I am trying to explore command line argument in C++ in Visual Studio 2010. It is working fine but it is adding one extra parameter at args[0] any reason why?
Here is the snapshot of how I am setting my arguments
the main method is as follows
int main(int argc, char *argv[])
{
cout<< endl << "total args passed in the command line - " << argc << endl;
for(int i = 0; i < argc; i++){
cout << i+1<< " argument - " << argv[i] << endl;
}
return 0;
}
The output on cmd is looking like this
As you can see that I am adding only two arguments but it is showing the count as 3.
The first entry in the list is the program name; that's how it's designed to work.
args[0] contains the name of your program and argc includes this in the count aswell.
If you would like to exclude this, try this:
int main(int argc, char *argv[])
{
cout<< endl << "total args passed in the command line - " << argc - 1 << endl;
for(int i = 1; i < argc; i++){
cout << i << " argument - " << argv[i] << endl;
}
return 0;
}
Related
I'm new to C++. I'm writing a simple calculator using command line. The command line should have this format:
programname firstNumber operator secondNumber
Here what I got so far:
#include <iostream>
#include <fstream>
using namespace std;
int main(int argc, char* argv[])
{
if (argc != 3)
{
cerr << "Usage: " << argv[0] << endl;
exit(0);
}
else
{
int firstNumber = atoi(argv[1]);
char theOperator = atoi(argv[2]);
int secondNumber = atoi(argv[3]);
switch (theOperator)
{
case'+':
{
cout << "The answer is " << firstNumber + secondNumber << endl;
break;
}
case '-':
{
cout << "The answer is " << firstNumber - secondNumber << endl;
break;
}
case '*':
{
cout << "The answer is " << firstNumber * secondNumber << endl;
break;
}
case '/':
{
if (secondNumber == 0)
{
cout << "Can not devide by a ZERO" << endl;
break;
}
else
{
cout << "The answer is " << firstNumber / secondNumber << endl;
break;
}
}
}
}
}
The program does not run. When I run it, it displays an appropriate usage message and end the program. Can anyone please help me?
Others have already given you the answer but you could have very easily figured this one out on your own. Just print what argc is at the point where you know the code is going into:
int main(int argc, char* argv[])
{
if (argc != 3)
{
cout << "argc is: " << argc << endl; // Debug output that you delete later
cerr << "Usage: " << argv[0] << endl;
exit(0);
}
else
And then come back with what argc is. When you find that argc is actually 4 and you want to know what is inside argc you should write some code to print it so that you can figure it out... Like this:
int main(int argc, char* argv[])
{
cout << "argc is: " << argc << endl; // Debug output that you delete later
for (int i = 0; i < argc; ++i)
{
// Print out all of the arguments since it's not working as you expect...
cout << "argv[" << i << "] = " << argv[i] << endl;
}
if (argc != 3)
{
cerr << "Usage: " << argv[0] << endl;
exit(0);
}
else
and you would have very quickly figured out what is wrong...
Please learn how to do this because it will save your but in the future and you won't have to wait for an answer here.
Additionally there is another error in your code.
Why on earth are you converting the + character from a string to an int?
else
{
int firstNumber = atoi(argv[1]);
char theOperator = atoi(argv[2]); // <<< WTF? Why?
int secondNumber = atoi(argv[3]);
switch (theOperator)
You probably want to get rid of the atoi part there and just go with:
char theOperator = argv[2][0]; // First character of the string
Provided that the second argument will always have only one letter... Which you might want to enforce/check. See strlen() and std::string and note that the type of argv[2] is char* (pointer to char).
I also recommend that you read How to debug small programs which is linked from the SO Howto-Ask Help Page. It may help a little. And no, I don't think your question is bad. Debugging small programs is a skill you'll need in the future if you intend to program so it will benefit you to learn it now.
Welcome to programming and C++ :)
The parameter argc also counts the program's name.
Try this:
if (argc != 4) // We expect 4 arguments: programname number operator number
{
cerr << "Usage: " << argv[0] << " <number> <operator> <number>" << endl;
exit(0);
}
In your code, running the program correctly (with all 3 parameters) displays the error message because argc equals 4.
If you type in
programname firstNumber operator secondNumber
You have 4 arguments, not 3.
argv[0] = programname
argv[1] = firstNumber
argv[2] = operator
argv[3] = secondNumber
Looks like your program is working correctly... at least as far as printing the usage message goes.
See also the other comments regarding your use of the operator argument.
I'm writing a program to familiarize myself with input and command line arguments.
