integers displayed instead of a sorted string - c

Question

i want to sort a string using counting sort but instead of a sorted string some integer values are displayed with a warning iteration 254u invokes undefined behavior pointing at the expression c[i]+=c[i-1] even though the iterations doesn't seem to exceed the signed int limit
the code is
#include <stdio.h>
#include <stdlib.h>
#include<string.h>
#define range 255
void count_sort(char arr[],int n)
{
int c[range];
int i;
memset(c,0,sizeof(c));
char b[strlen(arr)];
for(i=0;i<n;i++)
{
++c[arr[i]];
}
for(i=1;i<=range;++i)
{
c[i]+=c[i-1];
}
for(i=n-1;i>=0;--i)
{
b[c[arr[i]]-1]=arr[i];
--c[arr[i]];
}
for(i=0;i<n;i++)
{
arr[i]=b[i];
}
}
void print(char arr[],int n)
{
int i;
for(i=0;i<n;i++)
{
printf("%d ",arr[i]);
}
}
int main()
{
char arr[]="november";
int n=sizeof(arr)/sizeof(arr[0]);
count_sort(arr,n);
print(arr,n);
return 0;
}
what changes are required to make the code run correctly

Answer 1

I checked you here link. The code is correct except for the lines below
(1) printf("%d ",arr[i]);
Change this line to the following since you are printing characters and not integers.
printf("%c ",arr[i]);
If you keep %d then you are casting the char to an integer, and you'll get the ascii value. So for %d specifier, you will get a sorted array of integer values which are ascii values for the corresponding characters.
(2) int c[range];
Change the line to the following
int c[range+1];
Now in your second for loop, it will not exceed the bounds of the array c and you will not get the undefined behavior.
Hope this will help you.

Answer 2

Here are some points that I'd like to share:
You're using %d as format specifier in printf() in print function so change that to %c like this:
printf("%c ", arr[i]);
If your string is properly \0 terminated and you can do without printing space between characters then you can directly use printf() with %s it:
printf("%s", arr);
You don't need the custom print() function in this case.
This loop
for ( int i = 1; i <= range; ++i ) // <== out-of-bounds
{
c[i] += c[i-1];
}
is causing out-of-bounds access resulting in Undefined Behavior. As range represents 255 but the loop condition is i <= range whereas it should be i < range to keep the iterations in the valid range i.e. 0-254.
You are calculating the length of the string like this:
int n = sizeof(arr) / sizeof(arr[0]);
sizeof also calculates the terminating '\0' character so the length of the string will be:
sizeof( arr ) = 9
sizeof( arr[0] ) = 1
n = 9 / 1 = 9
But, the length of the string november is 8. strlen() would be a better choice here.
int n = strlen( arr );
Or,
int n = sizeof( arr ) - 1; // don't include `\0`
In the count_sort function, you're declaring the arrayc like this:
int c[range];
And, then using memset to set it. You could initialize it like this:
int c[range] = {0};
More on memset here.
The array b is declared like this:
char b[strlen(arr)];
^^^^^^^^^^^
But, you already have n, use that. You don't need to calculate the length again.

Answer 3

You should use %c instead of %d to print the characters.
You should also use for( i=1; i < range; i++ ), i <= range will give fault as last index of array is range-1.

Answer 4

there are some problem in you code first you are using wrong format specifier for printing chars.You have to use %c instead of %d in your printing function.
also you should remove = from this loop for(i=1;i<=range;++i) , otherwise you will pass boundaries of array c which will lead to undefined behavior , or if you need that element you should declare c[range+1] instead of c[range].
also note that in your sorting since ASCII code of \0 is smallest , it will become first element of your array and so neither b nor arr will be terminated.
I suggest this loop:
for (i = n - 2; i >= 0; --i)
{
b[c[arr[i]] - 2] = arr[i];
--c[arr[i]];
}
b[n-1] = '\0';