Test: Strings - 2 - Back-End Programming MCQ

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.

Scope resolution (::) is not permitted to be overloaded.

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

No, as both the functions arguments is same and compiler ignores return type to consider overloading though different in return type.

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

The gets(str1) read the input string from user and store in variable str1.

The gets(str2) read the input string from user and store in variable str2.

The code i = strcmp(str1, str2); The strcmp not only returns -1, 0 and +1, but also other negative or positive values. So the value of i is “unpredictable integer value”.

printf(“%d\n”, i); It prints the value of variable i.

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

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.

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 “not empty string”.

No output, as we are comparing both base addresses and are not same.

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

char str[] = “Code\0Quiz\0″; The variable str is declared as an array of characters and initialized with value “Code”

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

The output of the program is “Code”.

option (a) & (b) are valid. There is no such syntax or provision as in option (c).