在main（）内传递C ++命令行参数，但执行相同的工作

Question

Is there a way to keep the command line argument as standard and yet pass the argv internally inside the main?

Like, change this:

int main(int argc, char **argv) {
    App app(argc,argv);
    return app.exec();
}

to something like this:

int main(int argc, char **argv) {
    vector<string> argv ="arguments";
    int argc = X;
    App app(argc,argv);
    return app.exec();
}

When I have:

./a.out -a A -b B -c C

I know this is weird. I just want to find a workaround not to change the source code of a huge project and just run my command line arguments every time with only ./a.out .

Answer 1

Put each of your arguments in a char array, and then put pointers to those arrays into an array of pointers.

char arg1[] = "./a.out";
... 
char argN[] = "whatever";
char* argv[] = { arg1, ..., argN}

App app(N, argv);

Answer 2

You may be looking for

const char *myArgv[]={"-a","A","-b","B"};
int myArgc=4;
App app(myArgc,myArgv);
return app.exec();

Answer 3

std::vector<std::string> args(argv, argv+argc);

Answer 4

You may rename arguments passed to app as you wish to avoid a name conflict, for ex.

int main(int argc, char **argv) {
    vector<string> app_argv = /* contents */;
    int app_argc = app_argv.size() ;
    App app(app_argc, app_argv);
    return app.exec();
}