c++ - 如何从swift将一组 float 传递给C++函数

Question

我在 Bridging-Header.h 中声明了一个 C++ 函数

struct MyFloat3{
    float x;
    float y;
    float z;
};
struct MyFloat3 ExtCurl(const float** triangle);

我已按照 http://www.swiftprogrammer.info/swift_call_cpp.html 中的说明进行操作包装器.cpp

#include "Curl.h"
extern "C" MyFloat3 ExtCurl(const float** triangle){
    return Curl(triangle);
}

curl .h

struct MyFloat3{
    float x;
    float y;
    float z;
};
MyFloat3 Curl(const float** triangle);

curl .cpp

#include "Curl.h"
MyFloat3 Curl(const float** triangle){
  MyFloat3 curl;
  curl.x = (triangle[1][1] - triangle[0][1])*(triangle[2][2]-triangle[0][2]) - (triangle[1][2] - triangle[0][2])*(triangle[2][1]-triangle[0][1]);
  curl.y = (triangle[1][2] - triangle[0][2])*(triangle[2][0]-triangle[0][0]) - (triangle[1][0] - triangle[0][0])*(triangle[2][2]-triangle[0][2]);
  curl.z = (triangle[1][0] - triangle[0][0])*(triangle[2][1]-triangle[0][1]) - (triangle[1][1] - triangle[0][1])*(triangle[2][0]-triangle[0][0]);
  return curl;
}

我试图从我的 swift 代码中调用它

var triangle:[[Float]] = [
            [1.0, 0.0, 0.8],
            [0.0, 0.5, 0.0],
            [4.0, 0.0, 6.0]
        ]
var normal:MyFloat3
normal = ExtCurl(triangle)

编译器责骂

Node.swift:127:38: Cannot convert value of type '[[Float]]' to
expected argument type 'UnsafeMutablePointer<UnsafePointer<Float>?>!'

我的 C++ 库的另一个函数返回 void 但通过给它的指针填充数组工作正常

Answer 1

let triangle0: [[Float]] = [
    [1,2,3],
    [10,20,30],
    [100,200,300]]

let normal = triangle0.flatMap{ $0 }.withUnsafeBufferPointer { (buffer) -> float3 in
    var p = buffer.baseAddress
    return ExtCurl(&p)
}

更新以查看差异(针对 OOPer)

let triangle:[Float] = [
    1.0, 0.0, 0.8,
    0.0, 0.5, 0.0,
    4.0, 0.0, 6.0
] 
let normal = triangle.withUnsafeBufferPointer { (buffer) -> float3 in
    var p = buffer.baseAddress
    return ExtCurl(&p)
}

更新工作示例(Swift、C++)//

主.swift

var triangle:[[Float]] = [
    [1.0, 0.0, 0.8],
    [0.0, 0.5, 0.0],
    [4.0, 0.0, 6.0]
]

triangle.flatMap{ $0 }.withUnsafeBufferPointer {(buffer)->() in
    var p = buffer.baseAddress
    let normal = fc(&p)
    print("from Swift:", normal)
}

tempt-Bridging-Header.h

struct float3 {
    float x;
    float y;
    float z;
};

struct float3 fc(const float **);

测试2.hpp

#ifndef test2_hpp
#define test2_hpp

#include <stdio.h>

struct float3CPP {
    float x;
    float y;
    float z;
};

struct float3CPP fcpp(const float **);

extern "C" struct float3CPP fc(const float ** p) {
    return fcpp(p);
};

#endif /* test2_hpp */

测试2.cpp

#include "test2.hpp"

struct float3CPP fcpp(const float ** triangle) {
    float3CPP curl;
    curl.z = (triangle[1][0] - triangle[0][0])*(triangle[2][1]-triangle[0][1]) - (triangle[1][1] - triangle[0][1])*(triangle[2][0]-triangle[0][0]);
    curl.y = (triangle[1][2] - triangle[0][2])*(triangle[2][1]-triangle[0][1]) - (triangle[1][0] - triangle[0][0])*(triangle[2][2]-triangle[0][0]);
    curl.x = (triangle[1][1] - triangle[0][1])*(triangle[2][2]-triangle[0][2]) - (triangle[1][2] - triangle[0][2])*(triangle[2][1]-triangle[0][1]);
    return curl;
}

它打印

from Swift: float3(x: 111856.898, y: -3.65359902e+24, z: 3.86805511e+24)
Program ended with exit code: 0

正如预期的那样......

