Non-Printable Char to int Conversion Issue in C

Current post is on non-printable character conversion to int issue in C language.

By the topic you will assume, what is this? Even a C student who has just started will know that, there is no harm, as we are upgrading the type from character (char) to integer (int).

However, many people including the advanced programmer do not know is that: “conversion of character having non-printable value to int is machine dependent”.

It means that if you take a variable:

char mynonprintablechar = 129;

And then you try to upgrade to int implicitly, the results which you get, will make you scratch your hair.

Example 1: If - Else Statement

if( mynonprintablechar == 129)
     printf(“You expected me”);

else

     printf(“You may get me”);

Example 2: Switch Statement

switch(mynonprintablechar)
{
    case 129:
        printf(“You expected me”);
        break;
    default:
        printf(“You may get me”);
}

The solution:

1. Have unsigned char as variable type.
2. downcast the int against which you compare to char.
3. cast as unsigned char.

Do drop me a mail, if I saved you from scratching your hair :)

Order the array of 0 & 1

Question:

Order the list with ‘0’ first and ‘1’ second, which has ‘0’ and ‘1’ stored in irregular fashion.

Constraints:
-> Time & Space complexity to be minimal

Example:

Array: 0-1-0-1-1-0-1

Answer: 0-0-0-1-1-1-1

Answer:

-> Start with one pointer from start and another pointer from back.
-> Stop if you find ‘1’ for first pointer
-> Stop if you find ‘0’ for second pointer
-> Interchange till both the pointers are not crossed over.

Complexity:

Time: O(2N) -> O(N) , where N represents longest link list length

Example:

Array: 0-1-0-1

Answer: 0-0-1-1

Tracing:

Step 1:

-> Let Pointer 1 points to head of array
-> Let Pointer 2 points to tail of array

Step 2:

-> Pointer 1 will stop at node 2 as it has '1'

Step 3:

-> Pointer 2 will stop at node 3 as it has '0'

Step 4:

-> Interchange the values of node 2 & 3 with 0 & 1 respectively

Step 5:

-> Pointer 1 crosses over Pointer 2. Stop

Answer:

0-0-1-1

Reach i-th position from tail of link list

Question:

To reach i-th position from tail of link-list by traversing only once.

Example:

1-2-3-4-5-6-7-8-9

To reach 3rd node from tail. i.e Node 7

Answer:

-> Take two pointers.

-> Take first pointer to i-th position from head

-> Start incrementing both the pointers

-> When the first pointer reaches the tail, at that point second pointer will be pointing to i-th node from tail

Example:

1-2-3-4-5-6-7-8-9, to reach 3rd node from tail i.e. Node 7.

Tracing:

Step 1:

-> Pointer 1 & Pointer 2 points to node 1

Step 2:

-> Increment Pointer 1 '3' times, as we want 3rd node from tail

Step 3:

-> Increment both the pointers

Movement of Pointer 1: 3-4-5-6-7-8-9
Movement of Pointer 2: 1-2-3-4-5-6-7

Step 4:

Stop when Pointer 1 points to 9, at that time the Pointer 2 points to Node 7

Answer:

Node 7

Test whether two link list have common nodes or not?

Question:

To test whether two link list have common nodes or not?

Hints:
-> There may be more then 1 common nodes
-> The length of both the link list might not be same

Constraints:
-> Time & Space complexity to be minimal


Example 1:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-6-7

Answer: Common nodes from "Node 6"

Example 2:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-d-e

Answer: No common nodes

Answer:

-> Take two pointers, each pointing to head of each list
-> Take two counter variables for each list to store the length of list

-> Traverse both the list and increment the count on each node traversal
-> Get the length difference by subtracting the counter with higher value to lower one and put to some difference variable

-> Increment the pointer "difference variable" times whose length is greater

-> Increment both the pointers and compare at each iteration, till any node reaches the tail

If they reach tail and both pointers didn't got common value -> No common nodes
Else -> Common nodes from that node

Complexity:

Time: O(2N) -> O(N) , where N represents longest link list length
Space: O(3) -> O(1)

Example 1:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-6-7

Answer: Common nodes from "Node 6"

Tracing:

See the previous post.

Example 2:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-d-e

Answer: No common nodes

Tracing:

Step 1:

-> Let Pointer 1 points to head of List 1 and counter is Length 1
-> Let Pointer 2 points to head of List 2 and counter is Length 2

Step 2:

-> The Length 1 value will be: 7
-> The Length 2 value will be: 5
-> Difference variable value will be: 2

Step 3:

-> List 1 length is bigger, hence Pointer 1 will be incremented two time to point to Node 2

Step 4:

Movement of Pointer 1: 3-4-5-6-7
Movement of Pointer 2: a-b-c-d-e

We reached the tail, hence no common node

Answer:

No common nodes

Find common meet point node of two link lists?

Question:

There are two link list, which meet at some node X,find the meet point?

Hints:
-> There may be more then 1 common nodes
-> The length of both the link list might not be same

Constraints:
-> Time & Space complexity to be minimal

Example:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-6-7

Here "Node 6" is the meet point.

Answer:

-> Take two pointers, each pointing to head of each list
-> Take two counter variables for each list to store the length of list

-> Traverse both the list and increment the count on each node traversal
-> Get the length difference by subtracting the counter with higher value to lower one and put to some difference variable

-> Increment the pointer "difference variable" times whose length is greater

-> Increment both the pointers and compare at each iteration, till any node reaches the tail
-> The point at which both the pointers get same value is the common node meet point

Complexity:

Time: O(2N) -> O(N) , where N represents longest link list length
Space: O(3) -> O(1)

Example:

List 1: 1-2-3-4-5-6-7

List 2: a-b-c-6-7

Here "Node 6" is the meet point.

Tracing:

Step 1:

-> Let Pointer 1 points to head of List 1 and counter is Length 1
-> Let Pointer 2 points to head of List 2 and counter is Length 2

Step 2:

-> The Length 1 value will be: 7
-> The Length 2 value will be: 5
-> Difference variable value will be: 2

Step 3:

-> List 1 length is bigger, hence Pointer 1 will be incremented two time to point to Node 2

Step 4:

Movement of Pointer 1: 3-4-5-6
Movement of Pointer 2: a-b-c-6

Node 6 is the common meet point.

Answer:

Node 6

Difference between "Struct" of "C" & "C++" ?

Difference:

C Struct :
-> Cannot have member functions.
-> Cannot exhibit polymorphic behaviour.
-> "Struct MyStruct a". "Struct" Keyword required while declaration.

C++ Struct:
-> Can have member functions.
-> Can exhibit polymorphic behaviour.
-> "MyStruct a". "Struct" Keyword not required while declaration.




.