C error: pointer expected - c

Question

I made a simple program in C that finds if a number is prime. I am new to C and decided to try to use scanf instead of hardcoded numbers to check. When I run my code:
#include <stdio.h>
typedef int bool;
#define true 1
#define false 0
int main(){
//I am going to check if the "checking" variable is a prime
int checking;
scanf("%d",checking);
//"stopper" is where the for loop will stop
int stopper = checking**0.5 + 1;
//"found" will show if I have found something bad
bool found = false;
for (int i = 2; i < stopper; i++)
{
if (checking % i == 0){found = true; break;}
}
if (!found) {printf("it is prime");}
else {printf("it is not prime");}
}
when I try to compile this with TCC it gives the error (primes.c is the name of the document)
primes.c:12 error: pointer expected
I don't know how to fix this.
EDIT: I just made stopper = checking/2 and the program crashes

Answer 1

int stopper = checking**0.5 + 1;
line 12... what do you expect the ** operator to do?
A * typically performs a multiply, or dereferences a pointer.
The compiler could be interpreting it as follows:
int stopper = checking * (*0.5) + 1;
Of course, trying to dereference a float (*0.5) is bad / impossible, hence the error.
Did you mean:
instead of **, you meant * (multiply)
instead of ** (not a C operator), you meant pow() (raise to the power of)
You also need to be specific - even if you're an expert at precedance, the reader may not be, and you may well be wrong.
If you're not sure what's going on, use braces to be specific, which of the following do you mean?
int stopper = checking * (0.5 + 1);
int stopper = (checking * 0.5) + 1;
int stopper = pow(checking, 0.5) + 1;
int stopper = pow(checking, 0.5 + 1);
If you are indeed after the square root, then as #JeremyP says, invert your thinking - multiply is much less costly than pow():
for (int i = 2; i * i <= checking; i++)

Answer 2

You have two problems in your program:
(1) Replace
int stopper = checking**0.5 + 1;
with
int stopper = checking*0.5 + 1;
(2) Replace
scanf("%d",checking);
with
scanf("%d",&checking);
You should be good to after these corrections.

Answer 3

There's no built in operator to raise a number to the power of another number. There is a function to do that - also a square root function but you don't need them (see below).
x**y
is parsed as
x* (*y)
which is where your pointer error comes from
Instead of trying to find the square root, come at it from the other direction, change your for loop like this
for (int i = 2; i * i <= checking; i++)
Also, as others have said, the scanf is wrong.
scanf("%d",&checking);
// ^- scanf needs a pointer to an int.

Answer 4

Firstly, the scanf is wrong.
Use scanf("%d",&checking);
For finding the square root use math.h library for sqrt function.
int stopper = sqrt(checking) + 1;
There were a lot of typing mistakes in your code. Please, correct your syntax before asking a question.
Ideone link for code