CODEWITHSUNDEEP
cpointersgcctime
SHLOK SAXENA
SHLOK SAXENA

Reputation: 11

Why is that in c language when changing time in srand(time(0)); to srand(time(1)); does not give output

#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main() {
    int n, m;
    printf("Enter the lower bound (n): ");
    scanf("%d", &n);
    printf("Enter the upper bound (m): ");
    scanf("%d", &m);
    if (n > m)
    {
        printf("Invalid range. Ensure that n <= m.\n");
        return 1;
    }
    srand(time(0)); //(here)
    int randomInRange = (rand() % (m - n + 1)) + n;
    printf("Random number between %d and %d: %d\n", n, m, randomInRange);
    return 0;
}

when i tried changing the value in srand(time(0)) to srand(time(1)) the code runs, takes the user input of n and m but in the end it doesn't give any output. Why is this happening?

Upvotes: -4

Views: 83

Answers (2)

omni-drft
omni-drft

Reputation: 21

The time function, actually expects a pointer as an argument. Passing 0 is the same thing as passing NULL. That's why you see people use either srand(time(0) or srand(time(NULL). Passing 1 as an argument causes the program inproperly, because 1 is not a valid memory address (pointer).

Upvotes: 1

Sourav Ghosh
Sourav Ghosh

Reputation: 134316

To make is superfluous, the call

srand(time(0));

can also be written as

srand(time((void *)0));

which is equivalent to

srand(time(NULL));

In other words, the 0 here holds a special meaning, it is not considered as an integer argument, it is passed to the function as a null pointer constant. Changing that to integer value 1 breaks the function call (as opposed to the function signature, see the man page).

Further reading: Chapter 6.3.2.3, Pointers:

An integer constant expression with the value 0, or such an expression cast to type void *, is called a null pointer constant. ...

Upvotes: 1

Related Questions