Dealing with char ** argv

Question

How do I assign a sequence of character strings to a char ** argv variable in a program? Its a command line argument. I'm currently trying to convert an .exe app to a dll.

For example:

{ "string1", "string2", "string3" } --- > char ** argv variable

My problem is somehow realted to this: How does an array of pointers to pointers work? but I can't get it to work using the snippet shown there. Help!

Are you trying to assign the three command-line arguments to three separate char* variables in your program, or are you trying to assign three variables into a char** you have declared yourself? — Paul Stephenson, Jan 12 '10 at 08:42
I'm trying to assign 3 strings to the char** I have declared. I might also reassigned a different sequence of strings on the later part of the program. What is the correct way to do this? — Rock, Jan 12 '10 at 08:51
Why on Earth would you want to modify the passed-in parameters? — John Dibling, Jan 12 '10 at 15:37

Andreas Brinck · Answer 1 · 2010-01-12T08:51:39.887

8

const char* argv[] = {"string1", "string2", "string3", 0};

If the arguments aren't compile time constants I would do something like:

std::vector<const char*> arguments;
arguments.push_back(somePointer);
arguments.push_back(someOtherPointer);
arguments.push_back(0);
const char** argv = &arguments[0];

EDIT: Using PaxDiablos information that an argv-array should be null terminated.

edited Jan 12 '10 at 08:51

answered Jan 12 '10 at 08:41

Andreas Brinck

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@Andreas, if it's a true argv (as in what main expects), it should have a fourth pointer set to NULL. That may not, however, be part of the requirements for this situation. – paxdiablo Jan 12 '10 at 08:54
That's better, deleting mine since it adds nothing new now (and because it uses malloc - ugh!). – paxdiablo Jan 12 '10 at 09:00
I can't remember how they're stored internally, but in this example would std::vector be usable? – Mr. Boy Jan 12 '10 at 11:44
@John No, there's no way to directly pass a `std::vector` to a function expecting a `char**`. – Andreas Brinck Jan 12 '10 at 13:20

score 0 · Answer 2 · answered Jan 12 '10 at 09:37

0

what about getopt?

answered Jan 12 '10 at 09:37

teZeriusz

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score 0 · Answer 3 · answered Jan 12 '10 at 11:25

Note that Andreas Brincks answer is really what you want to do, where vector does all the heavy lifting of allocation and provides exception safety. I strongly suggest that you look into changing whatever reason there is for not being able to use vector. But if you really really cannot do so, I guess you could do something along the lines of the below code:

int numArgs = YOUR_NUMBER_HERE;

char **myArgv = new char*[numArgs+1];

for (int i=0; i<numArgs; ++i) {
    myArgv[i] = new char[at least size of your argument + 1];
    strncpy(myArgv[i], whereever your string is, buffer size);
    myArgv[buffer size] = '\0';
}
myArgv[numArgs] = NULL;


// use myArgv here


// now you need to manually free the allocated memory
for (int i=0; i<numArgs; ++i) {
    delete [] myArgv[i];
}
delete [] myArgv;

Dealing with char ** argv

3 Answers3