I'm trying to scan the user's input for either 'a' or 'b', and if it matches print the next argument given out. For some reason no matter what the user enters it always outputs as "Invalid." Can anyone see what I might be doing wrong?
int main(int argc, char* argv[])
{
if(argc != 5)
{ //checks if input is blank
cout << "Usage: <Function>, String, Usage: <Function>, String," << endl;
}
else
{
for(int i = 1; i<argc; i++)
{
cout << argv[i] << endl;
if(argv[i][1] == 'a')
{
cout << argv[i] << "ASCII" << endl;
}
if(argv[i][1] == 'b')
{
cout << argv[i] << "BINARY" << endl;
}
else
{
cout << "incorrect format" << endl;
}
}
}
}
argv[i][1] is the second character of the string argv[i] because arrays in C++ are zero-based.
I think you may want to use argv[i][0] instead, the first character.
See the following code for a sample:
#include <iostream>
int main(int argc, char *argv[]) {
for (int i = 1; i < argc; i++) {
std::cout << " Argument: " << argv[i] << '\n';
std::cout << " First: " << argv[i][0] << '\n';
}
return 0;
}
Running that as per the following transcript, gives the expected output:
pax> testprog alpha beta gamma delta epsilon
Argument: alpha
First: a
Argument: beta
First: b
Argument: gamma
First: g
Argument: delta
First: d
Argument: epsilon
First: e
My post is organized in three sections:
1. My code
2. Example input and output
3. My three questions
MY CODE:
#include <iostream>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <fstream>
#include <sstream>
#include <string>
#include <cstring>
using namespace std;
void deleteTrash(char*, char*);
const int kStr = 2;
const int kStrLen = 3;
int main(int argc, char* argv[])
{
if (argc < 4) {
cout << "Incorrect argument given." << endl;
cout << "Try again." << endl;
return 0;
}
cout << "PRINT argv[2]" << endl;
cout << "-----" << endl;
for (int i = 0; i < sizeof(argv[2]); i++) {
cout << "Iterator: " << i << endl;
cout << argv[2][i] << endl;
}
char* inputString;
deleteTrash(argv[kStr], inputString);
cout << "PRINT inputString" << endl;
cout << "-----" << endl;
for (int i = 0; i < sizeof(inputString); i++) {
cout << i << endl;
cout << inputString[i] << endl;
}
int strLen;
stringstream num;
num << argv[kStrLen];
num >> strLen;
if ( num.fail() ) {
cout << "Incorrect argument given." << endl;
cout << "Try again." << endl;
return 0;
}
if ( strLen < sizeof(inputString) ) {
cout << "Incorrect argument given." << endl;
cout << "Try again." << endl;
return 0;
}
return 0;
}
void deleteTrash(char* tempString, char* inputString)
{
int tempStringLen = sizeof(tempString);
int newSize = 0;
while (tempString[newSize] != '\0')
newSize++;
char newString[newSize + 1];
int iterator = 0;
while (tempString[iterator] != '\0') {
newString[iterator] = tempString[iterator];
iterator++;
}
newString[newSize] = '\0';
cout << "PRINT newString" << endl;
cout << "-----" << endl;
for (int i = 0; i < sizeof(newString); i++) {
cout << newString[i] << endl;
}
inputString = newString;
cout << "PRINT inputString" << endl;
cout << "-----" << endl;
for (int i = 0; i < sizeof(inputString); i++) {
cout << "Iterator: " << i << endl;
cout << inputString[i] << endl;
}
return;
}
EXAMPLE INPUT:
./hw1q5 4 W# 3
OUTPUT:
PRINT argv[2]
-----
Iterator: 0
W
Iterator: 1
#
Iterator: 2
Iterator: 3
3
Iterator: 4
Iterator: 5
T
Iterator: 6
E
Iterator: 7
R
PRINT newString
-----
W
#
PRINT inputString
-----
Iterator: 0
W
Iterator: 1
#
Iterator: 2
Iterator: 3
Iterator: 4
Iterator: 5
Iterator: 6
Iterator: 7
PRINT inputString
-----
0
Segmentation fault: 11
MY QUESTIONS:
Why does argv contain more than 3 elements (M, #, and \0). It prints out 8 elements (print statement iterates 0 - 7), which, after W, #, \0, are garbage. Should it not be printing only the 3 elements (M, #, and \0). Why is this happening? How may I fix it?
Why is it that when I set newString (type char) to inputString (type char*), by doing inputString = newString, inputString iterates 8 time in the print statement, printing blanks after printing M, #, and \0.
Why is the seg fault happening in the third statement?
sizeof() does not return the length of a null-terminated character array string. Instead you need something like strlen().
