Thread: help with a simple programe :)

hi everybody
here is the question

write a program using functios to find out the average of three marks(m1,m2,m3)of n number of students , the function should read the 3 marks and calculate the average and anther function to take the average and print the grade depending on the following category .
the main call the two functions

60-70 pass
71-80 good
81-90 v.good
91-100 excellent

this is what i've wrote so far

Code:

#include <stdio.h>
#include <stdlib.h>
double avg (int m1 ,int m2, int m3);
void cate (void);
int main()
{
  int m1,m2,m3 ;
    printf("Enter the first mark\n");
    scanf("%d",&m1);
    printf("Enter the second mark\n");
    scanf("%d",&m2);
    printf("enter the third mark\n");
    scanf("%d",&m3);
    printf("the average is %f\n",avg(m1,m2,m3));
    cate();
    return 0;
}
double avg (int m1 ,int m2,int m3)
{
    return((m1+m2+m3)/3);
}
void cate (void)
{
    double avg ();
 if ( avg >=60 && avg <=70){
 printf("pass");
 }
 else if ( avg >=71 &&  avg  <= 80){
 printf ("good");
 }
 else if ( avg  >=81 &&  avg <= 90){
 printf("v.good");
 }
 else
 printf("excellent");
 }

the problem is that i keep taking this messege

in function 'cate'
warning: a comparison between a pointer and integer (line 25)
warning: a comparison between a pointer and integer (line 25)
warning: a comparison between a pointer and integer (line 28)
warning: a comparison between a pointer and integer (line 28)
warning: a comparison between a pointer and integer (line 29)

so any help about this?? please

Because you have the following line in your cate function:

Code:

double avg ();

Which makes avg effectively a pointer to the function avg above. Try passing the result of avg into cate and using that to produce your output:

Code:

int main(void)
{
    double a;
    ...
    a = avg(m1, m2, m3);
    printf("the average is %f\n",a);
    cate(a);
    return 0;
}

You don't pass a into cate like I suggested, so the avg variable in there is uninitialized and has who knows what value in it.

when i pass (a)into cate functions ; many errors ocurrs :S

As opposed to the many errors you have now.

You need to pass the average number into the cate function; that's not an option. You will need to fix whatever errors you have when you do that (i.e., make the function prototype/declaration correct).

Code:

void cate ( double );
...
cate ( a );
...
void cate ( double avg)
{
 if ( avg >=60 && avg <=70){
...

When I read the functions in the example program, I'm not seeing that they are reading right. Functions are just like a formula for an algebra problem when you're working word problems.

I suggest reading functions and writing functions in a pattern similar to this:

When declaring that a function exists, read as:

Code:

" . . . answer type formula name (answer type variable 1, answer type variable 2)

When getting prepared to call a function, read the need for that function as:

• formula to be used
• value to go in to the formula
• value to come out of the formula
• e.g. formula-name (x,y)

When calling a function, read the code as:

Code:

" . . . formula-name (apply to variable 1, apply to variable 2) . . . "

When defining or describing a function, read the code as:

Code:

" . . . answer type formula-name (variable 1, variable 2){
    declare 
    work variable 1 against some math
   return an answer to fit variable 2
    }
    . . ."

Notice that in and around the function, the names would be so that "apply to variable 1" and "variable 1" would be similar, but not the same; this would help keep the reading of the program smoother.

In the example program, I see that the functions are there, but I don't read the code as though they are a part of the program. The whole program should be able to be read aloud as a set of instructions.

If you cannot read the program aloud so that it makes sense, and it's breaking, then that spot where things stop flowing: that's the kink in the logic.

I apologize if that was confusing, but that is how I think about it.

Code:

return((m1+m2+m3)/3);

... this won't return the correct result since all your variables are ints. Use ((m1+m2+m3)/3.0.