-4

我必须在一些库中工作，无论我做什么，这段代码都会出现以下错误。

传递 `const amko::problem::launch' 作为 'this'argument of 'const double amko::problem::launch::ratio(double, double)' 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;
}

而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！

4

2 回答 2

2

launch::objfun_impl是一个const成员函数，它不能改变成员或调用其他函数。这意味着它不能调用const非非static成员函数，例如launch::ratio.

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

static /* <- ADDED static HERE */ double launch::ratio(const double a, const double b);
于 2012-03-16T18:00:35.163 回答
0

问题是您的ratio成员函数不是const，即使您没有修改对象的任何成员（为什么它根本是成员函数？）。你在里面objfun_impl打电话ratio。现在，objfun_implis const，因此承诺不会修改对象，但调用ratio会破坏该承诺。

于 2012-03-16T18:04:10.410 回答