>我有一个代码,在函数结束时,我需要从 int 转换为数组的所有元素的双倍,以便能够在退出函数之前进行最后的push_back。我现在的代码是:
template <class T, size_t dims> class A {
typedef typename std::array<int, dims> ArrayInt;
typedef typename std::array<double, dims> ArrayDouble;
typedef typename std::vector <ArrayDouble> VectorDouble;
/* ...*/
foo() {
/* ...*/
ArrayInt myArrayInt;
ArrayDouble myArrayDouble;
VectorDouble myVectorDouble;
/* Initialize myArrayInt
Do some other stuff */
for (int i = 0; i < dims; ++i)
myArrayDouble[i] = static_cast<double>(myArrayInt[i]);
myVectorDouble.push_back(myArrayDouble);
}
}
它工作正常,但我对这些台词感到不舒服:
for (int i = 0; i < dims; ++i)
myArrayDouble[i] = static_cast<double>(myArrayInt[i]);
有没有更好的方法?
谢谢。
您可以使用算法中的函数。
copy_n :
std::copy_n( myArrayInt.begin(), dims, myArrayDouble.begin() );
或带副本:
std::copy( myArrayInt.begin(), myArrayInt.end(), myArrayDouble.begin() );
这可以用更少的代码编写,但它是明确的。
ArrayInt myArrayInt;
ArrayDouble myArrayDouble;
VectorDouble myVectorDouble;
/* Initialize myArrayInt
Do some other stuff */
using std::transform;
using std::copy;
using std::begin;
using std::end;
// with explicit conversion
auto to_double = [](const int i) { return double{i}; };
transform(begin(myArrayInt), end(myArrayInt), begin(myArrayDouble),
to_double);
// with implicit conversion
copy(begin(myArrayInt), end(myArrayInt), begin(myArrayDouble));