不确定如何创建MakeFIle

Unsure how to create a MakeFIle

本文关键字：创建 MakeFIle 何创建 不确定      更新时间：2023-10-16

我在为一个包含3个c++文件的项目创建makefile时遇到了问题。

在包含文件(tree.h、helperMethods.h和main.cpp)的目录中，我首先使用命令emacs makefile.make。然后，在emacs中，我将makefile写入如下:

all: tree.h helperMethods.h main.cpp
[tab]g++ -std=c++0x -o project3 tree.h helperMethods.h main.cpp

我已经手动验证了命令g++ -std=c++0x -o project3 tree.h helperMethods.h main.cpp确实编译了这三个文件，并生成了一个可执行文件。

在此之后，我保存makefile，退出emacs，并尝试运行它。

首先，我使用make，但这返回:

make: *** No targets specified and no makefile found.  Stop.

然后我尝试使用make -f makefile。Make ，但这也不起作用，返回:

make: Nothing to be done for `all'.
在这一点上，我不确定为什么没有正确构造makefile。我没有很多制作文件的经验;我很确定我在makefile的主体中有正确的命令，但是当我写它时，我没有正确地设置它。

如果是相关的，这里有3个文件:

tree.h: 

#ifndef tree_h
#define tree_h
#include <vector>
using namespace std;
template<class T>
class tree
{
public:
    tree<T> * parent;
    T element;
    vector<tree<T> > nodes;
    tree(const T& theElement)
    {
        element = theElement;
        nodes = vector<tree<T> >();
    }
    tree()
    {
        nodes = vector<tree<T> >();
    }
};
#endif

