Why won't my C code complie when thier is no error shown?

Question

Ok, so i was working on a code that would calauate the penalties for intoxicated driving having 3-90 days of peanlity for # of peanlites commited and the fine from $200-500. But i wanted to ask why is my code not compliing below? Any explanations on what i got wrong? Because an error wont show up...

//al = alcohol and n = warning number
void process_breathalyzer_results(double al, double n);


#include <stdio.h>

int main() {
process_breathalyzer_results(0.08, 1);  
return 0;
}

void process_breathalyzer_results(double al, double n){

if (al < 0.05){
    printf("PASS\n");
    }else
    
    if ((al >= 0.05) && (al < 0.08)){
         if (n == 0){
             printf("Invaild Input\n");
        }else
         if (n == 1){
             printf("WARNING penalty: $200, driving suspension: 3 
        days\n");
        }else
         if (n == 2){
             printf("WARNING penalty: $300, driving suspension: 7 
        days\n");
        }else
         if (n >= 3){
             printf("WARNING penalty: $400, driving suspension: 30 
         days\n");
         }else
    
              if (al >= 0.08){
                   printf("FAIL penalty: $500, driving suspension: 
         90 days\n");
             
              }
           }

         }

the PASS and WARNING works but the FAIL portion won't work....

Answer 1

Neither the first if nor the second if conditions are evaluated to logical true

if (al <= 0.05){
    printf("PASS\n");
    }else
    
    if ((al > 0.05) && (al < 0.08))

for the passed arguments. As a result inner if statements do not get the control.

So the program outputs nothing.

Answer 2

in process_breathalyzer_results(0.08, 1); change values for the ones that prints something to console. example process_breathalyzer_results(0.06, 2); .