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Implemented 16-bit version of the code for more flexibility #99

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65 changes: 32 additions & 33 deletions chapter_2/exercise_2_08/rightrot.c
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Original file line number Diff line number Diff line change
@@ -1,48 +1,47 @@
#include <math.h>
#include <limits.h>
#include <stdio.h>

void printbin(unsigned int x);
unsigned int rightrot(unsigned int x, unsigned int n);
/* Exercise 2.8: Write a function rightrot(x, n) that returns the value
* of the integer x rotated to the right by n bit positions.
*/

int main(void) {
// Using hexadecimal instead of binary literals (0b...) for C89/C99
// compatibility 0xF5 = 0b11110101
unsigned int x = 0xF5;
void to_binary(unsigned short n) {
unsigned long i = 0;
int binary[32];
printf("0b");

printbin(x);
printbin(rightrot(x, 5));
if (n == 0) {
printf("0\n");
return;
}

return 0;
while (i < sizeof(unsigned short) * CHAR_BIT) {
binary[i++] = n % 2;
n = n / 2;
}
for (int j = i - 1; j >= 0; j--) {
printf("%d", binary[j]);
}
printf("\n");
}

void printbin(unsigned int x) {
unsigned int n = sizeof(unsigned int);
unsigned short rightrot(unsigned short x, int n) {
unsigned w = sizeof(unsigned short) * CHAR_BIT;
n %= w;

printf("0b");
return (x >> n) | (x << (w - n));
}
Comment on lines +28 to +33
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@coderabbitai coderabbitai bot Dec 15, 2025

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⚠️ Potential issue | 🔴 Critical

Undefined behavior when n is 0.

When n is 0 (or a multiple of w), line 32 evaluates to (x >> 0) | (x << w). Shifting by the bit width or more is undefined behavior in C.

Apply this diff to fix the issue:

 unsigned short rightrot(unsigned short x, int n) {
     unsigned w = sizeof(unsigned short) * CHAR_BIT;
     n %= w;
-
+    if (n == 0)
+        return x;
     return (x >> n) | (x << (w - n));
 }
🤖 Prompt for AI Agents 
In chapter_2/exercise_2_08/rightrot.c around lines 28 to 33, the code does a
left shift by (w - n) which becomes a shift by w when n is 0 (or a multiple of
w) causing undefined behavior; fix by after computing n %= w, check if n == 0
and simply return x to avoid any shift by w, otherwise perform the existing
rotate expression.


int i;
for (i = n * 8 - 1; i >= 0; --i) {
(x & (unsigned int)pow(2, i)) ? putchar('1') : putchar('0');
}
int main(void) {
unsigned short x;
x = 55;

putchar('\n');
}
printf("x = %d -->", x);
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@coderabbitai coderabbitai bot Dec 15, 2025

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⚠️ Potential issue | 🟡 Minor

Use correct format specifier for unsigned short.

The format specifier %d is for signed integers. For unsigned short, use %hu or cast to unsigned and use %u.

Apply this diff:

-    printf("x = %d -->", x);
+    printf("x = %hu -->", x);
📝 Committable suggestion

‼️ IMPORTANT
Carefully review the code before committing. Ensure that it accurately replaces the highlighted code, contains no missing lines, and has no issues with indentation. Thoroughly test & benchmark the code to ensure it meets the requirements.

Suggested change
printf("x = %d -->", x);
printf("x = %hu -->", x);
🤖 Prompt for AI Agents 
In chapter_2/exercise_2_08/rightrot.c around line 39, the printf uses the signed
integer specifier "%d" for an unsigned short variable x; change the format
specifier to "%hu" (or cast x to unsigned and use "%u") so the value prints
correctly, and update the printf call to use the chosen unsigned specifier.

to_binary(x);

unsigned int rightrot(unsigned int x, unsigned int n) {
unsigned int msb_1 = ~(~(unsigned)0 >> 1);

// Using unsigned int for i to match the unsigned n parameter
// This prevents sign comparison warnings
unsigned int i;
for (i = 0; i < n; ++i) {
if (x & 1) {
x = (x >> 1) | msb_1;
} else {
x = (x >> 1);
}
}
to_binary(rightrot(x, 3));

return x;
return 0;
}

// NOTE: The rightrot function rotate the entire unsigned int var and if we
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