Pass a lambda into the second clause of a ternary operator - c++

I am trying make ternary operator work with lambda. Following is the logic that I want to cast into a ternary operation
if (boolean_condition_var == false) {
RandomMethod();
}
// From here some logic to execute
int i = 0;
// blah blah logic
Following is how I am trying to make it work after reading through this link.
boolean_condition_var == false ? RandomMethod() : std::function<void(void)>([this](){
// execute some logic here
int i = 0;
//blah blah logic
});
But I am getting the following error:
Issue:
error: left operand to ? is void, but right operand is of type 'std::function<void ()>'
boolean_condition_var == false ? RandomMethod() : std::function<void(void)>([this](){ int i = 0; });
Question:
How can I pass a lambda into the second clause of my ternary operator?

Since you are not calling the lambda, it makes little sense to create it. What you are trying to do is equivalent to
boolean_condition_var ? void(0) : RandomMethod();
On the other hand, if you want to call the lambda, use the call operator:
boolean_condition_var? ([this](){...your code...}) ( ) : RandomMethod();
// ^
// |
//--------------------------------------------------+
There's no need to cast anything to std::function.
A plain old if-then-else statement would be more welcome here.
On the third hand, if you want your ternary operator to return a callable, you probably need something like
using voidf = std::function<void(void)>;
auto result = boolean_condition_var?
voidf([this](){...your code...}):
voidf([this](){RandomNethod();});
.... result();
Capturing boolean_condition_var by value and moving the condition into the (now single) lambda could be more readable and more efficient though.

How can I pass a lambda into the second clause of my ternary operator?
Using comma operator ?
boolean_condition_var == false
? (RandomMethod(), 0)
: (std::function<void(void)>([this](){ /*...*/ })(), 0);
Or, I suppose better,
std::function<void(void)> logicElse { [this](){ /*...*/ } };
boolean_condition_var == false
? (RandomMethod(), 0)
: (logicElse(), 0);

Related

C++: Always-Throw function in conditional expression

Conditional expressions allow throw expressions as operands. I would like to have a conditional expression where one of the operands is an always-throw function, however it appears that this is not possible.
#include<exception>
[[ noreturn ]] void foo() {
throw std::exception();
}
int main() {
int a = true ? 1 : throw std::exception();
int b = true ? 1 : foo(); // This will not compile
}
I've tried inlining foo as well, but it wouldn't compile either. The error is the same:
test.cc: In function 'int main()':
test.cc:9:18: error: third operand to the conditional operator is of type 'void', but the second operand is neither a throw-expression nor of type 'void'
int b = true ? 1 : foo(); // This will not compile
Is there a way to achieve calling that function inside the expression? My use case is a transpiler from a language to C++ and that other language supports case expressions where if no case is hit in the expression, an exception is thrown. Case expressions I'm trying to support are similar to this:
http://zvon.org/other/haskell/Outputsyntax/caseQexpressions_reference.html
You can use the comma operator like so:
int a = true ? 1 : (throw std::exception(), 0);
int b = true ? 1 : (foo(), 0);
See it working on godbolt.org.

In C++17 can an if statement with an initializer be used to unpack an optional?

