Instead of random, it generates sequence of numbers [closed] - c

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I want to generate x number of random numbers and everything works except it gives me sequenced numbers, not random numbers.
#include <stdio.h>
#include <stdlib.h>
#include <time.h>
int main(){
int x;
printf("How many elements do you want?\n");
scanf("%d", &x);
int *array = (int *)malloc(x * sizeof(int));
srand(time(NULL));
for(int y = 0; y < x; y++){
*(array + y) = 2 + rand() % 99;
}
for(int y = 0; y <x; y++){
printf("Array element %d: %d\n", y, (*array+ y));
}
return 0;}
It gives the numbers like: 27 28 29 30 31, only the first number is random in each execution, others are sequential.

Answer 1

Most likely you're printing a wrong value.
You need to change (*array+ y) to *(array+ y) in printf() statement, otherwise, you end up printing the value of the first random number, plus the iteration counter (which explains the output you get).
Just to add some opinionated view: The array indexing operator [] is there, some chose to prefer using it to avoid mistakes just like this. Your statements can be easily re-written to make use of that, like
for(int y = 0; y < x; y++){
array[y] = 2 + rand() % 99;
}
for(int y = 0; y <x; y++){
printf("Array element %d: %d\n", y, array[y]);
}
That said,
Please see do I cast the result of malloc?
Always check for the success of the called function before using the return value. Check against NULL return of the malloc() call, otherwise you may end up dereferencing NULL, which invokes undefined behavior