random number generator that doesn't repeat it self (cpp/c++) - c++

Question

I want to make a function that generates numbers but doesn't repeat it self. If every number is generated the array can be emptied and it can start over again.
This is the code I made but it doesn't work.
The comments in the code explains the code a little.
The largest number that is allowed is "howManyWords".
This is used to display words which are stored in an array
I want to use it like this: array\[random()\]
#include <stdio.h>
#include <iostream>
#include <string>
#include <stdlib.h>
using namespace std;
//public scope
int howManyWords; // how many words you have enter
int random(){
int random;
int numbers[howManyWords];
srand(time(0)); //changing the algorithm
random = rand() % howManyWords;
numbers[random] = random; // store the digit in the position in the array equal to the digit that is generated
for(int i=0; i<howManyWords; i++){ // going through every element in the array
if(numbers[i] == random){ // if the number is already generated, generate a different number
random = rand() % howManyWords;
}
}
return random;
}

Answer 1

Rather than your function, which discards the state of which numbers it has returned each time it is called, you should use a function object.
struct random_t {
random_t(int max) : values(max), current(max) {
std::iota(values.begin(), values.end(), 0);
}
template<typename URBG = std::random_device &>
int operator()(URBG&& urbg = default_random) {
if (current == values.size()) {
shuffle(std::forward<URBG>(urbg));
}
return values[current++];
}
private:
template<typename URBG>
void shuffle(URBG&& urbg) {
std::shuffle(values.begin(), values.end(), std::forward<URBG>(urbg));
current = 0;
}
std::vector<int> values;
std::vector<int>::size_type current;
static thread_local std::random_device default_random;
};
See it live