How to count numbers

[wangerman]

#include <stdio.h>
int main()
{
float a[4]={.02, .05, .055, .01};

int i;
for (i=0; i<=300; i++)
if (a[i]>0.4 && a[i]<0.6) count++
printf("There are %d numbers between .4 and .6\n", some code???);
return 0;
}


I have to count how many numbers there are between .4 and .6 using c++ code and this is how far I got. There are actually 500 numbers in the program that I have to count, but to make things simple I put 4 numbers in there. Can anyone please tell me how to complete the program? Thanks!!

[risby]

#include <stdio.h>

int main()
{
float a[4]={.02, .05, .055, .01};
 
int i;
for (i=0; i<=300; i++)
if (a[i]>0.4 && a[i]<0.6) count++
printf("There are %d numbers between .4 and .6\n", some code???);
return 0;
 

}


I have to count how many numbers there are between .4 and .6 using c++ code and this is how far I got. There are actually 500 numbers in the program that I have to count, but to make things simple I put 4 numbers in there. Can anyone please tell me how to complete the program? Thanks!!

You just have to declare the variable "count" as an integer just as you have with the variable "i" and you must initialize to zero before you start incrementing it in your loop. Then you just reference that variable where you currently have "some code???". Also, why is your loop condition i < 300 when you have 500 numbers to check?

[gasfusion]

printf("There are %d numbers between .4 and .6\n", count);

everything else should work fine

[wangerman]

Thanks everybody for your help! I got the program to work now. And sorry, I meant 300 numbers, not 500. Heres the program in case anyone will need it for their homework!

#include <stdio.h>
int main()
{
float a[300]={300 numbers in here};

int i;
int count=0;
for (i=0; i<=300; i++)
if (a[i]>0.4 && a[i]<0.6) count++;
printf("There are %d numbers between .4 and .6\n", count);
return 0;
}