helperMethods.h 

#ifndef helperMethods_h
#define helperMethods_h
#include "tree.h"
#include <tuple>
#include <string>
#include <iostream>
#include <vector>
using namespace std;
string printName(tuple<string, string, string> myTuple){
    string ret = "";
    if(std::get<0>(myTuple).length() > 0)
        ret += get<0>(myTuple) + " ";
    if(std::get<1>(myTuple).length() > 0 || std::get<2>(myTuple).length() > 0)
        ret += "(";
    if(std::get<1>(myTuple).length() > 0)
        ret += get<1>(myTuple);
    if(std::get<1>(myTuple).length() > 0 && std::get<2>(myTuple).length() > 0)
        ret += ", ";
    if(std::get<2>(myTuple).length() > 0)
        ret += get<2>(myTuple);
    if(std::get<1>(myTuple).length() > 0 || std::get<2>(myTuple).length() > 0)
        ret += ")";
    return ret;
}
bool tupleContain(tuple<string, string, string> myTuple, string myString){
    return (std::get<0>(myTuple).compare(myString) == 0) || (std::get<1>(myTuple).compare(myString) == 0);
}
void findElement(tree<tuple<string, string, string> > myTree, string myString, bool* found, vector<int> *ret)
{
    if(tupleContain(myTree.element, myString))
        *found = true;
    if(! * found)
    {
        for(int counter = 0; counter < (int)myTree.nodes.size(); counter ++)
        {
            if(!* found)
            {
                (*ret).push_back(counter);
                findElement(myTree.nodes.at(counter), myString, found, ret);
                if(!* found)
                    (*ret).pop_back();
            }
        }
    }
}
void getLineage(tree<tuple<string, string, string> > myTree, string myString){
    bool dummyForFound = false;
    bool * found = & dummyForFound;
    vector<int> lineage = vector<int>();
    vector<int> * pointer = & lineage;
    findElement(myTree, myString, found, &lineage);
    if(lineage.size() == 0)
    {
        cout << "Species not present" << endl;
        return;
    }
    vector<string> printString = vector<string>(lineage.size() + 1);
    tree<tuple<string, string, string> > * currentNodePointer = & myTree;
    for(int counter = 0; counter <= (int) lineage.size(); counter ++)
    {
        string currentLine = "";
        for(int counter2 = 0; counter2 < 2*((int) lineage.size() - counter); counter2 ++)
            currentLine += ">";
        if(counter != lineage.size())
            currentLine += " ";
        tree<tuple<string, string, string> > currentNode = * currentNodePointer;
        currentLine += printName(currentNode.element);
        if(counter < (int) lineage.size())
        {
            int foo = lineage.at(counter);
            tree<tuple<string, string, string> > currentNodeDummy = currentNode.nodes.at(foo);
            *currentNodePointer = currentNodeDummy;
        }
        printString.at(counter) = currentLine;
    }
    for(int counter = 0; counter < (int) printString.size(); counter ++)
        cout << printString.at(printString.size() - counter - 1) << endl;
    cout << endl;
}
void getCommonLineage(tree<tuple<string, string, string> > myTree , string name1, string name2)
{
    bool dummyForFound = false;
    bool * found = & dummyForFound;
    vector<int> lineage1 = vector<int>();
    vector<int> * pointer1 = & lineage1;
    vector<int> lineage2 = vector<int>();
    vector<int> * pointer2 = & lineage2;
    findElement(myTree, name1, found, pointer1);
    * found = false;
    findElement(myTree, name2, found, pointer2);
    if(lineage2.size() == 0 || lineage1.size() == 0)
    {
        cout << "At least one species not present." << endl;
        return;
    }
    bool stillSame = lineage1.at(0) == lineage2.at(0);
    cout << "Level[0] Common Ancestor: ROOT (ROOT, ROOT)" << endl;
    tree<tuple<string, string, string>> * lastSharedNode = & myTree;
    int finalCounter = 0;
    for(int counter = 0; counter < (int) min(lineage1.size(), lineage2.size()) && stillSame; counter ++)
    {
        tree<tuple<string, string, string> > dummyNode = * lastSharedNode;
        tree<tuple<string, string, string> > currentNode = dummyNode.nodes.at(lineage1.at(counter));
        *lastSharedNode = currentNode;
        if(counter < (int) min(lineage1.size(), lineage2.size()) - 1 && lineage1.at(counter + 1) != lineage2.at(counter + 1))
            stillSame = false;
        tuple<string, string, string> currentElement = currentNode.element;
        cout << "Level[" << counter + 1 << "] Commont Ancestor: " << printName(currentElement) << endl;
        finalCounter ++;
    }
    cout << endl;
    cout << "Ancestry unique to " << name1 << endl;
    tree<tuple<string, string, string> > savedNode = *lastSharedNode;
    tree<tuple<string, string, string> > * currentUnsharedNode = lastSharedNode;
    for(int counter = finalCounter; counter < (int) lineage1.size(); counter ++)
    {
        tree<tuple<string, string, string> > dummyNode = * currentUnsharedNode;
        tree<tuple<string, string, string> > currentNode = dummyNode.nodes.at(lineage1.at(counter));
        tuple<string, string, string> currentElement = currentNode.element;
        *currentUnsharedNode = currentNode;
        cout << "Level[" << counter + 1 << "] ";
        if(counter == lineage1.size() - 1)
            cout << "Species of interest: ";
        cout << printName(currentElement) << endl;
    }
    cout << endl;
    currentUnsharedNode = &savedNode;
    cout << "Ancestry unique to " << name2 << endl;
    for(int counter = finalCounter; counter < (int) lineage2.size(); counter ++)
    {
        tree<tuple<string, string, string> > dummyNode = * currentUnsharedNode;
        tree<tuple<string, string, string> > currentNode = dummyNode.nodes.at(lineage2.at(counter));
        tuple<string, string, string> currentElement = currentNode.element;
        *currentUnsharedNode = currentNode;
        cout << "Level[" << counter + 1 << "] ";
        if(counter == lineage2.size() - 1)
            cout << "Species of interest: ";
        cout << printName(currentElement) << endl;
    }
    cout << endl;
}
#endif

main.cpp 

#include "rapidxml.h"
#include "tree.h"
#include "helperMethods.h"
#include <string>
#include <string.h>
#include <stdio.h>
#include <iostream>
#include <vector>
#include <queue>
#include <tuple>
using namespace rapidxml;
int main(int argc, const char * arv[]){
    tuple<string, string, string> human ("Human", "Homo sapiens", "Species");
    tuple<string, string, string> apes ("Apes", "", "");
    tuple<string, string, string> dogs ("Dogs", "", "");
    tuple<string, string, string> root ("Root", "", "");
    tuple<string, string, string> bears ("Bears", "", "");
    tuple<string, string, string> cat ("Cat", "", "");
    tuple<string, string, string> horse ("Horse", "", "");
    tree<tuple<string, string, string> > myTree = tree<tuple<string, string, string>>(root);
    tree<tuple<string, string, string> > b = tree<tuple<string, string, string>>(dogs);
    tree<tuple<string, string, string> > c = tree<tuple<string, string, string>>(apes);
    tree<tuple<string, string, string> > d = tree<tuple<string, string, string>>(bears);
    tree<tuple<string, string, string> > e = tree<tuple<string, string, string>>(horse);
    tree<tuple<string, string, string> > f = tree<tuple<string, string, string>>(human);
    tree<tuple<string, string, string> > h = tree<tuple<string, string, string>>(cat);
    d.nodes.push_back(f);
    e.nodes.push_back(h);
    b.nodes.push_back(d);
    b.nodes.push_back(e);
    myTree.nodes.push_back(b);
    myTree.nodes.push_back(c);
    cout << printName(myTree.nodes.at(0).element);
    cout << "Welcome to my Tree of Life program!" << endl << endl;
    int choice = 1;
    while(choice == 1 || choice == 2) {
        cout << "Please choose from the following options:" << endl;
        cout << "   1.Get the lineage of a species" << endl;
        cout << "   2.Get the commmon lineage of two species" << endl;
        cout << "   3.Exit program" << endl << endl;
        cin >> choice;
        cout << endl;
        if(choice == 1)
        {
            cout << "Please enter the name of the species of interest:" << endl;
            string name;
            cin >> name;
            cout << endl;
            getLineage(myTree, name);
        }
        if(choice == 2)
        {
            string name1, name2;
            cout << "Please enter the name of the first species: " << endl;
            cin >> name1;
            cout << "Please enter the name of the second species: " << endl;
            cin >> name2;
            getCommonLineage(myTree, name1, name2);
        }
    }
    return 0;
}