I'm writing some code using std::optional's and am wondering if C++17's 'if statements with initializers' will be able to help unpack values?
std::optional<int> optionalInt = GetOptionalInt();
I'm making up the function Unpack here:
if( auto [value, has_value] = optionalInt.Unpack(); has_value )
{
// Use value here.
}
But, my question is. Will C++17 'if statement with initializer' help here? If so, how would it be coded?
Update, this is actually mainly an issue when using optional which is extremely easy to misuse because the optional and *optional both return bools and you don't get any compiler warning when somebody trys to access the value and forgets the *.
There is not, and cannot possibly be, such an Unpack() function.
But you could certainly do:
if (std::optional<int> o = GetOptionalInt(); o) {
// use *o here
}
though the extra o check is kind of redundant.
This is one of those places where it'd be nice if optional<T> modeled a container of at most one element, so that you could do:
for (int value : GetOptionalInt()) {
// possibly not entered
}
but we don't have that interface.
In order for this to work, there has to be a value for the unpacked value if it isn't there.
So
template<class T, class U>
std::pair< T, bool > unpack_value( std::optional<T> const& o, U&& u ) {
return { o.value_or(std::forward<U>(u)), (bool)o } )
}
would do what you wanted.
But as an optional already returns if it is engaged in a bool context you really should just:
if (auto i = get_optional())
then use *i within the body.
...
Now if optional stated that operator* returned a reference, and that return value was defined but accessing it was not defined when it was not engaged, then you could write an Unpack method or function that doesn't require a default value.
As far as I am aware this is not true. And as it doesn't really add anything, I don't see why it should be true.
Maybe this would work:
auto optValue = getOptional();
if (auto value = *optValue; optValue) { ...use value here... }

Defining const "variable" inside if block

I have the following code:
Foo a;
if (some_fairly_long_condition) {
a = complicated_expression_to_make_foo_1();
} else {
a = complicated_expression_to_make_foo_2();
}
I have two issues with this:
a is a const and should be declared so
the "empty" constructor, Foo() is called for no reason (maybe this is optimised away?)
One way to fix it is by using the ternary operator:
const Foo a = some_fairly_long_condition?
complicated_expression_to_make_foo_1():
complicated_expression_to_make_foo_2();
Is this good practice? How do you go about it?
To answer the second part of your question:
I usually put the initialization code into a lambda:
const Foo a = [&]()->Foo{
if (some_fairly_long_condition) {
return complicated_expression_to_make_foo_1();
} else {
return complicated_expression_to_make_foo_2();
}
}();
In most cases you should even be able to omit the trailing return type, so you can write
const Foo a = [&](){ ...
As far as the first part is concerned:
I'd say that greatly depends on how complex your initialization code is. If all three parts are really complicated expressions (and not just a function call each) then the solution with the ternary operator becomes an unreadable mess, while the lambda method (or a separate named function for that matter) allows you to break up those parts into the respective sub expressions.
If the problem is to avoid ternaty operator and your goal is to define the constant a, this code is an option:
Foo aux;
if (some_fairly_long_condition) {
aux = complicated_expression_to_make_foo_1();
} else {
aux = complicated_expression_to_make_foo_2();
}
const Foo a(aux);
It is a good solution, without any new feature ---as lambdas--- and including the code inline, as you want.

ternary operator doesn't work with lambda functions

I am assigning to a std::function<double()> a lambda expression. This snippet works
if(fn_type==exponential)
k.*variable = [=,&k](){ return initial*exp(-k.kstep*par); };
else
k.*variable = [=,&k](){ return initial*pow(k.kstep, par); };
whereas if I want to use the ternary operator
k.*variable = (fn_type==exponential ? [=,&k](){ return initial*exp(-k.kstep*par); } : [=,&k](){ return initial*pow(k.kstep, par); });
I get the following error:
error: no match for ternary ‘operator?:’ in <awfully long template error, because this whole thing is in a class defined in a function...>
Is this a gcc bug (I'm using 4.7.2)? Otherwise why is there this limit in the standard?
The second and third operands of the conditional operator must have the same type or there must be some common type to which they can both be converted that the compiler can figure out. There are only a handful of conversions that the compiler will consider.
Your two lambda expressions have different types, and there is no common type to which they can both be converted (conversions to user-defined types, like std::function<double()>, cannot be considered because there are potentially an infinite number of valid target types).
You can directly convert each of the operands to std::function<double()>:
k.*variable = fn_type==exponential
? std::function<double()>([=,&k](){ return initial*exp(-k.kstep*par); })
: std::function<double()>([=,&k](){ return initial*pow(k.kstep, par); });
But really, it's cleaner with the if/else.
Also faced this issue - won't compile!
'if/else' not good for me, I would like auto type deducing feature turn on.
auto memcpy_traits =
[&](uint8_t* line_dst, const uint8_t* curr_src, const size_t bytes_to_copy) {
std::memcpy(line_dst, curr_src, bytes_to_copy);
line_dst += bytes_to_copy;
curr_src += bytes_to_copy;
}
:
[&](uint8_t* line_dst, const uint8_t* curr_src, const size_t bytes_to_copy) {
std::memcpy(line_dst, curr_src, bytes_to_copy);
line_dst += bytes_to_copy;
curr_src += bytes_to_copy;
};

