C Q&A #4: Calling main recursively

What do you think about the following C Code ? What will be the output.

#include <stdio.h>

int main (void)
  static int a = 3;

  if (a == 0)
    return 0;

  printf ("going down\n");

  main ();

  printf ("coming up\n");

  return 0;

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Detect Endianness of a System

In a previous post “Little and Big Endian conversion” i have briefly discussed about the Big-Endian and the Little-Endian representations. It is the ordering of the bytes (elementary addressable elements) within the representation of a larger basic data type. These are the two byte orderings in the memory within a larger datatype. For example if the size of integer is 4 bytes in a system, then will the least significant byte be stored in the lower memory address or in the higher memory address. In short: if the most significant byte is stored in higher memory address then it is called the Little Endian representation, and if the most significant byte is stored in lower memory address than the least significant byte then it is called the Big-Endian representation. For a diagram see the post “Little and Big Endian conversion”. Also refer the Wikipedia Entry on Endianness.

So how to determine if your system a Little Endian system or a Big Endian system by running a piece of code? Continue reading “Detect Endianness of a System”

C Q&A #3: When ((x == x + 5) && (y != y)) is true.

The post heading gives an expression which is a contradiction and not possible. In general yes it is a contradiction, but can you write a code such that the below code prints “Hello World.” ?

/* Add source code in the following code such that "Hello World." is printed. */

#include <stdio.h>
#include <math.h>

int main (void)

  if (x == x + 5)
    printf ("Hello ");
  if (y != y)
    printf ("World.\n");

  return 0;

Continue reading “C Q&A #3: When ((x == x + 5) && (y != y)) is true.”

Get sorted index orderting of an array

Yesterday i was translating some code i wrote in R to C++. I had some calculation to do which required the list of index of the top n values in a list. Therefore what i needed was not a the list sorted itself or get a sorted copy of a given list, but actually the sorted order of the index into the actual array. To describe the problem, for example consider the following list:

arr = 20 50 40 80 10

The sorted (non-decreasing) ordering of the values is obviously

sorted_arr = 10 20 40 50 80

The sorted (non-decreasing) ordering of the index into the original array is (index starts at 1 in this example)

sorted_index_into_arr = 5 1 3 2 4

Therefore the smallest value in the list arr could be found by indexing into arr using the first value of sorted_index_into_arr, which stores the index of the array arr holding the smallest value.

arr[ sorted_index_into_arr[1] ]

The sorted index ordering is very easy to get in languages like R and Octave or Matlab. For example in R we can do the following to get the sorted index order using the order function:

> arr <- c (20, 50, 40, 80, 10)
> arr
[1] 20 50 40 80 10
> order (arr)
[1] 5 1 3 2 4

In the case of Octave or Matlab you can get the index order using the sort function, which will return two lists (1xn matrix), the first list is the sorted array itself, and the next one is the sorted index order into the original array.

octave:1> a = [20 50 40 80 10]
a =

   20   50   40   80   10

octave:2> [sorted index] = sort (a)
sorted =

   10   20   40   50   80

index =

   5   1   3   2   4

Although these functional languages provides gives these features in C/C++ it is not immediately available using the builtin sort library functions. But the sorting routines accepts a function pointer to a comparison function, writing appropriate code for which will do the trick.
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C Q&A #2: Watch your shift!

What do you this should be the output of the below code? Is there any problem in the code?

#include <stdio.h>

int main(void)
  int x = -1, y = 1, a, b, c, d, e, f;

  a = x << 4;
  b = x >> 4;
  c = y << 4;
  d = y >> 4;
  e = y << sizeof (int) * 8;
  f = y << -4;

  printf("a = %x \nb = %x \nc = %x \nd = %x \ne = %x \nf = %x",a, b, c, d, e, f);

  printf ("\n");
  return 0;

Continue reading “C Q&A #2: Watch your shift!”

Find Sum (i … j) in a list

The problem is to sum the ith to jth element of a given list. The easiest solution runs in \mathcal{O}(n) time, which starts are the ith index and adds up the numbers in the list until it reaches the jth index. In this post i will show the process of calculating the sum of elements i through j both inclusive (where i \le j), which takes \mathcal{O}(1) time for finding the sum, and \mathcal{O}(n) time and space for preprocessing the given list of length n.
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Find a String in an Alphabet Grid

Everybody of us has definitely played the word search game in which a grid of characters are given, and a list of words are given which are to be found in the grid horizontally, vertically or diagonally. In this post i will show a simple solution to a more generalized version of the search which finds the strings in the grid with backtracking.
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