I was recently working on a project for a class and was having a lot of problems with passing in command line arguments. I decided to test out with a really simple code I found online from geeksforgeeks to see if I could get any sort of command line stuff to work and it still is not working. It will not print any argv values and when I debug it, it says that argc is 1 despite me putting in 4 command line arguments. I have been trying to find answers to this online for hours and have no idea what is going on especially when using this really simple code. I attached the code I was testing below. It only prints out "You have entered 1 arguments:" I am relatively new to coding but very confused.
#include <iostream>
using namespace std;
int main(int argc, char** argv)
{
cout << "You have entered " << argc
<< " arguments:" << "\n";
for (int i = 0; i < argc; ++i)
cout << argv[i] << "\n";
return 0;
}
I tried passing two parameters from bash to c++ program to be used but cant seem to get it to work exactly right. I would use the command ./bash "Turtle" "Cat" in linux command line.
#!/bin/bash
./main.out $1 $2
But the C++ file would only read ./main.out from argv. The cout would just print cut off versions of ./main.out such as ./main.out then /main.out and then main.out. Am I incorrectly using the parameters in the placement of the bash file?
int main(int argc, char *argv[]) {
cout << argv+0 << endl;
cout << argv+1 << endl;
cout << argv+2 << endl;
return 0;
}
This is because you are not printing the arguments as strings, your printing their memory locations. In clang++-9 (what I tested it in) this is what happens when a pointer is passed to std::cout, in many compilers (MSVC, correct me if im wrong) this will simply print nothing.
What you need to do is reference it as an array index and print that
int main(int argc, char **argv) {
cout << argv[0] << endl;
cout << argv[1] << endl;
cout << argv[2] << endl;
return 0;
}
The above code works fine for me (passing arguments manually when executing) when compiled with clang++-9. If you are going to use this code you should also check that there are at least 3 arguments (value of argc) otherwise you may point at invalid memory when indexing argv
Also try and avoid std::endl and use "\n" instead, std::endl needlessly flushes the buffer and is not required 99% of the time
Your bash file is fine. The problem is in your C++. Try something like this:
#include <iostream>
int main(int argc, char **argv) {
std::cout << argv[1] << "\n";
std::cout << argv[2] << "\n";
}
[warning: this doesn't check for errors, so if you don't pass any parameters, it'll misbehave badly.]
char *argv[] is a pointer of pointers (same as char **argv), you need
cout << argv[1] << endl;
I have a project in C++ (using Visual Studio 2013), and I know that if i want to pass command arguments I have to go to Project > Properties > Configuration Properties > Debugging and then type the command in "Command Arguments".
But I'd like to see those command arguments in the actual program (after I click "Start without debugging"), since I can only see the output, without the command.
int main (int argc, char *argv[])
{
for (int i = 1; i < argc; i++)
std::cout << "argument " << i << " = " << argv[i] << std::endl;
}
#include <iostream>
int main(int argc, char* argv[])
{
for(int i = 0; i < argc; ++i)
{
std::cout << "arg[" << i << "]: " << argv[i] << std::endl;
}
return 0;
}
Output:
arg[0]: C:\VS2015\PrintCmdArgs\Debug\PrintCmdArgs.exe
arg[1]: here
arg[2]: are
arg[3]: some
arg[4]: arguments
Press any key to continue . . .
There are several ways you can "see" the command arguments once the program has started running.
Simply print them from main.
If you need to display them in a form or print them while the application is running then you'll need to copy the arguments to some global/static variable and then access that when you need to.
Here's my code:
int _tmain(int argc, _TCHAR* argv[])
{
for (int i = 1;i<argc;i++) printf("Argument %d:%s\n",i,argv[i]);
// output = Argument 1:param
for (int k = 1; k < argc; k++) cout << "Argument " << k << ": " << argv[k];
// output = Argument 1: 00BF5878
return(0);
}
My question is: why do I see different outputs on cout and printf?
