如何从Swift将float数组传递给C ++函数 [英] How to pass an array of array of floats to C++ function from swift
问题描述
我有一个在Bridging-Header.h
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 上的说明进行操作一个> wrapper.cpp
I have followed the instructions at http://www.swiftprogrammer.info/swift_call_cpp.html wrapper.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;
}
我试图从我的快捷代码中调用它
I tried to call it from my swift code
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空值,但通过给它的指针填充数组可以正常工作
An another function of my C++ library which returns void but fills an array through the pointer given to it is working fine
推荐答案
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(对于OOPer)
update to see the difference (for 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)
}
使用工作示例(Swift,C ++)进行更新//
UPDATE with working example (Swift, C++)//
main.swift
main.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
tempt-Bridging-Header.h
struct float3 {
float x;
float y;
float z;
};
struct float3 fc(const float **);
test2.hpp
test2.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 */
test2.cpp
test2.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
符合预期...
请检查三角形[x] [y]中的值...那里没有您想要看到的内容:-)
please check the values in triangle[x][y] ... there are not what you would like to see there :-)
所以您唯一的麻烦是如何将float ** p转换为t [3] [3]
so your only trouble is how to convert float **p to 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;
}
将返回您想要的
from Swift: float3(x: 2.5999999, y: 5.0, z: -1.5)
Program ended with exit code: 0
更新2 如果您将c ++函数定义为
UPDATE 2 if you define c++ function as
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;
};
您的c ++头文件将是
your c++ header file will be
struct float3CPP fcpp2(float t[3][3]);
extern "C" struct float3CPP fc2(float t[3][3]) {
return fcpp2(t);
};
带有桥接头
struct float3 {
float x;
float y;
float z;
};
struct float3 fc2(float [][3]);
和主要迅速
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
更新3 的最佳"方法是 将C ++函数声明为
UPDATE 3 the "best" way, how to do it, is to declare c++ function as
// parameter is pointer to array of array 3 of const float
struct float3CPP fcpp3(const float (* const t)[][3]) {
return fcpp2(*t);
}
更新标题....并在Swift中使用它
update your headers .... and use it from Swift
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
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