c++ - 如何将 STL vector 传递给采用 const [] 的函数 (c++)

Question

我有一个 3d STL vector ，

vector<vector<vector<double> > > mdata;

我还有一个功能

myfun(const double ya[]);

更准确地说，它是 GNU 科学图书馆的一个函数，

gsl_spline_init(gsl_spline * spline, const double xa[], const double ya[], size_t size);

但这与我的问题无关。

所以现在我想将数据的“最后”维度传递给 myfun。我一直在尝试这个:

for (int s = 0; s < msize; s++) {
    accelerators = new gsl_interp_accel*[msize];
    splines = new gsl_spline*[msize];
    for (int i = 0; i < msize; i++) {
        accelerators[i] = gsl_interp_accel_alloc();
        splines[i] = gsl_spline_alloc(gsl_interp_cspline_periodic, msize+1);
        gsl_spline_init(splines[i], &(*mgrid.begin()), &(*mdata[s][i].begin()), msize+1);
    }
}

但是编译器(g++、64 位、Ubuntu)提示:

In member function ‘std::vector<std::vector<std::vector<double,
std::allocator<double> >,
std::allocator<std::vector<double,
std::allocator<double> > > >,
std::allocator<std::vector<std::vector<double,
std::allocator<double> >,
std::allocator<std::vector<double,
std::allocator<double> > > > > >
SimpleAmfCalculator::interp_m(int)’: Calculator.cpp:100: error: cannot convert ‘std::vector<double,
std::allocator<double> >*’ to ‘const
double*’ for argument ‘3’ to ‘int
gsl_spline_init(gsl_spline*, const
double*, const double*, size_t)’ make: *** [Calculator.o] Error 1

非常感谢任何帮助!

Answer 1

您可以传递第一个元素的地址，例如:

#include <vector>

void fun(const double data[])
{

}

int main()
{
    std::vector<std::vector<std::vector<double> > > data3d;
    ....
    fun(&data3d[0][0][0]);
}

---

vector 的元素是连续存储的。所以我希望这种方式是标准的:)

23.2.4 Class template vector

1 A vector is a kind of sequence that supports random access iterators. In addition, it supports (amortized) constant time insert and erase operations at the end; insert and erase in the middle take linear time. Storage management is handled automatically, though hints can be given to improve efficiency. The elements of a vector are stored contiguously, meaning that if v is a vector where T is some type other than bool, then it obeys the identity:

&v[n] == &v[0] + n for
all 0 <= n < v.size().