写一段C++代码，实现两个坐标为整数的矩形求交

好的，以下是代码：

#include <stdio.h>

typedef struct {
    int x;
    int y;
} Point;

typedef struct {
    Point tl; // top-left point
    Point br; // bottom-right point
} Rect;

// 判断两个矩形是否有交集
int has_intersect(Rect rect1, Rect rect2) {
    if (rect1.tl.x > rect2.br.x || rect2.tl.x > rect1.br.x) {
        return 0;
    }
    if (rect1.tl.y < rect2.br.y || rect2.tl.y < rect1.br.y) {
        return 0;
    }
    return 1;
}

// 计算两个矩形的交集
Rect get_intersect(Rect rect1, Rect rect2) {
    Rect rect = {0};
    if (!has_intersect(rect1, rect2)) {
        return rect;
    }
    rect.tl.x = rect1.tl.x > rect2.tl.x ? rect1.tl.x : rect2.tl.x;
    rect.tl.y = rect1.tl.y < rect2.tl.y ? rect1.tl.y : rect2.tl.y;
    rect.br.x = rect1.br.x < rect2.br.x ? rect1.br.x : rect2.br.x;
    rect.br.y = rect1.br.y > rect2.br.y ? rect1.br.y : rect2.br.y;
    return rect;
}

// 测试代码
int main() {
    Rect rect1 = {{0, 0}, {4, 4}};
    Rect rect2 = {{2, 2}, {6, 6}};
    Rect rect3 = {{5, 5}, {7, 7}};
    Rect rect_intersect;

    rect_intersect = get_intersect(rect1, rect2);
    printf("(%d,%d)-(%d,%d)\n", rect_intersect.tl.x, rect_intersect.tl.y, rect_intersect.br.x, rect_intersect.br.y);

    rect_intersect = get_intersect(rect1, rect3);
    printf("(%d,%d)-(%d,%d)\n", rect_intersect.tl.x, rect_intersect.tl.y, rect_intersect.br.x, rect_intersect.br.y);

    return 0;
}

这段代码实现了两个坐标为整数的矩形求交的功能，具体的实现过程就是判断两个矩形是否有交集，如果有，再计算出它们的交集矩形。值得注意的是，我们这里使用了 C 语言中的一些结构体和指针的技巧，来使代码的结构更加清晰和简洁。