我在为包含3个c ++文件的项目创建makefile时遇到问题。
在包含文件的目录（tree.h，helperMethods.h和main.cpp）中，我首先使用命令 emacs makefile.make 。然后，在emacs中，我将makefile写为：
all: tree.h helperMethods.h main.cpp
[tab]g++ -std=c++0x -o project3 tree.h helperMethods.h main.cpp
我已手动验证命令 g ++ -std = c ++ 0x -o project3 tree.h helperMethods.h main.cpp 确实编译了这三个文件，并制作了一个可执行的可执行文件应该。
在此之后，我保存makefile，退出emacs，然后尝试运行它。
首先我使用 make ，但这返回了：
make: *** No targets specified and no makefile found. Stop.
然后我尝试使用 make -f makefile.make ，但这也不起作用并返回：
make: Nothing to be done for `all'.
此时，我不确定为什么makefile构造不正确。我对makefile没有太多经验;我很确定我的makefile正文中的命令是正确的，但是当我编写它时我没有正确设置它。
如果完全相关，这里有3个文件：
tree.h中：
#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
的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;
}
答案 0 :(得分：3)
Makefile通常被称为Makefile（虽然makefile也可以）但没有任何扩展名。如果您使用扩展名（不推荐），则必须告诉make makefile的名称。这很乏味也没必要。
其次，您不需要将头文件放在编译命令行中，也不应该这样做。您在文件中使用#include行来指定头文件。所以你的编译命令可能如下所示：
g++ -std=c++0x -o project3 main.cpp
现在，在那个命令中：
main.cpp是来源
project3是目标（即将要创建的文件。）
此外：
tree.h和helperMethods.h;编译取决于这个文件。 （你知道，但它们并没有出现在命令行中，所以它并不明显。）此外，如果它们发生了变化，你必须重新编译你的文件。makefile解释了如何从源创建目标，还列出了目标的依赖项 >。 （从技术上讲，make不区分 sources 和依赖;它认为它们都是先决条件。但是识别差异很方便。）
因此对于上面的命令，整个make配方可能如下所示：
project3: main.cpp tree.h helperMethods.h
g++ -std=c++0x -o project3 main.cpp
通常，我们不使用main.cpp之类的文件名;相反，我们将调用project3 project3.cpp的主文件，就像目标（输出）文件是project3一样。事实上，如果你这样做，并且你没有任何其他依赖项（头文件），你可以输入：
make project3
根本没有makefile ，而make会提供命令：
g++ -o project3 project3.cpp
在这种情况下，该命令会出错，因为它没有指定正确的C ++标准。但它通常有效，并且使用相同的基本名称命名目标和来源是一个很好的理由。
还有一件事：您的文件tree.h是一个仅限标题的库，这很好。但是helperMethods.h只是伪装成头文件。这真的是一个完整的实现。你应该解决这个问题。
此外，将using namespace std放入头文件中确实是个坏主意。头文件应该在需要它的所有内容上明确使用std::命名空间前缀。通常不建议在任何地方使用using namespace std，但它在头文件中特别糟糕，因为它会静默地污染包含标头的任何文件的默认命名空间。这可能会导致非常模糊的错误或编译错误。
答案 1 :(得分：1)
学校项目，嗯？对于你自己正在做的事情，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，因为严格来说，它们永远不需要独立编译。你的makefile需要一个规则来构建main.co中的main.o，然后是一个用于从main.o构建可执行文件的规则。头文件应该作为main.o的依赖项添加，这样如果其中任何一个发生更改，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
为此目的，其他人可能会生成一个截然不同但同样有效的makefile。这个过程有一定的自由度。
你在这里有一个规则来构建整个事物（请记住，如果没有提供目标，makefile中的第一个是自动选择的），清理任何中间文件的规则，构建来自目标文件的可执行文件，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，手动管理所有依赖项会很不愉快，但在这个非常简单的情况下它很好。