Check if type can be an argument to boost::lexical_cast<string>

Check if type can be an argument to boost::lexical_cast<string>

By : AĦmệd Kặŝệm
Date : November 22 2020, 01:01 AM
hop of those help? Your expression is not sufficient, as the lexical_cast function template takes everything and only reports errors via an internal static_assert. Instead test whether inserting the object into an std::ostream is valid:
code :
template <typename T, typename=void>
struct IsLexCastable : std::false_type {};

// Can be extended to consider std::wostream as well for completeness
template <typename T>
struct IsLexCastable<T,
            decltype(void(std::declval<std::ostream&>() << std::declval<T>()))>
  : std::true_type {};

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How do I use boost::lexical_cast and std::boolalpha? i.e. boost::lexical_cast< bool >("true")

How do I use boost::lexical_cast and std::boolalpha? i.e. boost::lexical_cast< bool >("true")

By : cCabo
Date : March 29 2020, 07:55 AM
I wish this help you I'm posting the answer to my own question here for others who may be looking for something like this:
code :
struct LocaleBool {
    bool data;
    LocaleBool() {}
    LocaleBool( bool data ) : data(data) {}
    operator bool() const { return data; }
    friend std::ostream & operator << ( std::ostream &out, LocaleBool b ) {
        out << std::boolalpha << b.data;
        return out;
    friend std::istream & operator >> ( std::istream &in, LocaleBool &b ) {
        in >> std::boolalpha >> b.data;
        return in;
#include <boost/lexical_cast.hpp>
#include <iostream>
#include "LocaleBool.hpp"

int main() {
    bool b = boost::lexical_cast< LocaleBool >("true");
    std::cout << std::boolalpha << b << std::endl;
    std::string txt = boost::lexical_cast< std::string >( LocaleBool( b ) );
    std::cout << txt << std::endl;
    return 0;
How does boost::lexical_cast take only one template type?

How does boost::lexical_cast take only one template type?

By : irintch3
Date : March 29 2020, 07:55 AM
it helps some times I've looked through the mess that is lexical_cast.hpp and this continues to escape me. , Template arguments can be deduced from the function arguments:
code :
template <typename T>
void foo(const T& x)
    // T is the type of X

foo(5); // T is be int
foo("hello"); //T is const char[6]
Boost compile error on converting UUID to string using boost::lexical_cast

Boost compile error on converting UUID to string using boost::lexical_cast

By : tygr
Date : March 29 2020, 07:55 AM
it fixes the issue You're missing the include, I guess:
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code :
#include <boost/lexical_cast.hpp>
#include <boost/uuid/uuid_io.hpp>
#include <boost/uuid/uuid.hpp>
#include <boost/uuid/random_generator.hpp>

int main() {
    boost::uuids::uuid uuid = boost::uuids::random_generator()();
    auto uuidString = boost::lexical_cast<std::string>(uuid);
When does boost::lexical_cast to std::string fail?

When does boost::lexical_cast to std::string fail?

By : user2718606
Date : March 29 2020, 07:55 AM
Hope that helps I'm writing unit tests and trying to have all my code covered. , It can fail for example if a user-defined conversion throws:
code :
enum class MyType {};

std::ostream& operator<<( std::ostream&, MyType const& )
    throw "error";

int main()
        boost::lexical_cast< std::string >( MyType{} );
        std::cout << "lexical_cast exception";
Using boost::lexical_cast<double>(string)

Using boost::lexical_cast<double>(string)

By : Bernardo
Date : March 29 2020, 07:55 AM
I wish this help you Since the code appears fine, here's an educated guess. myArray[9] is the tenth array element. myArray[0] is the first. You actually wanted the ninth element, myArray[8].
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