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digitalmars.D.learn - static if and early exit from function doesn't seem to work?

reply aliak <something something.com> writes:
Alo! I'm making a recursive concat function that is similar to 
chain. The following code works:

import std.range: isInputRange;

auto concat(R, V...)(R range, V values) if (isInputRange!R) {
     import std.range: chain, ElementType;
     static if (V.length) {
         static if (isInputRange!(V[0])) {
             return range.chain(values[0]).concat(values[1..$]);
         } else static if (is(V[0] == ElementType!R)) {
             return range.chain([values[0]]).concat(values[1..$]);
         } // add an else assert here.
     } else {
         return range;
     }
}

But the following does not:

auto concat(R, V...)(R range, V values) if (isInputRange!R) {
     import std.range: chain, ElementType;
     static if (!V.length) {
         return range;
     }
     static if (isInputRange!(V[0])) {
         return range.chain(values[0]).concat(values[1..$]);
     }
     static if (is(V[0] == ElementType!R)) {
         return range.chain([values[0]]).concat(values[1..$]);
     }
}

You get a:

Error: tuple index 0 exceeds 0
Error: template instance range.concat.concat!(Result) error 
instantiating

So it seems it tries to compile the statements below the check on 
V.length even though it's guaranteed to be true and there's a 
return statement inside the if.

Is it a current limitation of static if? or a bug? or is 
something like this just not possible because of something I'm 
not seeing?

Cheers
Dec 31 2017
next sibling parent Colin <grogan.colin gmail.com> writes:
On Sunday, 31 December 2017 at 13:32:03 UTC, aliak wrote:

 Alo! I'm making a recursive concat function that is similar to 
 chain. The following code works:

 [...]


I suspect it's because you've no 'else static if'.
Dec 31 2017
prev sibling parent reply Adam D. Ruppe <destructionator gmail.com> writes:
On Sunday, 31 December 2017 at 13:32:03 UTC, aliak wrote:

 So it seems it tries to compile the statements below the check 
 on V.length even though it's guaranteed to be true and there's 
 a return statement inside the if.


Yeah, static if includes or excludes code independently at 
compile time.

So what you wrote there would be like, assuming the first to 
static ifs pass:

auto concat(R, V...)(R range, V values) if (isInputRange!R) {
     import std.range: chain, ElementType;
         return range;
         return range.chain(values[0]).concat(values[1..$]);
}


The code is still there, even if it isn't reached due to an early 
return, and thus still must compile.


Using else static if means it won't be generated.
Dec 31 2017
parent aliak <something something.com> writes:
On Sunday, 31 December 2017 at 13:47:32 UTC, Adam D. Ruppe wrote:

 On Sunday, 31 December 2017 at 13:32:03 UTC, aliak wrote:

 So it seems it tries to compile the statements below the check 
 on V.length even though it's guaranteed to be true and there's 
 a return statement inside the if.


 Yeah, static if includes or excludes code independently at 
 compile time.

 So what you wrote there would be like, assuming the first to 
 static ifs pass:

 auto concat(R, V...)(R range, V values) if (isInputRange!R) {
     import std.range: chain, ElementType;
         return range;
         return range.chain(values[0]).concat(values[1..$]);
 }


 The code is still there, even if it isn't reached due to an 
 early return, and thus still must compile.

 Using else static if means it won't be generated.


Ah ok, thanks! So it is intended behavior. I wonder if treating a 
return like a static else would be a good idea though. I at least 
can't see how it would break anything at this time.
Jan 01 2018