我有一个可变模板函数:
template<typename T, typename ArgType>
vector<T>
createVector(const int count, ...)
{
vector<T> values;
va_list vl;
va_start(vl, count);
for (int i=0; i < count; ++i)
{
T value = static_cast<T>(va_arg(vl, ArgType));
values.push_back(value);
}
va_end(vl);
return values;
}
这适用于T和argtype的某些(对我来说,很奇怪)配置,但不是我期望的那样:
// v1 = [0.0, 1.875, 0.0]
vector<float> v1 = createVector<float, float>(3, 1.0f, 2.0f, 3.0f);
// v2 = [0.0, 1.875, 0.0]
vector<float> v2 = createVector<float, float>(3, 1.0, 2.0, 3.0);
// v3 = [1.0, 2.0, 3.0]
vector<float> v3 = createVector<float, double>(3, 1.0, 2.0, 3.0);
// v4 = [1.0, 2.0, 3.0]
vector<float> v4 = createVector<float, double>(3, 1.0f, 2.0f, 3.0f);
// v5 = [1.0, 2.0, 3.0]
vector<double> v5 = createVector<double, double>(3, 1.0, 2.0f, 3.0);
当argtype是double(即使传递float)时,为什么这会起作用,而当它是float时,为什么不起作用?