Can I make the ternary operator treat my class like a bool?

I've recently been doing a huge refactoring where I was changing a lot of my code to return booleans instead of an explicit return code. To aid this refactoring I decided to lean on the compiler where possible by getting it to tell me the places where my code needed to be changed. I did this by introducing the following class (see here for the lowdown on how this works):
///
/// Typesafe boolean class
///
class TypesafeBool
{
private:
bool m_bValue;
struct Bool_ {
int m_nValue;
};
typedef int Bool_::* bool_;
inline bool_ True() const { return &Bool_::m_nValue; }
inline bool_ False() const { return 0; }
public:
TypesafeBool( const bool bValue ) : m_bValue( bValue ){}
operator bool_() const { return m_bValue ? True() : False(); }
};
Now, instead of using a normal bool type as the return type, I used this class which meant that I couldn't compile something like this any more:
TypesafeBool SomeFunction();
long result = SomeFunction(); // error
Great: it has made the refactoring manageable on a huge codebase by letting the compiler do a lot of the hard work for me. So now I've finished my refactoring and I'd quite like to keep this class hanging around and carry on using it since it affords us an extra level of safety that the built-in bool type doesn't.
There is however one "problem" which is preventing me from doing this. At the moment we make heavy use of the ternary operator in our code, and the problem is that it is not compatible with this new class without explicit casts:
TypesafeBool result = ( 1 == 2 ? SomeFunction() : false ); // error: different types used
TypesafeBool result = ( 1 == 2 ? SomeFunction() : (TypesafeBool)false );
If I could "solve" this issue so that I could use my class in a seamless manner I would probably carry on using it throughout the codebase. Does anyone know of a solution to this issue? Or is it just impossible to do what I want?
In the context of the conditional operator, the type of the expression is the common type of the last two operands. The complete rules to determine this common type are a bit complex, but your case happens to be trivial: if one of the two possible return values is a class type, the other value must have the same class and the common type is obviously also that class.
That means that if one of the operands is a TypesafeBool, then the other must be as well.
Now the problem you're really trying to solve has been solved before. The trick is not providing a class; instead use a typedef. See for instance safe bool.
class CCastableToBool
{
public:
// ...
operator bool() const
{
//...
{
return true;
}
//...
return false;
}
private:
// ...
};
but beware, in C++ it is considered really dangerous to have a class that can be casted to bool. You are warned :-)
you can read this there, SafeBool
You should explicitely call TypesafeBool::True() in all your ternary tests.
TypesafeBool result = ( 1 == 2 ? SomeFunction().True() : false );
I don't know about a seamless manner, the ternary operator has some restrictions on its use...
However, why don't you define two constants ?
TypesafeBool const True = TypesafeBool(true);
TypesafeBool const False = TypesafeBool(false);
And then:
TypesafeBool result = ( 1 == 2 ? SomeFunction() : False );
Of course, it's a bit unorthodox since I play on the capitalization to avoid reusing a reserved word :)
Is it a possibility to make the constructor of TypesafeBool explicit? Of course, now the usage has to be
TypesafeBool result( 1 == 2 ? b : false );
Could you use an assignment operator that takes in a bool as the external argument, as well as one that takes a TypesafeBool? It might be something to try out...
Nice try, but if your code base is large, you are probably better off using a static checker such as PC-Lint to look for implicit bool<->int conversions instead.