smithy | A framework for building WebAssembly apps in Rust | Binary Executable Format library
kandi X-RAY | smithy Summary
kandi X-RAY | smithy Summary
Smithy is a framework for writing WebAssembly applications entirely in Rust. Its goal is to allow you to do so using idiomatic Rust, without giving up any of the compiler's safety guarantees.
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Trending Discussions on smithy
QUESTION
I get a java.lang.NullPointerException when I compile my code and I don't know why. It says:
...ANSWER
Answered 2021-Jun-04 at 18:26In your Store
class the goods array is not initialized. It will work if you change it to
QUESTION
I have the following data:
...ANSWER
Answered 2020-Sep-17 at 20:36There's quite a lot of ways to go about this, but here is one. Note that since you have the same study in multiple facets, and the TypeTwo
is not consistent, we have to do some trickery to be able to sort within each facet.
I also sorted by effect size, since that's pleasing and quite commonly done.
There is a mistake in your data for Hunter 2015 which has two effect sizes for Death, so that's why there's a red bar in with the green bars.
Using some random data for a mean effect:
QUESTION
I have a dataset like this that I have turned into a massive dendrogram using time series clustering:
...ANSWER
Answered 2020-Aug-14 at 14:42This is mostly a data reshaping problem. First convert DF
from wide to long format, then merge with DF2
, summarise by groups of time and cluster. Finally, plot the result.
In order to have matching names in DF
and DF2
, I have changed the posted data.
QUESTION
I have the following and I can print out each property. Easy enough, but is there an easy way to print the key and value of each item in the JSON string? Simply looking to print in console the key and the value.
...ANSWER
Answered 2020-Jul-10 at 06:52String userJson = "{\"name\":\"smithy\",\"email\":\"blah@gmail.com\",\"age\":21,\"isDeveloper\":true}";
JsonObject convertedObject = new Gson().fromJson(userJson, JsonObject.class);
for(String key:convertedObject.keySet()){
System.out.println("Key - " + key);
System.out.println("Value - " + convertedObject.get(key));
}
QUESTION
I have xml in following format
...ANSWER
Answered 2018-May-28 at 08:43DECLARE @XML AS XML= N'
John
Johnny
Johnson
Smith
Smithy
Schmit
'
INSERT INTO @tblTest(firstName,LastName)
SELECT t1.c.value('../name[1]','varchar(100)') As FirstName,t1.c.value('alias[1]','varchar(50)') as SecondName
FROM @xml.nodes('/Entity/aliases') t1(c)
QUESTION
I have a collection of addresses, with postal codes. I want to get all the records by using an array of postal codes
so this query works fine
...ANSWER
Answered 2019-Nov-12 at 06:04You can use regular expression (regex) like in the following example.
Sample documents:
QUESTION
I have some data that I'm trying to store in a nested dictionary.
I would then like to be able to iterate through this nested dictionary and output the data somehow. (Ultimately, in cells, but for now I'm just trying to get the "most nested" keys).
The data is set up like this:
...ANSWER
Answered 2019-Sep-26 at 17:26Try this traversal instead:
QUESTION
I have this set of data using Microsoft SQL Server Management Studio
...ANSWER
Answered 2017-Sep-12 at 17:08After walking away with some time to refresh my brain i had a eureka moment and solved this. It was actually easy when I thought about it.
so
the code to get the desired table goes something like this:
QUESTION
I'm quite new in AngularJs and I'm trying to get data from 2 different JSON returning services.
I have a list, obtained from one service, where I'm showing the user ID.
This is working, but what I need is, given that user ID, obtain the user's first name and last name from the other JSON.
How can I do it? Any help is welcome.
This is the AngularJs (1.6) controller:
...ANSWER
Answered 2017-Nov-09 at 04:01You can chain a second then function onto your first one. Inside that function, you can make your other http request using your now available userid.
QUESTION
I have an image gallery with some hover affects that I want to refine. When the user hovers on an image, the other pictures in the gallery get dimmed out. But I also want some text to slide in from the left on hover as well. Something like this website has http://gugroppo.com/projects.
I have the text appear on hover but I can't get it to transition in from the left smoothly; or have my overlay effect appear smoothly as well. Everything just appears. Here's my codepen.
...ANSWER
Answered 2017-Oct-04 at 20:55What you want to do is add overflow: hidden to your container
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