution would be to use a boolean:
void test(T)(T t)
{
bool isTrue = true;
static if (is(T == Foo))
isTrue = t.isTrue;
if (isTrue)
check();
}
But that kind of defeats the purpose of static if (avoiding runtime
overhead). Does anyone have a trick up their sleeve for these types of
situations? :)
Have you checked the generated code? When the static if check fails, it
should be reduced to:
void test(T)(T t)
{
bool isTrue = true;
if (isTrue)
check();
}
And the compiler should be able to tell that isTrue is always true.
Otherwise,
void test(T)(T t)
{
enum doCheck = is(T == Foo);
bool isTrue = true;
static if (is(T == Foo))
auto isTrue = t.isTrue;
if (!doCheck || isTrue)
check();
}
The compiler takes care of it because doCheck is known at compile time.
Re: How to avoid code duplication in static if branches?
Andrej Mitrovic wrote:
...snip...
I want to avoid writing check() twice. I only have to statically
check a field of a member if it's of a certain type (Foo).
One solution would be to use a boolean:
void test(T)(T t)
{
bool isTrue = true;
static if (is(T == Foo))
isTrue = t.isTrue;
if (isTrue)
check();
}
But that kind of defeats the purpose of static if (avoiding runtime
overhead). Does anyone have a trick up their sleeve for these types of
situations? :)
There's always this:
void test(T)(T t)
{
static if (is (T == Foo))
if (!t.isTrue)
return;
check();
}
Jerome
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