On Friday, 26 August 2016 at 08:21:14 UTC, Alex wrote:
On Friday, 26 August 2016 at 07:20:00 UTC, Daniel Kozak wrote:
auto j2 = j.toString.parseJSON;
ha! cool! thanks! :)
found a bug...
https://issues.dlang.org/show_bug.cgi?id=16432
not very serious... but not found yet? ;)
On Friday, 26 August 2016 at 07:46:13 UTC, Daniel Kozak wrote:
Another way is to implement deepCopy by yourself (something
like below)
import std.json;
import std.stdio;
JSONValue deepCopy(ref JSONValue val)
{
JSONValue newVal;
switch(val.type)
{
case JSON_TYPE.STRING:
On Friday, 26 August 2016 at 07:20:00 UTC, Daniel Kozak wrote:
auto j2 = j.toString.parseJSON;
ha! cool! thanks! :)
On Friday, 26 August 2016 at 06:56:06 UTC, Alex wrote:
Hi everybody,
I'm little at loss: as documented, a JSONValue is only shallow
copied:
...
So the effect that the code of "j" is altered was expected.
The question is: how to make a deep copy of a JSONValue? Is
there a simple way without
Dne 26.8.2016 v 08:56 Alex via Digitalmars-d-learn napsal(a):
void main()
{
import std.json;
import std.stdio;
string s = "{ \"language\": \"D\", \"rating\": 3.14, \"code\": 42 }";
JSONValue j = parseJSON(s);
writeln("code: ", j["code"].integer);
auto j2 = j;
j2["c
Hi everybody,
I'm little at loss: as documented, a JSONValue is only shallow
copied:
void main()
{
import std.json;
import std.stdio;
string s = "{ \"language\": \"D\", \"rating\": 3.14,
\"code\": 42 }";
JSONValue j = parseJSON(s);
writeln("code: ", j["code"].integer);
