Generating random numbers in conditions - c

Question

I'm trying to generate three random numbers in three conditions and the three numbers must be from 0 to 100:
an odd number
an even number
a number larger than 50
Here is my code:
#include <stdio.h>
#include <time.h>
int main (void) {
int num1 = 0, num2 = 0, num3 = 0;
srand(time(NULL));
num1 = rand() % 100;
while (num1 % 2 != 0) {
num1 = rand() % 100;
}
num2 = rand() % 100;
while (num2 % 2 == 0) {
num2 = rand() % 100;
}
num3 = rand() % 100;
while (num3 > 50) {
num3 = rand() % 100;
}
printf("your numbers\n%d\n%d\n%d\n", num1, num2, num3);
return 0;
}
the compiler answers me:
warning: implicit declaration of function 'srand' [-Wimplicit-function-declaration]
srand(time(NULL));
and I do have stand in the requirements of the school:
it must be in loops
I cannot use break nor TRUE nor FALSE

Answer 1

Your srand() issue is simply because you haven't included stdlib.h, where that call (and rand() for that matter) is declared.
In any case, there's no real need to use a loop while discarding "invalid" numbers, you can use math for this :-) (a)
Assuming 0 to 100 inclusive:
num1 = rand() % 50 * 2 + 1 // 1, 3, 5, ..., 99
num2 = rand() % 51 * 2 // 0, 2, 4, ..., 100
num3 = rand() % 50 + 51 // 51, 52, ..., 100
For num1, the % gives a value 0..49 which, when doubled and incremented, gives you an odd number in the desired range. The second one is similar but with the range expanded slightly since there are even numbers at both ends. The third simply gives a number 0..49 which maps to 51..100 when 51 is added.
Similar results can be obtained if the range is only half-open (0..99 inclusive):
num1 = rand() % 50 * 2 + 1 // 1, 3, 5, ..., 99
num2 = rand() % 50 * 2 // 0, 2, 4, ..., 98
num3 = rand() % 49 + 51 // 51, 52, ..., 99
If, for some bizarre reason it has to use a loop (despite the inefficiencies), you have some problems with the conditions you use - they're basically all the wrong sense. In other words, you want (for example) the first loop to run while the number is even so that it will eventually produce an odd number.
You should be able to use something like the following. Each block consists of initialising the variable to a value that will force the loop to start then the loop that continues until a value with the desired properties is found:
int num1 = 0; // even forces loop entry
while ((num1 % 2) == 0) // wait for odd
num1 = rand() % 100;
int num2 = 1; // odd forces loop entry
while ((num2 % 2) == 1) // wait for even
num2 = rand() % 100;
int num3 = 1; // 50 or less forces loop entry
while (num3 <= 50) // wait for 51+
num3 = rand() % 100;
All of these limit the range of potential values to 0..99 inclusive, If you want to include 100, simply change the expressions to be rand() % 101.
(a) I'm actually not a fan of these sorts of limitations when set by educators. They're actually teaching inefficient ways to code. It would be far better if it was something that was a lot harder to do with a simple mathematical operation (like ensuring the number was neither 11, 17, 43 nor 97).
I suspect you could find a mathematical way to detect that but it would be far easier just to use a series of conditionals. Now I'm just waiting for someone to show me up by providing the mathematical formula for detecting those four :-)

Answer 2

There are some mistakes in the posted code:
a missing #include <stdlib.h>
the first test should be while (num1 % 2 == 0)
the second test should be while (num2 % 2 != 0)
the third test should be while (num3 <= 50)
Here is a modified version with loops but without looping :)
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main (void) {
int num1, num2, num3;
srand(time(NULL)); /* or better randomness with srand(clock()); */
num1 = 1 + rand() % 50 * 2; /* 1, 3, ..., 99 */
while (num1 % 2 == 0) {
num1 = rand() % 100;
}
num2 = rand() % 50 * 2; /* 0, 2, ..., 98 */
while (num2 % 2 != 0) {
num2 = rand() % 100;
}
num3 = 51 + rand() % 49; /* 51, 52, ..., 99 */
while (num3 <= 50) {
num3 = rand() % 100;
}
printf("your numbers\n%d\n%d\n%d\n", num1, num2, num3);
return 0;
}
For those interested, both gcc and clang detect that the first 2 loops can be eliminated, but not the third as they do not make the assumption that rand() return a non negative number: https://godbolt.org/z/ucpV-y

Answer 3

You need to #include <stdlib.h> to get rid of this warning.
However, there's a bigger problem with your code: you have all the conditions inverted.
This will loop until the number is odd:
while (num1 % 2 != 0) {
This will loop until the number is even:
while (num2 % 2 == 0) {
And this will loop until the number is less than or equal to 50:
while (num3 > 50) {
To fix that, simply invert the conditions:
while (num1 % 2 == 0) { /* until we get an even number */
/* ... */
while (num2 % 2 != 0) { /* until we get an odd number */
/* ... */
while (num3 <= 50) { /* until num3 is greater than 50 */