Question: At line 45, we are printing only one byte of the 32-bit variable, myInt. myInt as defined to be 0x11223344. Why is the last byte shown first?

 

CODE PROVIDED BELOW

 

/* 

 * Module name: debugging_demo_main.c

 * Author: The teacher

 * Date: August 22, 2017, 4:13 PM

 * Description: Demonstration of debugging (there is a bug in the code).

 */

 

#include <stdio.h>

#include <stdlib.h>

#include <string.h>

 

/* Description: main function

 * Receives: Nothing

 * Returns: Exit status

 * Requires: Nothing

 */

int main(void) {

  char *myString = "This is my string!";

  char *buffer;

  int myInt = 0x11223344;

  int strSize;

  int value;

 

  // Get the number of characters in the string

  strSize = strlen(myString);

  printf("The size of myString is %d.n", strSize);

  // Allocate memory for buffer, make it a bit larger than needed

  buffer = malloc(strlen(myString) + 10);

  // Fill the buffer with known values

  value = 0x41;

  for (int i = 0; i < (strlen(myString) + 10); i++) {

    buffer[i] = value++; // Copy value, then increment value

  }

  // Now copy myString

  strncpy(buffer, myString, strSize);

  // Print results

  printf("Original string: %sn", myString);

  printf("Copy of string: %sn", buffer);

  // Print the first byte of myInt

  printf("The first byte of myInt = %hhxn", myInt);

  // Free up the memory we allocated

  free(buffer);

  return (EXIT_SUCCESS);

}