Why doesn't this print a random number?

Question

What is the problem with this code?

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

void createNumb(int RandomNumber) {

    srand(time(NULL));

    int a, b, c, d;

    a = rand() % 10 + 0;
    b = rand() % 10 + 0;
    c = rand() % 10 + 0;
    d = rand() % 10 + 0;

    while (b == a) {
        b = rand() % 10 + 0;        
    }

    while (c == b || c == a) {
        c = rand() % 10 + 0;     
    }

    while (d == c || d == b || d == a) {
        d = rand() % 10 + 0;
    }

    RandomNumber = ("%d%d%d%d", a * 1000 + b * 100 + c * 10 + d * 1);   
}

int main() {   
    int x;
    createNumb(x);
    printf("%d", x);
}

When I run this program, it always give me the result 0.

I want it to print a random 4-digit number (each digit must be unique). What is the problem?

Answer 1

• Your createNumb function returns void instead of an int.
• You should use srand() only once, so it’s better move it to your main.

Your createNumb function should return a random integer, like this:

int createNumb(void) {
    /*
    ** Simplified for readability.
    ** Here you would use your function that generates 4 different digits
    */
    return rand();
}

int main(void) {
    srand(time(NULL))(
    int x = createNumb();
    printf("%d" , x);
}

Or, if you want createNumb to fill an existing integer, use a pointer:

void createNumb(int *randomNumber) {
    /*
    ** Simplified for readability.
    ** Here you would use your function that generates 4 different digits
    */
    *randomNumber = rand();
}

int main(void) {
    int x;
    srand(time(NULL))
    createNumb(&x); // Passing a pointer to x
    printf("%d" , x);
}

Answer 2

There are some problems in your C code, coming from a different language:

• use return to return a value from a function and specify the return type before the function name`.
• our attempt at formating the number is incorrect and unneeded. Incidentally, RandomNumber = ("%d%d%d%d", a * 1000 + b * 100 + c * 10 + d * 1); uses the operator , that ignores its left operand and evaluates to the right operand, whose value is stored into a local variable with automatic storage duration, and is lost upon function exit. Just use return a * 1000 + b * 100 + c * 10 + d * 1;.
• to compute a 4 digit pseudo-random value, you could just use rand() % 10000, but an approach such as your may be needed for more digits or for specific constraints such as unique digits.
• to make your results more random, use a faster changing source for srand() such as clock() and initialize the pseudo-random number generator just once int the main() function.
• you could use the printf format "%04d" to make display an initial 0 if the result is less than 1000.

Here is a modified version:

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

int createNumb(void) {
    int a, b, c, d;

    a = rand() % 10;
    b = rand() % 10;
    c = rand() % 10;
    d = rand() % 10;

    while (b == a) {
        b = rand() % 10;        
    }

    while (c == b || c == a) {
        c = rand() % 10;     
    }

    while (d == c || d == b || d == a) {
        d = rand() % 10;
    }

    return a * 1000 + b * 100 + c * 10 + d;   
}

int main() {
    int x;
    srand(clock());
    x = createNumb();
    printf("%04d\n", x);
    return 0;
}

Note that you can avoid the loops in creatNumb(). Here is a modified version that takes an argument from the command line to produce multiple results.

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

int createNumb(void) {
    int a, b, c, d, bb, cc, dd;

    a = rand() % 10;
    b = rand() % 9;
    c = rand() % 8;
    d = rand() % 7;

    bb = (b >= a);
    cc = (c >= a) + (c >= b);
    cd = (d >= a) + (d >= b) + (d >= c);

    return a * 1000 + (b + bb) * 100 + (c + cc) * 10 + (d + dd);
}

int main(int argc, char *argv[]) {
    int x, n;
    srand(clock());
    n = (argc > 1) ? strtol(argv[1], NULL, 0) : 1;
    while (n-- > 0) {
        x = createNumb();
        printf("%04d\n", x);
    }
    return 0;
}

Boolean operators evaluate to 1 when they are true and 0 otherwise. bb = (b >= a); is a short notation for:

bb = 0;
if (b >= a)
    bb = 1;

If you want to compute an array of 4 numbers with the given constraint, you should pass a pointer to the destination array as an argument to creatNumb(). Here is a modified version with this approach:

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

void createNumb(int *dest) {
    int a, b, c, d, bb, cc, dd;

    a = rand() % 10;
    b = rand() % 9;
    c = rand() % 8;
    d = rand() % 7;

    bb = (b >= a);
    cc = (c >= a) + (c >= b);
    cd = (d >= a) + (d >= b) + (d >= c);

    dest[0] = a;
    dest[1] = b + bb;
    dest[2] = c + cc;
    dest[3] = d + dd;
}

int main(int argc, char *argv[]) {
    int array[4];
    srand(clock());
    createNumb(array);
    printf("%d%d%d%d\n", array[0], array[1], array[2], array[3]);
    return 0;
}

Note that passing an array as a argument to a function actually passes a pointer to its first element: createNumb(array); is equivalent to createNumb(&array[0]);. This process is referred to as arrays decay into pointers in most expression contexts, it is somewhat confusing for beginners but allows for efficient argument passing.