You see an address in the output from cout because you have built the program with UNICODE or _UNICODE defined. Probably by way of a setting in a Visual Studio project. Then _tmain expands to MIcrosoft's non-standard wmain, and _TCHAR expands to wchar_t.
And cout doesn't know how that a pointer to wchar_t is supposed to point to a null-terminated string of wide characters.
I am not sure why that doesn't happen with printf. Lemme check.
OK I have checked and your printf is not printing "param" as you indicate it does.
Here's the corrected code I checked with:
#include <iostream>
#include <stdio.h>
#include <tchar.h>
using namespace std;
int _tmain( int argc, _TCHAR* argv[] )
{
for( int i = 0;i < argc; ++i )
{
printf("printf Argument %d:%s\n",i,argv[i]);
}
for( int i = 0; i < argc; ++i )
{
cout << "cout Argument " << i << ": " << argv[i] << endl;
}
}
And here's the result:
[D:\dev\test]
> cl foo.cpp /D _UNICODE
foo.cpp
[D:\dev\test]
> foo param
printf Argument 0:f
printf Argument 1:p
cout Argument 0: 004F9A9C
cout Argument 1: 004F9AA4
[D:\dev\test]
> _
In other words, the apparent conundrum is entirely caused by your inaccurate reporting of results.
Solution: instead of using Microsoft's non-standard wmain, and in particular instead of using the now totally meaningless Windows 9x support macros!, use a standard C++ main.
The difference is that printf takes the contents of the parameters and forces the data to match the format specifier. The content of argv[i] will be treated as a nul terminated string because of the %s format specifier.
The cout in conjunction with the stream insertion operator will output based on the type of the parameter. The compiler will search all the operator<<() methods for cout and choose the one the best matches the parameter.
In your case, the compiler is looking for the method operator<<(TCHAR *) and didn't find an exact match. Instead, the compiler recognized that the parameter was a pointer and chose operator<<(void *) which prints out a pointer.
Try defining operator<<(TCHAR*) and see what happens.
If I run my C++ application with the following main() method everything is OK:
int main(int argc, char *argv[])
{
cout << "There are " << argc << " arguments:" << endl;
// Loop through each argument and print its number and value
for (int i=0; i<argc; i++)
cout << i << " " << argv[i] << endl;
return 0;
}
I get what I expect and my arguments are printed out.
However, if I use _tmain:
int _tmain(int argc, char *argv[])
{
cout << "There are " << argc << " arguments:" << endl;
// Loop through each argument and print its number and value
for (int i=0; i<argc; i++)
cout << i << " " << argv[i] << endl;
return 0;
}
It just displays the first character of each argument.
What is the difference causing this?
_tmain does not exist in C++. main does.
_tmain is a Microsoft extension.
main is, according to the C++ standard, the program's entry point.
It has one of these two signatures:
int main();
int main(int argc, char* argv[]);
Microsoft has added a wmain which replaces the second signature with this:
int wmain(int argc, wchar_t* argv[]);
And then, to make it easier to switch between Unicode (UTF-16) and their multibyte character set, they've defined _tmain which, if Unicode is enabled, is compiled as wmain, and otherwise as main.
As for the second part of your question, the first part of the puzzle is that your main function is wrong. wmain should take a wchar_t argument, not char. Since the compiler doesn't enforce this for the main function, you get a program where an array of wchar_t strings are passed to the main function, which interprets them as char strings.
Now, in UTF-16, the character set used by Windows when Unicode is enabled, all the ASCII characters are represented as the pair of bytes \0 followed by the ASCII value.
And since the x86 CPU is little-endian, the order of these bytes are swapped, so that the ASCII value comes first, then followed by a null byte.
And in a char string, how is the string usually terminated? Yep, by a null byte. So your program sees a bunch of strings, each one byte long.
