c - Having trouble finding what is causing this error -

i trying create program output binary code 16 numbers. here have far:

#include <stdio.h> #include <stdlib.h> int i; int count; int mask;  int = 0xf5a2; int mask = 0x8000; int main() { printf("hex value= %x binary= \n", i); {     (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 have other mistakes, in advance!

the error referring expression:

count++1 

which makes no sense.

i assume want:

count++  

making line

for (count=0; count<15; count++) 

---

you have other strangeness in code such as:

int i;              // declare integer named "i" int mask;           // declare integer named "mask"  int = 0xf5a2;     // declare integer named "i".  did forget first one??? int mask = 0x8000;  // did forget declared integer named "mask"? 

---

printf("hex value= %x binary= \n", i); {     [...] }  // why did put bracket-scope under printf call?    // scopes typically follow loops , if-statements! 

---

 (mask = mask>>1);  // why put parens around plain-old expression?? 

---

after fixing weirdness in code, should like:

#include <stdio.h> #include <stdlib.h>  int main() {     int = 0xf5a2;     int mask = 0x8000;      printf("hex value= %x binary= \n", i);     (int count=0; count<15; ++count, mask>>=1)     {         printf("%d\n", (i&mask)? 1 : 0);     }     return 0; }