Dlang operator overloading = - operator-overloading

How i can use = operator overloading in a class in order to assign a value ??
int[4] users;
int someop(string op)(int j){
if(op == "="){
//example
users[j] = j
}
}

It is all in the documentation - https://dlang.org/spec/operatoroverloading.html#assignment ...
For details on how to overload other operators refer to the "Operator Overloading" section of the D Language Specification.

Related

Flutter : PieChart Error: The argument type 'RxInt' can't be assigned to the parameter type 'double?' [duplicate]

Very simple issue. I have the useless class:
class Useless{
double field;
Useless(this.field);
}
I then commit the mortal sin and call new Useless(0);
In checked mode (which is how I run my tests) that blows up, because 'int' is not a subtype of type 'double'.
Now, it works if I use new Useless(0.0) , but honestly I spend a lot of time correcting my tests putting .0s everywhere and I feel pretty dumb doing that.
As a temporary measure I rewrote the constructor as:
class Useless{
double field;
Useless(num input){
field = input.toDouble();
}
}
But that's ugly and I am afraid slow if called often. Is there a better way to do this?
Simply toDouble()
Example:
int intVar = 5;
double doubleVar = intVar.toDouble();
Thanks to #jamesdlin who actually gave this answer in a comment to my previous answer...
In Dart 2.1, integer literals may be directly used where double is expected. (See https://github.com/dart-lang/sdk/issues/34355.)
Note that this is syntactic sugar and applies only to literals. int variables still won't be automatically promoted to double, so code like:
double reciprocal(double d) => 1 / d;
int x = 42;
reciprocal(x);
would fail, and you'd need to do:
reciprocal(x.toDouble());
You can also use:
int x = 15;
double y = x + .0;
use toDouble() method.
For e.g.:
int a = 10
print(a.toDouble)
//or store value in a variable and then use
double convertedValue = a.toDouble()
From this attempt:
class Useless{
double field;
Useless(num input){
field = input.toDouble();
}
}
You can use the parse method of the double class which takes in a string.
class Useless{
double field;
Useless(num input){
field = double.parse(input.toString()); //modified line
}
}
A more compact way of writing the above class using constructor's initialisers is:
class Useless{
double _field;
Useless(double field):_field=double.parse(field.toString());
}
Since all divisions in flutter result to a double, the easiest thing I did to achieve this was just to divide the integer value with 1:
i.e.
int x = 15;
double y = x /1;
There's no better way to do this than the options you included :(
I get bitten by this lots too, for some reason I don't get any warnings in the editor and it just fails at runtime; mighty annoying :(
I'm using a combination:
static double checkDouble(dynamic value) {
if (value is String) {
return double.parse(value);
} else if (value is int) {
return 0.0 + value;
} else {
return value;
}
}
This is how you can cast from int to double
int a = 2;
double b = a*1.0;

How to declare a variable depending on if statement in Dart like in Kotlin?

How to declare a variable depending on if statement in Dart? In Kotlin it would look like this:
val max = if (a > b) {
a
} else {
b
}
Is it even possible in Dart?
#pskink's answer in the comment is correct but it didn't show how you would be able to do it in this scenario. Here is how you can do it in your scenario:
final max= a > b ? a : b;
The final keyword in Dart is the same as the val keyword in Kotlin. You will not be able to change the value of the variable. You could also use the var keyword in Dart which is the same as Kotlin's var keyword. You will be able to change the value of the variable after declaring it. You might be confused with the one-liner code since there isn't any if or else statements inside it. The code above is called a ternary operator.
Here is an explanation for it:
(condition/expresssion) ? val1(if true execute this) : val2(if false execute this)
For more than one statement, we can use a method by declaring its type as int.
void main() {
print(declareVariable());
}
int a = 10;
int b = 30;
int declareVariable() {
if(b < a){
return 1;
}
else if(b > a) {
return 2;
}
else {
return 0;
}
}
Edited :
We can declare more then one condition in single line in the same way.
var singleLine = b < a ? 1 : b > a ? 2 : 0;
This will print out the same answer as method.

In systemverilog is there a way to condition on a type?

