假设我有几个整数向量和一个指向整数向量的指针。如何将指针的一个元素的值更改为其他整数矢量之一的地址?
为此,我构建了一个类,使我可以将桌面引入虚幻引擎,但是在每个刻度上,它必须分配一种向量形式,该向量包含一个结构体来表示另一个数据类的值每次打勾时,我只想复制几个元素(像素颜色值)的内存地址,这样我就不必浪费时间去重复两次(对于台式机图像而言,它需要数百万次操作)
setTimeOut(function(){
GetGovernerate()
},150,function(){
setTimeOut(function(){
GetDistrict();
},150,function(){
GetTown();
});
});
奖金问题:
假设我有这个:
#include <iostream>
#include <vector>
using namespace std;
// Print function
void PrintVector(vector<int> v)
{
for( int i = 0; i < v.size(); i++ )
{
cout << v[i] << ", ";
}
cout << endl;
}
int main()
{
vector<int> vector1;
vector<int> vector2;
vector<int> *ptrvector;
//Do some assignment so the vectors have values
for( int i = 0; i<3; i++)
{
vector1.push_back(i);
vector2.push_back(2*i);
}
//Assign the pointer to the address of vector1.
ptrvector = &vector1;
//Print out:
PrintVector(vector1); // (1,2,3)
PrintVector(vector2); // (2,4,6)
PrintVector(*ptrvector); // (1,2,3)
// We should see that lines 1 and 3 are the same
//BROKEN BIT::
//Ideally want something like
ptrvector[0] = &vector2[2];
PrintVector(*ptrvector); // (6,2,3);
//Such that if I were to do this:
vector2[2] = 20;
PrintVector(*ptrvector); // It should change as a side effect: (20,2,3)
}
答案 0 :(得分:2)
这应该产生您想要的输出,但是要注意。如评论中所述,如果您以任何方式更改<HEAD>
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或std::vector<int*>
的大小并且对原始向量中的元素重新排序,则存储在vector1
中的地址应被视为无效。 ,您的指针将指向错误的值。我还删除了vector2
。参见:Why is “using namespace std” considered bad practice?。
using namespace std;