Let's just take one of the problems here:
// Wrong!
int main(int argc, char* argv[])
...
for (int i = 0; i < sizeof(argv[2]); i++) {
cout << "Iterator: " << i << endl;
cout << argv[2][i] << endl;
}
// Better
int main(int argc, char* argv[])
...
for (int i = 0; i < argc; i++) {
cout << "argv[" << i << "]: " << argv[i] << endl;
}
argv[] is an array of one or more "C" strings.
argc tells you how many strings are in the array.
You want to iterate through the strings in the array (argv[i]), not the characters in the string (for example, "argv[0][0]").
... AND ...
"sizeof(argv)" just gives you the size of a pointer (4 bytes, for a 32-bit CPU). It does NOT give you the #/elements in the array. That's what "argc" is for.
In answer to your first question, I'm going to refer you to another SO article: What does int argc, char *argv[] mean?
The first command line argument is always the command itself. So in your example:
./hw1q5 4 W# 3
There are four command line arguments: hw1q5, 4, W#, and 3.
In regards to your other questions, and the remainder of the first question, the majority of your problems stem from the assumption that sizeof(char*) returns the length of a null terminated string, which it does not (as has been pointed out both in comments and an earlier answer).
A good reference for understanding sizeof can be found here: http://en.cppreference.com/w/cpp/language/sizeof, or as I suspect this is a homework assignment based on your compiled program name, your C++ textbook.
I've got a problem:
When the number of input files is known, for example: 2 files, the declaration will be very simple:
int main(int argc, const char** argv) {
const char* inputfile1 = argv[1];
const char* inputfile2 = argv[2];
const char* outputfile = argv[3];
cout << "Appending "
<< inputfile1 << " and "
<< inputfile2 << " to "
<< outputfile << "..." << endl;
...
...
}
But now the NUMBER of input files is unknown, how to make the declaretion? Thx
Use loops, Luke!
int main(int argc, const char** argv) {
cout << "Appending ";
for(int i = 1; i < argc-1; i++)
cout << argv[i] << (i != argc-2 ? " and " : " to ");
cout << argv[argc-1] << "..." << endl;
...
...
}
Use an array, a vector, a list, or some other STL container, e.g. std::vector<std::string> for the names and std::vector<std::ostream> for the output streams.
What do I have to do to make my program use a file that has been dragged and dropped onto its icon as a parameter?
My current main method looks like this:
int main(int argc, char* argv[])
{
if (argc != 2) {
cout << "ERROR: Wrong amount of arguments!" << endl;
cout << "\n" << "Programm closed...\n\n" << endl;
exit(1);
return 0;
}
Converter a(argv[1]);
// ...
cout << "\n" << "Programm finished...\n\n" << endl;
// cin.ignore();
return 0;
}
What I'd really like to be able to do is select 10 (or so) files, drop them onto the EXE, and process them from within my application.
EDIT:
The incomming parameter is used as filename, constructed in the cunstructor.
Converter::Converter(char* file) {
// string filename is a global variable
filename = file;
myfile.open(filename.c_str(), ios_base::in);
}
The method where the textfile gets read:
string Converter::readTextFile() {
char c;
string txt = "";
if (myfile.is_open()) {
while (!myfile.eof()) {
myfile.get(c);
txt += c;
}
} else {
error("ERROR: can't open file:", filename.c_str());
}
return txt;
}
EDIT2:
deleted
Update:
I got again to this point.
Actual Main method:
// File path as argument
int main(int argc, char* argv[]) {
if (argc < 2) {
cout
<< "ERROR: Wrong amount of arguments! Give at least one argument ...\n"
<< endl;
cout << "\n" << "Programm closed...\n\n" << endl;
cin.ignore();
exit(1);
return 0;
}
vector<string> files;
for (int g = 1; g < argc; g++) {
string s = argv[g];
string filename = "";
int pos = s.find_last_of("\\", s.size());
if (pos != -1) {
filename = s.substr(pos + 1);
cout << "argv[1] " << argv[1] << endl;
cout << "\n filename: " << filename << "\n pos: " << pos << endl;
files.push_back(filename);
}
files.push_back(s);
}
for (unsigned int k = 0; k < files.size(); k++)
{
cout << "files.at( " << k << " ): " << files.at(k).c_str() << endl;
Converter a(files.at(k).c_str());
a.getATCommandsFromCSV();
}
cout << "\n" << "Programm finished...\n\n" << endl;
cin.ignore();
return 0;
}
Actually the console window apears for maybe 0.5 sec and closes again.
It doen't stop on any of my cin.ignore(); Maybe it doesn't get there?
Can anyone help?
