problem with reading program arguments in Visual Studio C++

Question

I'm running C++ program in VS2005, and I've set only one argument in project properties-> debug-> command line args, and it's named profile1.dll for example.

here's a code snippet

cout<<"number of arguments:" << argc<<endl;

for (int i=0; i<argc; i++)
       cout << "argument " << i << ": " << argv[i] << endl;

In the output I get

number of arguments:2
argument 0: c
argument 1: p

don't know why it doesn't print the name of the argument?

My guess is your main method has argv has type char* rather than char**. Can't know for sure without seeing your main method decl. — NG.
– NG., Commented Jul 28, 2010 at 11:39
I think Bob's answer is THE one. You are using UNICODE and 0 is preventing you from printing the whole string. Use UNICODE's functions to print the string. — Pablo Santa Cruz
– Pablo Santa Cruz, Commented Jul 28, 2010 at 11:54

Bob Moore · Accepted Answer · 2010-07-29 07:44:59Z

5

Does the name of your exe start with C? If you expect a string and you only get one character, it's usually because you've discovered that the Western alphabet in UTF-16 Unicode effectively puts a 0 between alternating ANSI chars. Are you compiling for Unicode ?

edited Jul 29, 2010 at 7:44

answered Jul 28, 2010 at 11:40

Bob Moore

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2 Comments

dragan.stepanovic Over a year ago

I'm using Unicode. I changed it to multi-byte, but still the same output

Sjoerd Over a year ago

@Kobac: When using Unicode, you should use std::wcout instead of std::cout.

Vicky · Accepted Answer · 2010-07-28 11:44:46Z

3

argv[0] is the name of your program. argv[1] is the first parameter. It sounds like you have declared the relevant parameter in main() as char* argv rather than char *argv[] or char **argv.

answered Jul 28, 2010 at 11:44

Vicky

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2 Comments

Bob Moore Over a year ago

But the OP already said in response to Pablo's answer that he had declared it as char* argv[]. Something's not right here.

Vicky Over a year ago

@kobac: thanks for accepting my answer, but as pointed out it wasn't the cause of your problem. Can you accept @Bob Moore's answer instead?

MaR · Accepted Answer · 2010-07-28 11:54:46Z

1

Leave TCHAR be, it's fine.

If you compile with unicode, use wcout for output:

int _tmain(int argc, _TCHAR* argv[])
{

  for (int i=0; i<argc; i++)
    wcout << "argument " << i << ": " << argv[i] << endl;

    return 0;
}

Otherwise compile without unicode and your code will just work as-is (and will not work with unicode parameters :o)

You can find setting in "Project Properties/General/Character Set"

answered Jul 28, 2010 at 11:54

MaR

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Comments

Pablo Santa Cruz · Accepted Answer · 2010-07-28 11:38:55Z

0

Can you put the prototype of your main function? What you are doing is apparently fine.

Make sure your main's function prototype is something similar to:

int main(int argc, char **argv)

Hope it helps.

answered Jul 28, 2010 at 11:38

Pablo Santa Cruz

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3 Comments

dragan.stepanovic Over a year ago

int _tmain(int argc, char* argv[])//_TCHAR* so by default it was _TCHAR* for the argv[], but if so, it print's an integer, so i changed it to char*

Sjoerd Over a year ago

@Kobac: change it back to TCHAR, and use std::wcout instead of std::cout

Pablo Santa Cruz Over a year ago

Now I see what's happening. You are using UNICODE and 0 are preventing you from printing the whole String. Check Bob's answer out.

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