makefile通常被称为Makefile(尽管makefile也可以)，没有任何扩展。如果您使用扩展名(不推荐)，您必须告诉make制作文件的名称。那既乏味又不必要。

第二，您不需要在编译命令行中放置头文件，也不应该这样做。在文件中使用#include行来指定头文件。所以你的编译命令可能看起来像这样:

g++ -std=c++0x -o project3 main.cpp

现在，在这个命令中:

  • main.cpp为源

  • project3是目标(即要创建的文件)

也

:

  • tree.h和helperMethods.h是编译工作所必需的;编译依赖于这个文件。(你知道这一点，但它们不会显示在命令行中，所以它不明显。)此外，如果它们改变了，你必须重新编译你的文件。

makefile解释了如何从源生成目标，并列出了目标的依赖项。(从技术上讲，make并不区分源和依赖;它认为它们都是先决条件。)

对于上面的命令，整个make recipe可能看起来像这样:

project3: main.cpp tree.h helperMethods.h 
        g++ -std=c++0x -o project3 main.cpp

通常，我们不使用main.cpp;相反，我们将项目3的主文件称为project3.cpp，就像目标(输出)文件是project3一样。实际上，如果您这样做了，并且您没有任何其他依赖项(头文件)，您可以键入:

make project3

根本没有makefile ， make会显示命令:

g++ -o project3 project3.cpp
在这种情况下，该命令将是错误的，因为它没有指定正确的c++标准。但它通常是有效的，这是一个很好的理由来命名您的目标和源具有相同的基本名称。

还有一件事:您的文件tree.h是一个仅限头文件的库，这很好。但helperMethods.h只是假装是一个头文件。这确实是一个完整的实现。你应该解决这个问题。

另外，把using namespace std放在头文件中真的是一个坏主意。头文件应该在所有需要它的地方显式地使用std::命名空间前缀。通常不建议在任何地方使用using namespace std，但它在头文件中尤其糟糕，因为它会无声地污染任何包含头文件的默认名称空间。这可能导致非常模糊的bug，或者编译错误。

学校作业，嗯?对于你自己正在做的事情来说，project3似乎是一个可怕的任意名称。

您不需要在makefile处理中包含.h文件，因为严格地说，它们不需要单独编译。你的makefile需要一个规则来构建main。然后是一个从main.o中构建可执行文件的规则。头文件应该作为main的依赖项添加。因此，如果它们中的任何一个改变，make将知道从那里重建。

基本上，它需要看起来像这样:

# http://stackoverflow.com/questions/15458126/unsure-how-to-create-a-makefile
all: project3
clean:
    -rm main.o project3
project3: main.o
    g++ -std=c++0x -o project3 main.o
main.o: main.cpp tree.h helperMethods.h
    g++ -std=c++0x -c main.cpp

其他人可能会为此目的生成一个完全不同但同样有效的makefile。这个过程有一定的自由度。

您在这里拥有的是构建整个东西的规则(请记住，如果没有提供目标，则自动选择makefile中的第一个)，清理任何中间文件的规则，从目标文件构建可执行文件的规则，以及从源文件构建目标文件的规则。这种方法不能很好地扩展;对于更大、更复杂的makefile来说，手工管理所有的依赖关系是很不愉快的，但在这个非常简单的例子中，这很好。

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