[英]Initializing struct vector with brace-enclosed initializer list
我初始化这样的普通类型向量:
vector<float> data = {0.0f, 0.0f};
但是当我使用结构而不是普通类型时
struct Vertex
{
float position[3];
float color[4];
};
vector<Vertex> data = {{0.0f, 0.0f, 0.0f}, {0.0f, 0.0f, 0.0f, 0.0f}};
我得到错误could not convert '{{0.0f, 0.0f, 0.0f}, {0.0f, 0.0f, 0.0f, 0.0f}}' from '<brace-enclosed initializer list>' to 'std::vector<Vertex>'
。 这有什么问题?
缺少一组{}
:
std::vector<Vertex> data =
{ // for the vector
{ // for a Vertex
{0.0f, 0.0f, 0.0f}, // for array 'position'
{0.0f, 0.0f, 0.0f, 0.0f} // for array 'color'
},
{
{0.0f, 0.0f, 0.0f},
{0.0f, 0.0f, 0.0f, 0.0f}
}
};
实际上你还需要一个{}
vector<Vertex> data = {{{0.0f, 0.0f, 0.0f}, {0.0f, 0.0f, 0.0f, 0.0f}}};
一个'{'用于向量,一个用于struct,一个用于struct member-arrays ...
也可以初始化具有矢量成员的对象。
#include <iostream>
#include <string>
#include <vector>
using namespace std;
class Test
{
public:
struct NumStr
{
int num;
string str;
};
Test(vector<int> v1,vector<NumStr> v2) : _v1(v1),_v2(v2) {}
vector<int> _v1;
vector<NumStr> _v2;
};
int main()
{
Test t={ {1,2,3}, {{1,"one"}, {2,"two"}, {3,"three"}} };
cout << t._v1[1] << " " << t._v2[1].num << " " << t._v2[1].str << endl;
return 0;
}
2 2两
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