请检查 triangle[x][y] 中的值...那里没有您希望看到的值:-)

所以你唯一的麻烦是如何将 float **p 转换为 t[3][3]

struct float3CPP fcpp(const float **t) {

    float triangle[3][3];
    for( int i = 0; i < 9; i++) {
        int row = i / 3;
        int col = i % 3;
        triangle[row][col] = *(*t + i);
    };


    float3CPP curl;

    curl.z = (triangle[1][0] - triangle[0][0])*(triangle[2][1]-triangle[0][1]) - (triangle[1][1] - triangle[0][1])*(triangle[2][0]-triangle[0][0]);
    curl.y = (triangle[1][2] - triangle[0][2])*(triangle[2][1]-triangle[0][1]) - (triangle[1][0] - triangle[0][0])*(triangle[2][2]-triangle[0][0]);
    curl.x = (triangle[1][1] - triangle[0][1])*(triangle[2][2]-triangle[0][2]) - (triangle[1][2] - triangle[0][2])*(triangle[2][1]-triangle[0][1]);


    return curl;
}

会给你想要的东西

from Swift: float3(x: 2.5999999, y: 5.0, z: -1.5)
Program ended with exit code: 0

更新 2如果将 C++ 函数定义为

struct float3CPP fcpp2(float triangle[3][3]) {

    float3CPP curl;

    curl.z = (triangle[1][0] - triangle[0][0])*(triangle[2][1]-triangle[0][1]) - (triangle[1][1] - triangle[0][1])*(triangle[2][0]-triangle[0][0]);
    curl.y = (triangle[1][2] - triangle[0][2])*(triangle[2][1]-triangle[0][1]) - (triangle[1][0] - triangle[0][0])*(triangle[2][2]-triangle[0][0]);
    curl.x = (triangle[1][1] - triangle[0][1])*(triangle[2][2]-triangle[0][2]) - (triangle[1][2] - triangle[0][2])*(triangle[2][1]-triangle[0][1]);

    return curl;
};

你的c++头文件将是

struct float3CPP fcpp2(float t[3][3]);
extern "C" struct float3CPP fc2(float t[3][3]) {
    return fcpp2(t);
};

带桥接头

struct float3 {
    float x;
    float y;
    float z;
};

struct float3 fc2(float [][3]);

和主 swift

let triangle:[[Float]] = [
    [1.0, 0.0, 0.8],
    [0.0, 0.5, 0.0],
    [4.0, 0.0, 6.0]
]

var t2 = triangle.reduce([]) { (r, row) -> [(Float, Float, Float)] in
    var r = r
    r.append((row[0], row[1], row[2]))
    return r
}

t2.withUnsafeMutableBufferPointer { (buffer) -> () in
    var p = buffer.baseAddress
    let normal = fc2(p)
    print("from Swift fc2:", normal)
}

打印正确的结果

from Swift fc2: float3(x: 2.5999999, y: 5.0, z: -1.5)
Program ended with exit code: 0

UPDATE 3“最好”的方法是将 C++ 函数声明为

// parameter is pointer to array of array 3 of const float
struct float3CPP fcpp3(const float (* const t)[][3]) {
    return fcpp2(*t);
}

更新你的标题....并在 Swift 中使用它

let triangle:[[Float]] = [
    [1.0, 0.0, 0.8],
    [0.0, 0.5, 0.0],
    [4.0, 0.0, 6.0]
]
triangle.flatMap{ $0 }.withUnsafeBufferPointer {(buffer)->() in
    let normal3 = fc3(OpaquePointer(buffer.baseAddress))
    print("from Swift fc3:", normal)
}

打印正确的结果:-)

from Swift fc3: float3(x: 2.5999999, y: 5.0, z: -1.5)
Program ended with exit code: 0