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C/C++ code

static unsigned int gi = 0;
int rand7()
{
   return (((rand() % 5 + 1) + (gi++ % 7)) % 7) + 1;
}

This is a very good one, because it call rand() only one time, and no loop thing, no expends extra memories.

Let me explain it: consider a integer array with size of 5:

1st get one number from 1 2 3 4 5 by rand5
2nd get one number from 2 3 4 5 6
3rd get one number from 3 4 5 6 7
4th get one number from 4 5 6 7 1
5th get one number from 5 6 7 1 2
5th get one number from 6 7 1 2 3
7th get one number from 7 1 2 3 4

So we got the TABLE, each one of 1-7 appears 5 times in it, and has all 35 numbers, so the probability of each number is 5/35=1/7. And next time,

8th get one number from 1 2 3 4 5
9th get one number from 2 3 4 5 6
......

After enough times, we can get the uniform distribution of 1-7.

So, we can allocate a array to restore the five elements of 1-7 by loop-left-shift, and get one number from array each time by rand5. Instead, we can generate the all seven arrays before, and using them circularly. The code is simple also, has many short codes can do this.

But, we can using the properties of % operation, so the table 1-7 rows is equivalent with (rand5 + i) % 7, that is : a = rand() % 5 + 1 is rand5 in C language, b = gi++ % 7 generates all permutations in table above, and 0 - 6 replace 1 - 7 c = (a + b) % 7 + 1, generates 1 - 7 uniformly. Finally, we got this code:

(((rand() % 5 + 1) + (gi++ % 7)) % 7) + 1 

Here is the full code to testing:

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

static unsigned int gi = 0;

//a = rand() % 5 + 1 is rand5 in C language,
//b = gi++ % 7 generates all permutations,
//c = (a + b) % 7 + 1, generates 1 - 7 uniformly.
//Dont forget call srand before rand7
int rand7()
{
   return (((rand() % 5 + 1) + (gi++ % 7)) % 7) + 1;
}

void main(void)
{
    unsigned int result[10] = {0};
    int k;

    srand((unsigned int)time(0)); //initialize the seed

    for (k = 0; k < 100000; k++)
        result[rand7() - 1]++;

    for (k = 0; k < 7; k++)
        printf("%d : %.05f\n", k + 1, (float)result[k]/100000);
}