Can I store a srand(time(0)) 1 + (rand() % 3) value to a variable?

Question

The title.

Answer 1

Why don't you test it out?

Answer 2

I did it,I tried many ways, but the was an eror bcs srand() is void type or rand() was the same everytime, and i mean somehow to store in a variable the result of the rand()

Answer 3

Error*

Answer 4

#include <iostream>
#include <stdlib.h>
#include <time.h>

using std::cout;
using std::endl;

int main() {
srand(time(0));
int num = rand();
cout << rand() << endl;
return 0;
}

^ That works

Answer 5

Yes you can store a random expression; in this case 1 + rand()% 3, in a variable... I did exactly that in the following code.

https://code.sololearn.com/cV4VWzSHaU1Q/?ref=app

Random expression: 1 + rand()% 3;

Variable: int Winner;