diff options
Diffstat (limited to 'd.html.markdown')
| -rw-r--r-- | d.html.markdown | 18 |
1 files changed, 10 insertions, 8 deletions
diff --git a/d.html.markdown b/d.html.markdown index 80c1dc65..4a362372 100644 --- a/d.html.markdown +++ b/d.html.markdown @@ -218,7 +218,7 @@ void main() { // from 1 to 100. Easy! // Just pass lambda expressions as template parameters! - // You can pass any old function you like, but lambdas are convenient here. + // You can pass any function you like, but lambdas are convenient here. auto num = iota(1, 101).filter!(x => x % 2 == 0) .map!(y => y ^^ 2) .reduce!((a, b) => a + b); @@ -228,7 +228,7 @@ void main() { ``` Notice how we got to build a nice Haskellian pipeline to compute num? -That's thanks to a D innovation know as Uniform Function Call Syntax. +That's thanks to a D innovation know as Uniform Function Call Syntax (UFCS). With UFCS, we can choose whether to write a function call as a method or free function call! Walter wrote a nice article on this [here.](http://www.drdobbs.com/cpp/uniform-function-call-syntax/232700394) @@ -238,21 +238,23 @@ is of some type A on any expression of type A as a method. I like parallelism. Anyone else like parallelism? Sure you do. Let's do some! ```c +// Let's say we want to populate a large array with the square root of all +// consecutive integers starting from 1 (up until the size of the array), and we +// want to do this concurrently taking advantage of as many cores as we have +// available. + import std.stdio; import std.parallelism : parallel; import std.math : sqrt; void main() { - // We want take the square root every number in our array, - // and take advantage of as many cores as we have available. + // Create your large array auto arr = new double[1_000_000]; - // Use an index, and an array element by referece, - // and just call parallel on the array! + // Use an index, access every array element by reference (because we're + // going to change each element) and just call parallel on the array! foreach(i, ref elem; parallel(arr)) { ref = sqrt(i + 1.0); } } - - ``` |