错误C2678 C A*路障

Question

我当前正在努力在C 中实现A* pathfinding算法。我试图运行代码以查看显示网格函数是否有效，但出现了C2678错误：binary'＆lt;'：找不到操作员，该操作员会采用'const coord'的左手操作数（或没有可接受的转换）。

我知道我的程序很混乱，可能根本不高，但是在优化之前，我试图使基本版本正常工作。错误是因为我试图输出坐标结构的布尔值吗？

代码：

#include <iostream>
#include <fstream>
#include <chrono>
#include <thread>
#include <vector>
#include <set>
using std::chrono::milliseconds;
using std::chrono::duration_cast;
using std::this_thread::sleep_for;
typedef std::chrono::steady_clock the_clock;
struct Location {
    int g = 0; // Distance covered so far 
    int h = 0; // Estimate of distance to goal
    float f = 0; // Estimated cost of the complete path
    bool walkable = 0; // 0 = Walkable, 1 = Wall
};
// Structure 
struct Coord {
    int x;
    int y;
    Location location;
};
// Declare size of grid
#define WIDTH 10
#define HEIGHT 10
typedef Location Array[HEIGHT][WIDTH];
Location grid[HEIGHT][WIDTH]; // Create an array of locations
void displayGrid() {
    /* Displays the Grid to the console! */
    system("CLS");
    for (int y = 0; y < HEIGHT; y++) {
        for (int x = 0; x < WIDTH; x++) {
            std::cout << grid[y][x].walkable;
        }
        std::cout << "n";
    }
    sleep_for(milliseconds(100)); // Visual delay
}
void initialiseGrid() {
    /* Fills the Grid array with values */
    srand((unsigned)time(0));
    for (int y = 0; y < HEIGHT; y++) {
        for (int x = 0; x < WIDTH; x++) {
            grid[y][x].walkable = 0; 
    }
}
/* Test grid */
grid[4][2].walkable = 1;
grid[5][2].walkable = 1;
grid[4][3].walkable = 1;
grid[5][3].walkable = 1;
grid[4][5].walkable = 1;
grid[5][5].walkable = 1;
grid[4][6].walkable = 1;
grid[5][6].walkable = 1;
}
void Astar(Coord startPoint, Coord endPoint) {
    /**/
    std::set<Coord> closedSet = {}; // Nodes that do not have to be considered again
    std::set<Coord> openSet = {}; // Nodes still to be considered to find the shortest path
    Coord currentNode; // Current node
    currentNode.x = startPoint.x;
    currentNode.y = startPoint.y;
    currentNode.location.g = 0; // 0 Distance from starting point
    openSet.insert(currentNode); // Insert starting node
    while (openSet.empty() == false) { // Loop while open list is not empty
        for (std::set<Coord>::iterator it = openSet.begin(); it != openSet.end(); it++) { // Iterate through each element in the open set to find the lowest F value
            if ((*it).location.f < currentNode.location.f) { // Check if iterator f value is smaller than the current value
                currentNode = *it; // Update the current node
            }
        }
        openSet.erase(currentNode); // Drop from the open set since been checked
        closedSet.insert(currentNode); // Add to the closed set
    }
}

int main(int argc, char *argv[]) {
    // Set start and end points
    Coord start;
    start.x = 3;
    start.y = 3;
    Coord end;
    end.x = 5;
    end.y = 6;
    initialiseGrid(); // Put -1 (empty) in
    // Start timing
    the_clock::time_point startTime = the_clock::now();
    // Stop timing
    the_clock::time_point endTime = the_clock::now();
    // Compute the difference between the two times in milliseconds
    auto time_taken = duration_cast<milliseconds>(endTime - startTime).count();
    displayGrid();
    std::cout << "That took: " << time_taken << " ms" << std::endl;
    return 0;
}

Answer 1

用std::set要求严格的订购和您的Coord类解决问题的最简单方法是提供operator <比较Coord中的x和y值，并返回一个Coord使用这些值小于另一个Coord。

您可以使用std :: tie

进行此操作

#include <tuple>
//...
struct Coord {
    int x;
    int y;
    Location location;
    bool operator <(const Coord& c) const 
    // returns true if this->x and this->y < c.x and c.y, false otherwise
    { return std::tie(x,y) < std::tie(c.x,c.y); }  
};

std::tie比较x组件，然后如果相等，则比较y组件。返回比较的结果（如果X，Y组件的第一组小于X，Y组件或false的第二组，则true否则）。

LIVE示例在这里