How to generate an array of 256 distinct numbers

Question

I have this:

#include     
using namespace std;   
int main()
{
    int a[256];
    int b;
    int k;
    for (int i = 0; i < 256; i ++){
             


        

Answer 1

std::random_shuffle is the way to go, as previously mentioned, but just in case you don't want to use it (maybe using ANSI C instead of C++), here's a quick-and-dirty implementation:

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

#define SIZE 256

static inline void
swap(int *a, int *b) {
    // Don't swap them if they happen to be the same element 
    // in the array, otherwise it'd be zeroed out
    if (a != b) {
        *a ^= *b;
        *b ^= *a;
        *a ^= *b;
    }
}

int main(void)
{
    int A[SIZE], i;
    // Initialize array with sequential incrementing numbers
    for (i = 0; i < SIZE; ++i)
        A[i] = i;

    // Initialize random seed
    srand(time(NULL));

    // Swap every element of the array with another random element
    for (i = 0; i < SIZE; ++i)
        swap(&A[i], &A[rand() % SIZE]);

    return 0;
}

Answer 2

#include <iostream>
#include <vector>
#include <algorithm>

int main(int argc, const char * argv[])
{
    std::vector<int> items(256);

    std::iota(items.begin(),items.end(),0);

    std::random_shuffle(items.begin(), items.end());

    for(auto i:items)
        std::cout<<i<<"  ";
}

Answer 3

As others mentioned, use std::random_shuffle:

std::vector<int> my_vec(256); //Reserve space for 256 numbers in advance.

for(int n = 0; n < 256; ++n)
{
  my_vec.push_back(n);
}

std::random_shuffle(my_vec.begin(), my_vec.end());

Answer 4

You could try something like this:

int main()
{
    std::vector<int> available(256);
    int a[256];

    for (int i = 0; i < 256; ++i)
        available.push_back(i);

    for (int i = 0; i < 256; ++i)
    {
        int idx = rand() % available.size();
        a[i] = available[idx];
        available.erase(available.begin()+idx);
    }

    // use a[] as needed...

    return 0;
}