Error in my sorting program - c

Question

#include <stdio.h>
#include <stdlib.h>
struct student{
char initials[2];
int score;
};
void sort(struct student* students, int n){
/*Sort the n students based on their initials*/
int i, j, replace;
for(i = 0; i < n; i++)
{
for(j = 0; j < n; j++)
{
if(students[j] > students[j+1])
{
replace = students[j];
students[j] = students[j+1];
students[j+1] = replace;
}
}
}
}
int main(){
/*Declare an integer n and assign it a value.*/
int n=10;
/*Allocate memory for n students using malloc.*/
struct student* students = (struct student *)malloc(n*sizeof(struct student));
/*Generate random IDs and scores for the n students, using rand().*/
int i;
for(i=0; i<10; i++){
(students + i)->initials[1] = rand( ) % 26 + 'A';
(students + i)->initials[2] = rand( ) % 26 + 'A';
}
/*Print the contents of the array of n students.*/
for(i=0; i<10; i++){
printf("%c%c\n", students[i].initials[1], students[i].initials[2]);
}
/*Pass this array along with n to the sort() function*/
sort(students, n);
/*Print the contents of the array of n students.*/
return 0;
}
I get the following errors when i compile this code,
Program5.c: In function ‘sort’:
Program5.c:23:23: error: invalid operands to binary > (have ‘struct student’ and ‘struct student’)
if(students[j] > students[j+1])
^
Program5.c:25:17: error: invalid operands to binary == (have ‘int’ and ‘struct student’)
replace == students[j];
^
Program5.c:27:23: error: incompatible types when assigning to type ‘struct student’ from type ‘int’
students[j+1] = replace;
Any help would be highly appreciated :)

Answer 1

The first two errors mean that the compiler can't find a > (greater than) operator or a == (equal to) operator that compares a student to a student. The compiler can't just make one up. You need to write your own > and == operators.
The third error means the compiler can't find an assignment operator (=) that takes a student and assigns it to an int. Again, you need to write that operator, because the compiler doesn't know what you want to happen.
You should be able to find the proper syntax for defining these operators by searching for something along the lines of "define c++ == operator" or "define c++ assignment operator".

Answer 2

Remember that arrays in C/C++ are zero-based, and that you're over-writing memory in the initials generation code.
Also watch your array indices in the inner sorting loop; at some point j+1 will be equal to n, and you'll accessing storage that doesn't belong to you.

Answer 3

Your get your first error because the compiler doesn't know how to compare your student structs, it can't tell how to sort them. You must define the operator for this. For example, if you want to order them by score, you can change your student struct to:
struct student {
char initials[2];
int score;
friend bool operator > (const student& lhs, const student& rhs) {
return lhs.score > rhs.score;
}
};
The other two error are caused because you are trying to assign a student struct to an int (students[j] to replace), and vice-versa. Changing the type of replaceto student should fix this problem.

Answer 4

You need to define the following function:
/** #return 0 - if the students are the same
value greater than 0 - if the first student compares greater than the second
value less than 0 - if the second student compares greater than the first
*/
int cmp_student(const struct student* std1, const struct student* std2);
Then in your sort function you can use this function like so:
void sort(struct student* students, int n){
/*Sort the n students based on their initials*/
int i, j, replace;
for(i = 0; i < n; i++)
{
for(j = 0; j < n; j++)
{
if(cmp_student(&students[j], &students[j+1]) < 0)
{
...
}
}
}
}
Note that there is a problem with the second nested loop in that function. What happens when j == (n - 1), what would the value of j + 1 be?
I would also recommend writing a swap function which takes two index positions and the array and swaps the contents of the array at those two positions. Something like this:
void swap(int id1, int id2, struct student* students);