Your program does not need to do anything special apart from handling command-line arguments. When you drag-drop a file onto an application in Explorer it does nothing more than to pass the file name as argument to the program. Likewise for multiple files.
If all you expect is a list of file names, then just iterate over all arguments, do whatever you want with them and be done. This will work for zero to almost arbitrarily many arguments.
Maybe you could write a test program like this:
int main(int argc, char* argv[])
{
// argv[0] is not interesting, since it's just your program's path.
for (int i = 1; i < argc, ++i)
cout << "argv[" << i << "] is " << argv[i] << endl;
return 0;
}
And see what happens after you throw different files at it.
EDIT: Just look at Joey's answer.
Answer to the main question
TO SEE THE ANSWER TO YOUR LAST PROBLEM SEE BOTTOM OF THIS ANSWER
All drag&dropped files are get-able as argv[orderOfTheFile] (orderOfTheFile is from 1-n),
however how does windows create that order, now that is a real mystery...
Anyway let's say I would create 26 plain text files ( *.txt ), from a.txt to z.txt on my Desktop,
now if I would drag&dropped them on my ArgsPrinter_c++.exe located directly on C:\ drive,
an output would be similar to this:
argc = 27
argv[0] = C:\ArgsPrinter_c++.exe
argv[1] = C:\Users\MyUserName\Desktop\c.txt
argv[2] = C:\Users\MyUserName\Desktop\d.txt
argv[3] = C:\Users\MyUserName\Desktop\e.txt
argv[4] = C:\Users\MyUserName\Desktop\f.txt
argv[5] = C:\Users\MyUserName\Desktop\g.txt
argv[6] = C:\Users\MyUserName\Desktop\h.txt
argv[7] = C:\Users\MyUserName\Desktop\i.txt
argv[8] = C:\Users\MyUserName\Desktop\j.txt
argv[9] = C:\Users\MyUserName\Desktop\k.txt
argv[10] = C:\Users\MyUserName\Desktop\l.txt
argv[11] = C:\Users\MyUserName\Desktop\m.txt
argv[12] = C:\Users\MyUserName\Desktop\n.txt
argv[13] = C:\Users\MyUserName\Desktop\o.txt
argv[14] = C:\Users\MyUserName\Desktop\p.txt
argv[15] = C:\Users\MyUserName\Desktop\q.txt
argv[16] = C:\Users\MyUserName\Desktop\r.txt
argv[17] = C:\Users\MyUserName\Desktop\s.txt
argv[18] = C:\Users\MyUserName\Desktop\t.txt
argv[19] = C:\Users\MyUserName\Desktop\u.txt
argv[20] = C:\Users\MyUserName\Desktop\v.txt
argv[21] = C:\Users\MyUserName\Desktop\w.txt
argv[22] = C:\Users\MyUserName\Desktop\x.txt
argv[23] = C:\Users\MyUserName\Desktop\y.txt
argv[24] = C:\Users\MyUserName\Desktop\z.txt
argv[25] = C:\Users\MyUserName\Desktop\a.txt
argv[26] = C:\Users\MyUserName\Desktop\b.txt
My ArgsPrinter_c++.exe source code:
#include <iostream>
using namespace std;
int main(int argc, char* argv[]) {
cout << "argc = " << argc << endl;
for(int i = 0; i < argc; i++)
cout << "argv[" << i << "] = " << argv[i] << endl;
std::cin.ignore();
return 0;
}
Your last problem
I have created a simple program that creates only a sceleton of your class so it can be used, and the program's main itself ran JUST FINE => if your program exits too soon, the problem will be in your class...
Tested source code:
#include <iostream>
#include <vector>
using namespace std;
class Converter{
public:
Converter(const char* f){ cout << f << endl; }
void getATCommandsFromCSV(){ cout << "called getATCommandsFromCSV" << endl; }
};
int main(int argc, char* argv[]) {
vector<string> files;
for (int g = 1; g < argc; g++) {
string s = argv[g];
string filename = "";
int pos = s.find_last_of("\\", s.size());
if (pos != -1) {
filename = s.substr(pos + 1);
cout << "argv[1] " << argv[1] << endl;
cout << "\n filename: " << filename << "\n pos: " << pos << endl;
files.push_back(filename);
}
files.push_back(s);
}
for (unsigned int k = 0; k < files.size(); k++)
{
cout << "files.at( " << k << " ): " << files.at(k).c_str() << endl;
Converter a(files.at(k).c_str());
a.getATCommandsFromCSV();
}
cout << "\n" << "Programm finished...\n\n" << endl;
cin.ignore();
return 0;
}