1

我已经阅读了所有内容Make: *** [] Error 1并且我在我的 main 方法中返回了 0。

我试过切换解析器,但这也不起作用。我已经尝试过内部构建,但它不起作用(即使它确实我想使用makefile)。

这是给出的错误示例。我相信这是阻止我创建运行我的应用程序所必需的二进制文件的原因。

错误图片链接-> http://upit.cc/i/c6b2db81.png

这是控制台位。我做了刷新然后构建。

23:52:19 **** Build of configuration Debug for project Lab1 ****
make all 
Building file: ../src/Run.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"src/Run.d" -MT"src/Run.d" -o "src/Run.o" "../src/Run.cpp"
Finished building: ../src/Run.cpp

Building file: ../shared-src/EndToken.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"shared-src/EndToken.d" -MT"shared-src/EndToken.d" -o "shared-src/EndToken.o" "../shared-src/EndToken.cpp"
Finished building: ../shared-src/EndToken.cpp

Building file: ../shared-src/ErrorToken.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"shared-src/ErrorToken.d" -MT"shared-src/ErrorToken.d" -o "shared-src/ErrorToken.o" "../shared-src/ErrorToken.cpp"
Finished building: ../shared-src/ErrorToken.cpp

Building file: ../shared-src/Token.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"shared-src/Token.d" -MT"shared-src/Token.d" -o "shared-src/Token.o" "../shared-src/Token.cpp"
Finished building: ../shared-src/Token.cpp

Building file: ../server-src/server-token/SToken.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"server-src/server-token/SToken.d" -MT"server-src/server-token/SToken.d" -o "server-src/server-token/SToken.o" "../server-src/server-token/SToken.cpp"
Finished building: ../server-src/server-token/SToken.cpp

Building file: ../server-src/server-token/StartSToken.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"server-src/server-token/StartSToken.d" -MT"server-src/server-token/StartSToken.d" -o "server-src/server-token/StartSToken.o" "../server-src/server-token/StartSToken.cpp"
Finished building: ../server-src/server-token/StartSToken.cpp

Building file: ../server-src/Message.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"server-src/Message.d" -MT"server-src/Message.d" -o "server-src/Message.o" "../server-src/Message.cpp"
Finished building: ../server-src/Message.cpp

Building file: ../server-src/Server.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"server-src/Server.d" -MT"server-src/Server.d" -o "server-src/Server.o" "../server-src/Server.cpp"
Finished building: ../server-src/Server.cpp

Building file: ../server-src/User.cpp
Invoking: GCC C++ Compiler
g++ -O0 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"server-src/User.d" -MT"server-src/User.d" -o "server-src/User.o" "../server-src/User.cpp"
Finished building: ../server-src/User.cpp

Building target: Lab1
Invoking: GCC C++ Linker
g++  -o "Lab1"  ./src/Run.o  ./shared-src/EndToken.o ./shared-src/ErrorToken.o ./shared-src/Token.o  ./server-src/server-token/SToken.o ./server-src/server-token/StartSToken.o  ./server-src/Message.o ./server-src/Server.o ./server-src/User.o   
./shared-src/Token.o:(.rodata._ZTV5Token[_ZTV5Token]+0x20): undefined reference to `Token::getNextToken(char*)'
./shared-src/Token.o:(.rodata._ZTV5Token[_ZTV5Token]+0x28): undefined reference to `Token::processToken(char*)'
collect2: error: ld returned 1 exit status
make: *** [Lab1] Error 1
23:52:20 Build Finished (took 904ms)

正如您从图片中看到的,Lab1 没有二进制文件,并且显示了 make 错误。

感谢评论的人找到了解决方案

我将这段代码作为对这个问题的评论的参考。

/*
 * Token.h
 *
 *  Created on: Sep 12, 2013
 *      Author: cam
 */
#pragma once

class Token {

public:
    Token();
    virtual ~Token();

    enum TOKEN_TYPE {START, MIDDLE, END, ERROR};
    /**
     * @pre processToken must be ran first!
     */
    virtual Token getNextToken(char*);
    virtual bool processToken(char*);

};

/* * Token.cpp * * 创建于:2013 年 9 月 12 日 * 作者:cam */

#include "Token.h"

Token::Token() {

}

Token::~Token() {

}

Token Token::getNextToken(char* buff) {

}

bool Token::processToken(char * buff) {

}
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1 回答 1

1

最终,我的基类中没有包含两个虚拟方法定义。虽然这可以编译,但我很想知道为什么这是必要的,因为它们被定义为虚拟的。

于 2013-09-13T14:16:08.653 回答