Having trouble finding what is causing this error - c

Question

I am trying to create a program that will output the binary code for 16 numbers. Here is what i have so far:
#include <stdio.h>
#include <stdlib.h>
int i;
int count;
int mask;
int i = 0xF5A2;
int mask = 0x8000;
int main()
{
printf("Hex Value= %x Binary= \n", i);
{
for (count=0; count<15; count++1)
{
if (i&mask)
printf("1\n");
else
printf("0\n");
}
(mask = mask>>1);
}
return 0;
}
The error:
|16|error: expected ')' before numeric constant|
Also let me know if I have any other mistakes, Thanks in advance!

Answer 1

The error is referring to this expression:
count++1
Which makes no sense.
I assume you want:
count++
Making the line
for (count=0; count<15; count++)
You have other strangeness in your code such as:
int i; // Declare an integer named "i"
int mask; // Declare an integer named "mask"
int i = 0xF5A2; // Declare another integer also named "i". Did you forget about the first one???
int mask = 0x8000; // Did you forget you already declared an integer named "mask"?
printf("Hex Value= %x Binary= \n", i);
{
[...]
} // Why did you put a bracket-scope under a PRINTF call?
// Scopes typically follow loops and if-statements!
(mask = mask>>1); // Why put parens around a plain-old expression??
After fixing weirdness in your code, it should look like:
#include <stdio.h>
#include <stdlib.h>
int main()
{
int i = 0xF5A2;
int mask = 0x8000;
printf("Hex Value= %x Binary= \n", i);
for (int count=0; count<15; ++count, mask>>=1)
{
printf("%d\n", (i&mask)? 1 : 0);
}
return 0;
}