请帮助，使用％echo参数

Question

I have this problem of how to use echo to supplement input and tell the program how many output to give.

Basically, I have the program logic sorted out, which is to randomly select a subset of x number from y inputs. My code is here:

The problem is, how can I forward the argc variable in the main argument to the program execution??

The context requires me to call: %echo ABCDEFG | Subset 2, which suppose to print 2 characters at random. But I can't do this, the number 2 seems cannot be forwarded inside here. And the call of %echo command does not seem to work as well. Anybody please help

int main(int argc){
RandomizedQueue<char> q;
char input;


while(cin.peek() != '\n'){
    cin >> input;
    q.enqueue(input);
}

for(int k = 0; k < argc; k++)
    cout << q.dequeue() << endl;

return 0;

}

Answer 1

I think the % in %echo is not something you're supposed to type, but rather signifies the command prompt itself. So you type this:

echo A B C D E F G | ./Subset 2

Assuming your program is named Subset and is in the current working directory. Then, Subset will receive argc==2 and argv=={"./Subset", "2"}. You'll need to declare main in the standard way (you're missing argv):

int main(int argc, char* argv[])

Once you do that, argv[1] will be "2" as a string (the program's first argument). You can do something like this to convert it to an int:

int count = atoi(argv[1]);

You should read up on the declaration of main, because argc is not the argument as a number, it is the number of arguments. And declaring main with argc but not argv is highly atypical.