C Programming Strings Online Test, C Programming Mock Test Series

char *strnset(char *s, int ch, size_t n); Sets the first n characters of s to ch

#include <stdio.h>
#include <string.h>

int main(void)
{
   char *string = "abcdefghijklmnopqrstuvwxyz";
   char letter = 'x';

   printf("string before strnset: %s\n", string);
   strnset(string, letter, 13);
   printf("string after  strnset: %s\n", string);

   return 0;
}

Output:

string before strnset: abcdefghijklmnopqrstuvwxyz

string after strnset: xxxxxxxxxxxxxnopqrstuvwxyz

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.

Example:

#include <string.h>
#include <stdio.h>

int main(void)
{
   char text[] = "I learn through IndiaBIX.com";
   char *ptr, c = 'i';

   ptr = strrchr(text, c);
   if (ptr)
      printf("The position of '%c' is: %d\n", c, ptr-text);
   else
      printf("The character was not found\n");
   return 0;
}

Output:

The position of ‘i’ is: 19

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.

Example:

#include <stdio.h>
#include <string.h>

int main(void)
{
   char *str1 = "IndiaBIX", *str2 = "ia", *ptr;

   ptr = strstr(str1, str2);
   printf("The substring is: %s\n", ptr);
   return 0;
}

Output: The substring is: iaBIX

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

#include <stdio.h>

int main(void)
{
   char string[80];

   printf("Enter a string:");
   gets(string);
   printf("The string input was: %s\n", string);
   return 0;
}

Output:

Enter a string: IndiaBIX

The string input was: IndiaBIX

Step 1: 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.

Step 2: printf(“%s\n”, 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”.

Step 1: char p[] = “%d\n”; The variable p is declared as an array of characters and initialized with string “%d”.

Step 2: p[1] = ‘c’; Here, we overwrite the second element of array p by ‘c’. So array p becomes “%c”.

Step 3: printf(p, 65); becomes printf(“%c”, 65);

Therefore it prints the ASCII value of 65. The output is ‘A’.

The function strlen returns the number of characters in the given string.

Therefore, strlen(“123456”) returns 6.

Hence the output of the program is “6”.

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’.

Step 1: char str[] = “India\0\BIX\0”; The variable str is declared as an array of characters and initialized with value “India”

Step 2: printf(“%s\n”, str); It prints the value of the str.

The output of the program is “India”.

Step 1: void fun(); This is the prototype for the function fun().

Step 2: fun(); The function fun() is called here.

The function fun() gets a character input and the input is terminated by an enter key(New line character). It prints the given character in the reverse order.

The given input characters are “abc”

Output: cba

printf(“India”, “BIX\n”); It prints “India”. Because ,(comma) operator has Left to Right associativity. After printing “India”, the statement got terminated.

Here str[] has declared as 7 character array and into a 8 character is stored. This will result in overwriting of the byte beyond 7 byte reserved for ‘\0’.

Step 1: char *names[] = { “Suresh”, “Siva”, “Sona”, “Baiju”, “Ritu”}; The variable names is declared as an pointer to a array of strings.

Step 2: int i; The variable i is declared as an integer type.

Step 3: char *t; The variable t is declared as pointer to a string.

Step 4: t = names[3]; names[3] = names[4]; names[4] = t; These statements the swaps the 4 and 5 element of the array names.

Step 5: for(i=0; i<=4; i++) printf(“%s,”, names[i]); These statement prints the all the value of the array names.

Hence the output of the program is “Suresh, Siva, Sona, Ritu, Baiju”.

The function strlen returns the number of characters int the given string.

Therefore, strlen(str) becomes strlen(“India”) contains 5 characters. A string is a collection of characters terminated by ‘\0’.

The output of the program is “5”.

Step 1:

printf(“%d, %d, %d”, sizeof(3.0f), sizeof(‘3’), sizeof(3.0));

The sizeof function returns the size of the given expression.

sizeof(3.0f) is a floating point constant. The size of float is 4 bytes

sizeof(‘3’) It converts ‘3’ in to ASCII value.. The size of int is 2 bytes

sizeof(3.0) is a double constant. The size of double is 8 bytes

Hence the output of the program is 4,2,8

Note: The above program may produce different output in other platform due to the platform dependency of C compiler.

In Turbo C, 4 2 8. But in GCC, the output will be 4 4 8.

The function printf() returns the number of charecters printed on the console.

Step 1: char a[] = “\0”; The variable a is declared as an array of characters and it initialized with “\0”. It denotes that the string is empty.

Step 2: if(printf(“%s”, a)) The printf() statement does not print anything, so it returns ‘0’(zero). Hence the if condition is failed.

In the else part it prints “The string is not empty”.

Step 1: char ch = ‘A’; The variable ch is declared as an character type and initialized with value ‘A’.

Step 2:

printf(“%d, %d, %d”, sizeof(ch), sizeof(‘A’), sizeof(3.14));

The sizeof function returns the size of the given expression.

sizeof(ch) becomes sizeof(char). The size of char is 1 byte.

sizeof(‘A’) becomes sizeof(65). The size of int is 4 bytes (as mentioned in the question).

sizeof(3.14f). The size of float is 4 bytes.

Hence the output of the program is 1, 4, 4

The following examples may help you understand this problem:

1. sizeof(“”) returns 1 (1*).

2. sizeof(“India”) returns 6 (5 + 1*).

3. sizeof(“BIX”) returns 4 (3 + 1*).

4. sizeof(“India\0BIX”) returns 10 (5 + 1 + 3 + 1*).
Here ‘\0’ is considered as 1 char by sizeof() function.

5. sizeof(“India\0BIX\0”) returns 11 (5 + 1 + 3 + 1 + 1*).
Here ‘\0’ is considered as 1 char by sizeof() function.

Step 1: char str[25] = “IndiaBIX”; The variable str is declared as an array of characteres and initialized with a string “IndiaBIX”.

Step 2: printf(“%s\n”, &str+2);

=> In the printf statement %s is string format specifier tells the compiler to print the string in the memory of &str+2

=> &str is a location of string “IndiaBIX”. Therefore &str+2 is another memory location.

Hence it prints the Garbage value.

The line char str = “IndiaBIX”; generates “Non portable pointer conversion” error.

To eliminate the error, we have to change the above line to

char *str = “IndiaBIX”; (or) char str[] = “IndiaBIX”;

Then it prints “IndiaBIX”.

The statement str = “Kanpur”; generates the LVALUE required error. We have to use strcpy function to copy a string.

To remove error we have to change this statement str = “Kanpur”; to strcpy(str, “Kanpur”);

The program prints the string “anpur”