【发布时间】:2011-09-17 22:43:53
【问题描述】:
我想要一个函数来返回基本类型对象的字节大小。我还希望它返回 STL 容器的总大小(以字节为单位)。 (我知道这不一定是内存中对象的大小,没关系)。
为此,我编写了一个带有bytes 函数的memorysize 命名空间,这样memorysize::bytes(double x) = 8(在大多数编译器上)。
我已经将它专门用于正确处理std::vector<double> 类型,但我不想为std::vector<ANYTHING> 形式的每个类编写不同的函数,那么如何更改模板以正确处理这种情况?
这是工作代码:
#include <iostream>
#include <vector>
// return the size of bytes of an object (sort of...)
namespace memorysize
{
/// general object
template <class T>
size_t bytes(const T & object)
{
return sizeof(T);
}
/// specialization for a vector of doubles
template <>
size_t bytes<std::vector<double> >(const std::vector<double> & object)
{
return sizeof(std::vector<double>) + object.capacity() * bytes(object[0]);
}
/// specialization for a vector of anything???
}
int main(int argc, char ** argv)
{
// make sure it works for general objects
double x = 1.;
std::cout << "double x\n";
std::cout << "bytes(x) = " << memorysize::bytes(x) << "\n\n";
int y = 1;
std::cout << "int y\n";
std::cout << "bytes(y) = " << memorysize::bytes(y) << "\n\n";
// make sure it works for vectors of doubles
std::vector<double> doubleVec(10, 1.);
std::cout << "std::vector<double> doubleVec(10, 1.)\n";
std::cout << "bytes(doubleVec) = " << memorysize::bytes(doubleVec) << "\n\n";
// would like a new definition to make this work as expected
std::vector<int> intVec(10, 1);
std::cout << "std::vector<int> intVec(10, 1)\n";
std::cout << "bytes(intVec) = " << memorysize::bytes(intVec) << "\n\n";
return 0;
}
如何更改模板规范以允许更一般的std::vector<ANYTHING> 案例?
谢谢!
【问题讨论】:
标签: c++ templates stl vector specialization