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#20927 #20937
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#20927 #20937
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Original file line number | Diff line number | Diff line change |
---|---|---|
@@ -0,0 +1,15 @@ | ||
import std.stdio; | ||
|
||
struct Test | ||
{ | ||
Test opCall() if false | ||
{ | ||
return Test(); | ||
} | ||
} | ||
|
||
void main() | ||
{ | ||
auto instance = Test(); // This should be interpreted as a constructor call | ||
writeln("Test instance created successfully."); | ||
} |
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what's with the
if false
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The line Test opCall() if false indicates that the opCall method is conditionally compiled based on the if false statement. This means that the method will not be available or compiled into the program, effectively making it a no-op.
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The main function creates an instance of Test, which will call the default constructor since opCall is not defined due to the if false condition.
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in which it should be
Test opCall() if (false)
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The opCall method is defined within the Test struct but is conditionally compiled based on the if (false) statement, meaning it will not be available in the compiled output.
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the code with the if (false) condition in the opCall method. Since the method is conditionally compiled and will not execute, can it will need be modify the condition to if (true) to allow the method to be compiled and executed.
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struct Test
{
Test opCall() if (true)
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Is it wright?