As i is with the length of string and hence in array rev the contents are copied as "\0 NOITANIMAXE C". But while displaying those contents with puts function it stops because of the first character '\0' and hence no output.

Concatenation means combining two strings together (appending the second at the end of first one).

printf(5+"Good Morning\n"); It skips the 5 characters and prints the given string.
Hence the output is "Morning".

A string is a collection of characters terminated by '\0'.

>> char str[] = "Exam\0\Veda\0"; The variable str is declared as an array of characters and initialized with value "Exam".

>> printf("%s", str); It prints the value of the str.
The output of the program is "Exam".

Array index always starts with 0.

strcmp(s1, s2);
returns int as follows:
== 0, when s1 == s2
< 0, when s1 < s2
> 0, when s1 > s2

In the fourth location of an array it contains 0 since default initialization will take place.

Declaration: strcmp(const char *s1, const char*s2);

The strcmp return an int value that is
if s1 < s2 returns a value < 0
if s1 == s2 returns 0
if s1 > s2 returns a value > 0

Declaration: char *strrchr(const char *s, int c);

It scans a string s in the reverse direction, looking for a specific character c.

The function strstr() Finds the first occurrence of a substring in another string
Declaration: char *strstr(const char *s1, const char *s2);

Return Value:
On success, strstr returns a pointer to the element in s1 where s2 begins (points to s2 in s1).
On error (if s2 does not occur in s1), strstr returns null.

gets(); collects a string of characters terminated by a new line from the standard input stream stdin

When an array is partially initialized at the time of declaration then the remaining elements of the array is initialized to 0 by default.

>> char str1[20] = "Hello", str2[20] = " World"; The variable str1 and str2 is declared as an array of characters and initialized with value "Hello" and " World" respectively.

>> printf("%sn", strcpy(str2, strcat(str1, str2)));

=> strcat(str1, str2) it append the string str2 to str1. The result will be stored in str1. Therefore str1 contains "Hello World".

=> strcpy(str2, "Hello World") it copies the "Hello World" to the variable str2.

Hence it prints "Hello World".

In c compiler does not check array with its bounds, value at the computed location is displayed.

>> int a[3][4] = {1, 2, 3, 4, 4, 3, 2, 1, 7, 8, 9, 0}; The array a[3][4] is declared as an integer array having the 3 rows and 4 colums dimensions.

>> printf("%u, %u\n", a+1, &a+1);
The base address(also the address of the first element) of array is 65472.
For a two-dimensional array like a reference to array has type "pointer to array of 4 ints". Therefore, a+1 is pointing to the memory location of first element of the second row in array a. Hence 65472 + (4 ints * 2 bytes) = 65480
Then, &a has type "pointer to array of 3 arrays of 4 ints", totally 12 ints. Therefore, &a+1 denotes "12 ints * 2 bytes * 1 = 24 bytes".
Hence, begining address 65472 + 24 = 65496. So, &a+1 = 65496
Hence the output of