So I am using a parameterized type in a common module.
Is there a way to say:
if( type == TYPE1 ) assign the struct one way
else if( type == TYPE2 ) assign another way
I was picturing this in a generate block.
Yes, you can use the type operator do a generate-if/case, or procedural if/case like:
real r;
if ( type(r) == type(real) ) ...
But unfortunately the code in all branches still must successfully compile, regardless of the condition. You will not be able to reference struct member that does not exist.
typedef struct {int a;} s1_t;
typedef struct {int a;int b;} s2_t;
s1_t s;
initial
#1 // procedural-if
if (type(s) == type(s1_t))
$display("%m s.a = %0d",s.a);
else if (type(s) == type(s2_t))
$display("%m s.b ==%0d",s.b); // this will not compile
There is type() operator described in IEEE1800-2012 § 6.23. Example usage from from the LRM:
bit[12:0] A_bus, B_bus;
parameter typebus_t = type(A_bus);
generate
case(type(bus_t))
type(bit[12:0]): addfixed_int #(bus_t) (A_bus,B_bus);
type(real): add_float #(type(A_bus)) (A_bus,B_bus);
endcase
endgenerate
There is also $typename() described in IEEE1800-2012 § 20.6.1. $typename() return s string of the type. Example usage from from the LRM:
// source code // $typename would return
typedef bitnode; // "bit"
node [2:0] X; // "bit [2:0]"
int signedY; // "int"
packageA;
enum{A,B,C=99} X; // "enum{A=32'sd0,B=32'sd1,C=32'sd99}A::e$1"
typedef bit[9:1'b1] word; // "A::bit[9:1]"
endpackage: A
importA::*;
moduletop;
typedef struct{node A,B;} AB_t;
AB_t AB[10]; // "struct{bit A;bit B;}top.AB_t$[0:9]"
...
endmodule

Unary negation operator overloading in D

Code
struct test
{
private real value;
this(real value)
{
this.value = value;
}
bool opUnary(string op)() if (op == "!")
{
return !value;
}
}
void main()
{
test a = 123.12345;
bool b = !a;
}
Compilation error
prog.d(19): Error: expression a of type test does not have a boolean value
http://ideone.com/Kec81
Also tested on dmd 2.053, 2.054
What's wrong with my code?
You cannot overload the ! operator in D - see http://www.d-programming-language.org/operatoroverloading.html#Unary for a list of overloadable unary operators. Without knowing what you're doing, it's hard to suggest a work around, it might be worth looking at alias this though - http://www.d-programming-language.org/class.html#AliasThis.

What does error conflicting types for '' mean?

i got an error that said "error: conflicting types for '____'. What does that mean?
Quickfix:
Make sure that your functions are declared once and only once before they are called. For example, change:
main(){ myfun(3.4); }
double myfun(double x){ return x; }
To:
double myfun(double x){ return x; }
main(){ myfun(3.4); }
Or add a separate function declaration:
double myfun(double x);
main(){ myfun(3.4); }
double myfun(double x){ return x; }
Possible causes for the error
Function was called before being declared
Function defined overrides a function declared in an included header.
Function was defined twice in the same file
Declaration and definition don't match
Declaration conflict in the included headers
What's really going on
error: conflicting types for ‘foo’ means that a function was defined more than once with different type signatures.
A file that includes two functions with the same name but different return types would throw this error, for example:
int foo(){return 1;}
double foo(){return 1.0;}
Indeed, when compiled with GCC we get the following errors:
foo.c:5:8: error: conflicting types for ‘foo’
double foo(){return 1.0;}
^
foo.c:4:5: note: previous definition of ‘foo’ was here
int foo(){return 1;}
^
Now, if instead we had a file with two function definitions with the same name
double foo(){return 1;}
double foo(){return 1.0;}
We would get a 'redefinition' error instead:
foo.c:5:8: error: redefinition of ‘foo’
double foo(){return 1.0;}
^
foo.c:4:8: note: previous definition of ‘foo’ was here
double foo(){return 1;}
^
Implicit function declaration
So why does the following code throw error: conflicting types for ‘foo’?
main(){ foo(); }
double foo(){ return 1.0; }
The reason is implicit function declaration.
When the compiler first encounters foo() in the main function, it will assume a type signature for the function foo of int foo(). By default, implicit functions are assumed to return integers, and the input argument types are derived from what you're passing into the function (in this case, nothing).
Obviously, the compiler is wrong to make this assumption, but the specs for the C (and thus Objective-C) language are old, cranky, and not very clever. Maybe implicitly declaring functions saved some development time by reducing compiler complexity back in the day, but now we're stuck with a terrible feature that should have never made it into the language. In fact, implicit declarations were made illegal in C99.
That said, once you know what's going on, it should be easy to dig out the root cause of your problem.
it's probably because your function "_" already exists in your library. It happened to me with this function:
I was using stdio.h
int getline (char s[ ] , int lim)
{
int c, i;
for (i=0; i < lim-1 && (c=getchar())!=EOF && c!='\n'; ++i)
s[i] = c;
if (c == '\n') {
s[i] = c;
++i;
}
s[i] = '\0';
return i;
}
When I changed "getline" to "getlinexxx" and gcc compiled it:
int getlinexxx (char s[], int lim)
{
int c, i;
for (i=0; i < lim-1 && (c=getchar())!=EOF && c!='\n'; ++i)
s[i] = c;
if (c == '\n') {
s[i] = c;
++i;
}
s[i] = '\0';
return i;
}
And the problem was gone
What datatype is '___'?
My guess is that you're trying to initialize a variable of a type that can't accept the initial value. Like saying int i = "hello";
If you're trying to assign it from a call that returns an NSMutableDictionary, that's probably your trouble. Posting the line of code would definitely help diagnose warnings and errors in it.