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

Question

how do i save int argc, char* argv in to int someting.

i am trying to get the arguments from a test program and save it into int ****;

#include <iostream> 
#include <cstdlib>
using namespace std;

int main(int argc, char* argv[]) {
 int limit = argc;
 cout<< limit <<endl;
 for (int candidate = 2; candidate < limit; candidate++) {
    int total = 1;
    for (int factor = 2; factor * factor < candidate; factor++) {
        if (candidate % factor == 0)
            total += factor + candidate / factor;

    }
    if (total == candidate) {
        cout << candidate << ' ';
    }
}
return 0;
}

and the program is pre-set the arguments is 100, and it just can't save in to int limit

it is looking for the Perfect Numbers the arguments is 100 and it need to print 6, 28 — Poo Po, Jan 09 '23 at 07:47
Read about [main function](https://en.cppreference.com/w/cpp/language/main_function): the `argc` is the number of arguments passed to the program. — heap underrun, Jan 09 '23 at 07:49

score 0 · Accepted Answer · answered Jan 09 '23 at 07:53

0

Something like this

int main(int argc, char* argv[]) {
    if (argc < 2) {
        cerr << "Not enough arguments\n";
        return -1;
    }
    int limit = atoi(argv[1]); // convert first argument to integer
    ...

answered Jan 09 '23 at 07:53

john

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thanks for ans the question, in the note it is using (const int limit = argc > 1 ? atoi(argv[1]) : 1000;) what is the different – Poo Po Jan 09 '23 at 08:01
`atoi` is possibly not the best choice due to lack of error handling – Alan Birtles Jan 09 '23 at 08:03
@PooPo Can you not follow the logic? `argc` is the number of arguments given to the program. In my version if there are not enough arguments I print an error message and quit. In your version if there are not enough arguments a default value of 1000 is used. – john Jan 09 '23 at 08:06

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

1 Answers1