传递命令行参数
问题描述：
好吧我已经更新了我的代码，但我仍然不完全确定如何使用我传递的命令行参数的向量。我试图设置它像我下面的代码，但它不会编译。它给了我找不到argc和argv的错误：错误C2065：'argc '：未声明的标识符 1> C：\用户\克里斯\文件\视觉工作室2008 \项目\ cplusplustwo \ cplusplustwo \ application.h（32）：错误C2065：的argv'：未声明的标识符传递命令行参数
的main.cpp
#include "application.h"
int main(int argc,char *argv[]){
vector<string> args(argv, argv + argc);
return app.run(args);
}
application.h
#include <boost/regex.hpp>
#include <iostream>
#include <string>
#include <fstream>
#include <sstream>
#include "time.h"
using namespace std;
class application{
private:
//Variables
boost::regex expression;
string line;
string pat;
string replace;
int lineNumber;
char date[9];
char time[9];
void commandLine(vector<string> args){
string expression=""; // Expression
string textReplace=""; // Replacement Text
string inputFile=""; // Input File
string outputFile=""; // Output Directory
int optind=1;
// decode arguments
for(vector<string>::iterator i = args.begin(); i != args.end(); ++i){
while ((optind < argc) && (argv[optind][0]=='-')) {
string sw = argv[optind];
if (*i == "-e") {
optind++;
expression = argv[optind];
}
else if (*i == "-t") {
optind++;
textReplace = argv[optind];
}
else if (*i == "-i") {
optind++;
inputFile = argv[optind];
}
else if (*i == "-o") {
optind++;
outputFile = argv[optind];
}
else{
cout << "Unknown switch: "
<< argv[optind] << "Please enter one of the correct parameters:\n"
<< "-e + \"expression\"\n-t + \"replacement Text\"\n-i + \"Input File\"\n-o + \"Onput File\"\n";
optind++;
}
}
}
}
//Functions
void getExpression(){
cout << "Expression: ";
getline(cin,pat);
try{
expression = pat;
}
catch(boost::bad_expression){
cout << pat << " is not a valid regular expression\n";
exit(1);
}
}
void boostMatch(){
//Define replace {FOR TESTING PURPOSES ONLY!!! REMOVE BEFORE SUBMITTING!!
replace = "";
_strdate_s(date);
_strtime_s(time);
lineNumber = 0;
//Files to open
//Input Files
ifstream in("files/trff292010.csv");
if(!in) cerr << "no file\n";
//Output Files
ofstream newFile("files/NEWtrff292010.csv");
ofstream copy("files/ORIGtrff292010.csv");
ofstream report("files/REPORT.dat", ios.app);
lineNumber++;
while(getline(in,line)){
lineNumber++;
boost::smatch matches;
copy << line << '\n';
if (regex_search(line, matches, expression)){
for (int i = 0; i<matches.size(); ++i){
report << "Time: " << time << "Date: " << date << '\n'
<< "Line " << lineNumber <<": " << line << '\n';
newFile << boost::regex_replace(line, expression, replace) << "\n";
}
}else{
newFile << line << '\n';
}
}
}
public:
void run(vector<string> args){
commandLine(vector<string> args);
getExpression();
boostMatch();
}
};
原帖
我想通过命令行参数出主。这是一个高级C++类的作业。我需要传递一个向量的命令行，我不知道我是否正确地做了一切。我会像我一样将它传递给一个矢量吗？还有一个copy（）命令可以用来将命令行参数复制到一个向量中而不是推回？
main.cpp
#include "application.h"
int main(int argc,char *argv[]){
vector<string> args;
application app;
for (int i=1;i<argc;i++){
args.push_back(argv[i]);
}
app.run(args);
return(0);
}
application.h
#include <boost/regex.hpp>
#include <iostream>
#include <string>
#include <fstream>
#include <sstream>
#include "time.h"
using namespace std;
class application{
private:
//Variables
boost::regex expression;
string line;
string pat;
string replace;
int lineNumber;
char date[9];
char time[9];
void commandLine(vector<string> args){
string expression=""; // Expression
string textReplace=""; // Replacement Text
string inputFile=""; // Input File
string outputFile=""; // Output Directory
int optind=1;
// decode arguments
for(vector<string>::iterator i = args.begin(); i != args.end(); ++i){
while ((optind < argc) && (argv[optind][0]=='-')) {
string sw = argv[optind];
if (*i == "-e") {
optind++;
expression = argv[optind];
}
else if (*i == "-t") {
optind++;
textReplace = argv[optind];
}
else if (*i == "-i") {
optind++;
inputFile = argv[optind];
}
else if (*i == "-o") {
optind++;
outputFile = argv[optind];
}
else{
cout << "Unknown switch: "
<< argv[optind] << "Please enter one of the correct parameters:\n"
<< "-e + \"expression\"\n-t + \"replacement Text\"\n-i + \"Input File\"\n-o + \"Onput File\"\n";
optind++;
}
}
}
}
//Functions
void getExpression(){
cout << "Expression: ";
getline(cin,pat);
try{
expression = pat;
}
catch(boost::bad_expression){
cout << pat << " is not a valid regular expression\n";
exit(1);
}
}
void boostMatch(){
//Define replace {FOR TESTING PURPOSES ONLY!!! REMOVE BEFORE SUBMITTING!!
replace = "";
_strdate_s(date);
_strtime_s(time);
lineNumber = 0;
//Files to open
//Input Files
ifstream in("files/trff292010.csv");
if(!in) cerr << "no file\n";
//Output Files
ofstream newFile("files/NEWtrff292010.csv");
ofstream copy("files/ORIGtrff292010.csv");
ofstream report("files/REPORT.dat", ios.app);
lineNumber++;
while(getline(in,line)){
lineNumber++;
boost::smatch matches;
copy << line << '\n';
if (regex_search(line, matches, expression)){
for (int i = 0; i<matches.size(); ++i){
report << "Time: " << time << "Date: " << date << '\n'
<< "Line " << lineNumber <<": " << line << '\n';
newFile << boost::regex_replace(line, expression, replace) << "\n";
}
}else{
newFile << line << '\n';
}
}
}
public:
void run(vector<string> args){
commandLine(vector<string> args);
getExpression();
boostMatch();
}
};
答
我只是写：
vector<string> args(argv + 1, argv + argc + !argc);
这样就可以排除argv[0]，但在某种程度上这是稳健的，即使argc == 0（可能在Linux下，和也许其他操作系统也是如此）。
答
application::commandLine()需要args作为参数，但它是指argc和argv，它们不在范围内。如果您查看编译器的实际错误消息，它应该包含一个文件名和行号，直接指向错误的位置。当要求提供有关错误消息的帮助时，请发布实际的错误消息而不是解释它。
+0
我现在发布了实际的错误消息 – shinjuo 2010-03-02 06:18:36
是你如何在传递参数之前将原始矢量在main中对齐？ – shinjuo 2010-03-02 05:48:40
@shinjuo：构建一个向量'args'，并用所有命令行参数初始化为一次。 （我很抱歉，我不知道如何阅读你的评论。） – 2010-03-02 05:51:56
只是为了确保我正确理解你，我可以删除所有主要的推回，只是使用它？ – shinjuo 2010-03-02 05:53:22