Question

我必须在一些库中工作，无论我做什么，我都会在这段代码中出现以下错误。

传递`const amko :: problem :: launch'作为'const double amko :: problem :: launch :: ratio（double，double）'的'this'参数'丢弃限定符

namespace amko { namespace problem {
launch::launch():base( 0.0, 20.0, 1 ) {}

base_ptr launch::clone() const
{
return base_ptr(new launch(*this));
}

const double launch::ratio( const double a, const double b)
{
const double area = a*b;
const double circumference = 2*a+2*b;
const double ratio = circumference/area;
return ratio;
}

void launch::objfun_impl(fitness_vector &f, const decision_vector &xv) const
{
amko_assert(f.size() == 1 && xv.size() == get_dimension());
const double x = xv[0];

const double y = launch::ratio(x,5);

f[0] =  y;
}

虽然下面的代码工作得很好。

namespace amko { namespace problem {

initialValueProblem::initialValueProblem():base( 0.0, 20.0, 1 ) {}

base_ptr initialValueProblem::clone() const
{
return base_ptr(new initialValueProblem(*this));
}

Eigen::VectorXd initialValueProblem::computeDerivative( const double time, const Eigen::VectorXd& state )
{
Eigen::VectorXd stateDerivative( 1 );
stateDerivative( 0 ) = state( 0 ) - std::pow( time, 2.0 ) + 1.0;
return stateDerivative;
}

void initialValueProblem::objfun_impl(fitness_vector &f, const decision_vector &xv) const
{
amko_assert(f.size() == 1 && xv.size() == get_dimension());
const double x = xv[0];

double intervalStart = 0.0;
double intervalEnd = 10.0;
double stepSize = 0.1;

Eigen::VectorXd initialState_;
initialState_.setZero( 1 );
initialState_( 0 ) = x;

numerical_integrators::EulerIntegratorXd integrator( boost::bind( &initialValueProblem::computeDerivative,
                                                                   const_cast<initialValueProblem*>( this ), _1, _2 ), intervalStart, initialState_ );
Eigen::VectorXd finalState = integrator.integrateTo( intervalEnd, stepSize );
f[0] =  fabs( finalState( 0 ) - 11009.9937484598 );
}

谢谢！

Answer 1

launch::objfun_impl是const成员函数，它不能更改成员或调用其他函数。这意味着它无法调用非const非static成员函数，例如launch::ratio。

因为launch::ratio似乎根本不访问成员，只是它的参数，最简单的解决方法是通过更改类定义中的原型使其成为static成员函数：

static /* <- ADDED static HERE */ double launch::ratio(const double a, const double b);

Answer 2

问题是你的ratio成员函数不是const，即使你没有修改对象的任何成员（为什么它根本就是成员函数？）。在objfun_impl内，您正在调用ratio。现在，objfun_impl为const，因此承诺不会修改对象，但调用ratio会破坏该承诺。

错误将`const`作为`const double'的`this`参数传递丢弃限定符

2 个答案: