Apple Mach-O Linker (Id)错误

Question

Xcode给了我3 "Apple Mach-O Linker (Id)错误“错误。但是，当我单击它们时，它并不会指向代码中的一行，因此我不知道问题出在哪里。我知道其他人也问过这个问题，但我能找到的所有解决方案都是针对每个人的代码的。我正在学习C++，所以这些错误都是我正在进行的初学者程序的一部分。

Apple Mach-O Linker (Id)错误"SensorNode::SensorNode(char*，float，int，float)"，引用来源: Apple O Linker (Id)错误“LOCATION()”，引用自: Apple O Linker (Id) Error Linker命令失败，退出代码1(使用-v查看调用)

如果有用的话，我会把我的代码放在这里:这是my "sensor_node.h"

#ifndef SENSORNODE_H
#define SENSORNODE_H

#include <iostream>

class LOCATION {
    float lat, longi, height;

public:
    LOCATION ();
    void setx(float xx);
    void sety(float yy);
    void setz(float zz);
    void print();
};

class SensorNode {
    char* NodeName;
    int NodeID;
    LOCATION Node1;
    float batt;
    int func;


public:
    SensorNode(char *n, float x, float y, float z, int i, float ah);
    void print();
    void setOK(int o);
    int getOK();
    void setLOC(float longi, float lat, float h);
};

#endif /* defined(__Project_3__sensor_node__) */

这是我的sensor_node.cpp：

#include "sensor_node.h"
//#include <iostream>
using namespace std;


void LOCATION::setx(float xx) {
    lat = xx;
    if (lat > 180.0 || lat < -180.0) {
        cout << "Latitude is not in the -180 to 180 degree range";
        lat = 0.0;
    }
}

void LOCATION::sety(float yy) {
    longi = yy;
    if (longi > 180.0 || longi < -180.0) {
        cout << "Latitude is not in the -180 to 180 degree range";
        longi = 0.0;
    }


}
void LOCATION::setz(float zz) {
    height = zz;
}

void LOCATION::print() {

    cout << "(LONGITUDE: " << longi << " ,LATITUDE: " << lat << " ,HEIGHT: " << height << " )";
}

这是我的main.cpp：

#include <iostream>
using namespace std;

#include "sensor_node.h"

int main() {

    LOCATION a; SensorNode s1("Pulse",15.9,-30.1,0,157,2.0);

    cout << "Beginning LOCATION tests.\n\n";
    cout << "  After initial construction:  ";
    a.print();
    cout << "\n";
    a.setx(-45.3);
    a.sety(27.6);
    a.setz(3.5);
    cout << "  After setting x/y/z to -45.3/27.6/3.5:  ";
    a.print();
    cout << "\n";

    cout << "  After attempting to set longitude to 180.1:  ";
    a.setx(180.1);
    a.print();
    cout << "\n";

    cout << "  After attempting to set longitude to -180.1:  ";
    a.setx(-180.1);
    a.print();
    cout << "\n";

    cout << "  After attempting to set latitude to 180.1:  ";
    a.sety(180.1);
    a.print();
    cout << "\n";

    cout << "  After attempting to set latitude to -180.1:  ";
    a.sety(-180.1);
    a.print();
    cout << "\n";
/*
    cout << "\n\n\n\nBeginning sensor node tests.\n\n";
    cout << "  After initial construction:";
    s1.print();
    cout << "\n  Printing the value returned by getOK: " << s1.getOK();
    cout << "\n  After changing location to 20/30/40:";
    s1.setLOC(20,30,40);
    s1.print();
    cout << "\n  After trying to set location illegally:";
    s1.setLOC(181, -181, 10);
    s1.print();
    cout << "\n  Node fails, then try to change location:";
    s1.setOK(0);
    s1.setLOC(5,10,15);
    s1.print();
    cout << "\n  Printing the value returned by getOK: " << s1.getOK();
    cout << "\n\n\n  End of tests.\n";

    cout << "Enter an integer to quit: ";
    cin >> hold;
*/
return 0;
}

Answer 1

您还没有为LOCATION类编写构造函数。您可以声明一个名为LOCATION的a和一个包含LOCATION的SensorNode，但是链接器无法确定LOCATION构造函数的代码在哪里，所以它无法链接。为LOCATION类编写一个构造函数，您应该很好。

Answer 2

您似乎忘记了实现SensorNode。在SensorNode.h中，您声明一个类，SensorNode具有数据和公共方法，但在SensorNode.cpp中，您没有提供SensorNode (构造函数)、print等的实现。链接器无法找到这些实现，因为它们尚未实现，因此出现链接器错误。

下面是一些样板，您可以从这里开始：

SensorNode::SensorNode(char *n, float x, float y, float z, int i, float ah)
{
}

void SensorNode::print()
{
}

void SensorNode::setOK(int o)
{
}

int SensorNode::getOK()
{
}

void SensorNode::setLOC(float longi, float lat, float h)
{
}