In general, you have three options when doing Windows programming:
Explicitly use Unicode (call wmain, and for every Windows API function which takes char-related arguments, call the -W version of the function. Instead of CreateWindow, call CreateWindowW). And instead of using char use wchar_t, and so on
Explicitly disable Unicode. Call main, and CreateWindowA, and use char for strings.
Allow both. (call _tmain, and CreateWindow, which resolve to main/_tmain and CreateWindowA/CreateWindowW), and use TCHAR instead of char/wchar_t.
The same applies to the string types defined by windows.h:
LPCTSTR resolves to either LPCSTR or LPCWSTR, and for every other type that includes char or wchar_t, a -T- version always exists which can be used instead.
Note that all of this is Microsoft specific. TCHAR is not a standard C++ type, it is a macro defined in windows.h. wmain and _tmain are also defined by Microsoft only.
_tmain is a macro that gets redefined depending on whether or not you compile with Unicode or ASCII. It is a Microsoft extension and isn't guaranteed to work on any other compilers.
The correct declaration is
int _tmain(int argc, _TCHAR *argv[])
If the macro UNICODE is defined, that expands to
int wmain(int argc, wchar_t *argv[])
Otherwise it expands to
int main(int argc, char *argv[])
Your definition goes for a bit of each, and (if you have UNICODE defined) will expand to
int wmain(int argc, char *argv[])
which is just plain wrong.
std::cout works with ASCII characters. You need std::wcout if you are using wide characters.
try something like this
#include <iostream>
#include <tchar.h>
#if defined(UNICODE)
#define _tcout std::wcout
#else
#define _tcout std::cout
#endif
int _tmain(int argc, _TCHAR *argv[])
{
_tcout << _T("There are ") << argc << _T(" arguments:") << std::endl;
// Loop through each argument and print its number and value
for (int i=0; i<argc; i++)
_tcout << i << _T(" ") << argv[i] << std::endl;
return 0;
}
Or you could just decide in advance whether to use wide or narrow characters. :-)
Updated 12 Nov 2013:
Changed the traditional "TCHAR" to "_TCHAR" which seems to be the latest fashion. Both work fine.
End Update
the _T convention is used to indicate the program should use the character set defined for the application (Unicode, ASCII, MBCS, etc.). You can surround your strings with _T( ) to have them stored in the correct format.
cout << _T( "There are " ) << argc << _T( " arguments:" ) << endl;
Ok, the question seems to have been answered fairly well, the UNICODE overload should take a wide character array as its second parameter. So if the command line parameter is "Hello" that would probably end up as "H\0e\0l\0l\0o\0\0\0" and your program would only print the 'H' before it sees what it thinks is a null terminator.
So now you may wonder why it even compiles and links.
Well it compiles because you are allowed to define an overload to a function.
Linking is a slightly more complex issue. In C, there is no decorated symbol information so it just finds a function called main. The argc and argv are probably always there as call-stack parameters just in case even if your function is defined with that signature, even if your function happens to ignore them.
Even though C++ does have decorated symbols, it almost certainly uses C-linkage for main, rather than a clever linker that looks for each one in turn. So it found your wmain and put the parameters onto the call-stack in case it is the int wmain(int, wchar_t*[]) version.
With a little effort of templatizing this, it wold work with any list of objects.
#include <iostream>
#include <string>
#include <vector>
char non_repeating_char(std::string str){
while(str.size() >= 2){
std::vector<size_t> rmlist;
for(size_t i = 1; i < str.size(); i++){
if(str[0] == str[i]) {
rmlist.push_back(i);
}
}
if(rmlist.size()){
size_t s = 0; // Need for terator position adjustment
str.erase(str.begin() + 0);
++s;
for (size_t j : rmlist){
str.erase(str.begin() + (j-s));
++s;
}
continue;
}
return str[0];
}
if(str.size() == 1) return str[0];
else return -1;
}
int main(int argc, char ** args)
{
std::string test = "FabaccdbefafFG";
test = args[1];
char non_repeating = non_repeating_char(test);
Std::cout << non_repeating << '\n';
}