Strings Questions and Answers

Q: What will be the output of the program ? #include <stdio.h> int main() { char *names[] = { "Suresh", "Siva", "Sona", "Baiju", "Ritu"}; int i; char *t; t = names[3]; names[3] = names[4]; names[4] = t; for(i=0; i<=4; i++) printf("%s,", names[i]); return 0; }

Correct answer: Suresh, Siva, Sona, Ritu, BaijuThe array holds pointers, and the code exchanges two of them through the temporary t - the strings themselves are never touched or copied. So elements 3 and 4 swap places, giving Suresh, Siva, Sona, Ritu, Baiju. Swapping pointers rather than character arrays is exactly why an array of pointers is preferred for sorting strings.

Q: The C library function _________ breaks string s1 into a series of tokens using the delimiter s2.

Correct answer: char *strtok(char *s1,const char *s2);strtok splits s1 into tokens separated by any of the characters in s2. The first call passes the string; every later call passes NULL to continue where it left off. Note that it modifies the string, writing a null over each delimiter, and keeps state between calls, which makes it unsuitable for reentrant or threaded code.

Q: The mem functions are meant for _______

Correct answer: manipulating objects as character arraysThe mem functions treat their arguments as blocks of bytes of a stated length, with no special meaning for the null character. That is what lets them copy, compare, search and fill arbitrary data - structures, integers, binary buffers - and not just text.

Q: What will be the output of the program ? #include <stdio.h> int main() { static char s[25] = "The cocaine man"; int i=0; char ch; ch = s[++i]; printf("%c", ch); ch = s[i++]; printf("%c", ch); ch = i++[s]; printf("%c", ch); ch = ++i[s]; printf("%c", ch); return 0; }

Correct answer: hhe!Take the four statements in turn, with s = "The cocaine man" and i = 0.s[++i] - i becomes 1 first, so this is s[1] = h.s[i++] - the old i, still 1, so h again; i becomes 2.i++[s] - the same as s[i++], so s[2] = e; i becomes 3.++i[s] - subscript binds tighter than the prefix ++, so this increments the character s[3], a space (32), to 33 = !.Output: hhe!

Q: Output of following program #include <stdio.h> int fun(char *p) { if (p == NULL || *p == '\0') return 0; int current = 1, i = 1; while (*(p+current)) { if (p[current] != p[current-1]) { p[i] = p[current]; i++; } current++; } *(p+i)='\0'; return i; } int main() { char str[] = "geeksskeeg"; fun(str); puts(str); return 0; }

Correct answer: gekskegThe function removes adjacent duplicates: current scans the string while i marks where to write, and a character is kept only when it differs from the one before it. On geeksskeeg the second e, the second s and the second e of ee are dropped, leaving gekskeg. The result is written back in place and terminated with a null.

Q: What does the function void *memmove(void *s1, const void *s2, size_t n); return?

Correct answer: returns the value of s1memmove returns its destination pointer, s1 - as memcpy, strcpy and strcat all do. Returning the destination lets a call be used directly as an argument to another function. Its distinctive property is that it works correctly even when the two regions overlap, which memcpy does not guarantee.

Q: Which among the given options compares at most n characters of string ch to string s?

Correct answer: int strncmp(ch, s, n)strncmp is the bounded form of strcmp: it compares at most n characters, stopping early at a difference or at a terminating null. The order of the first two arguments matters only for the sign of the result, but the question asks for ch compared to s, so ch comes first.

Q: What will be the output of the program ? #include <stdio.h> int main() { char str = "IndiaBIX"; printf("%s\n", str); return 0; }

Correct answer: Errorchar str declares a single character, and a string literal cannot be assigned to one - the initialiser is a pointer to an array of characters. The declaration would have to be char str[] = "IndiaBIX"; or char *str = "IndiaBIX";.

Q: What will be the output of the program ? #include <stdio.h> void swap(char *, char *); int main() { char *pstr[2] = {"Hello", "IndiaBIX"}; swap(pstr[0], pstr[1]); printf("%s\n%s", pstr[0], pstr[1]); return 0; } void swap(char *t1, char *t2) { char *t; t=t1; t1=t2; t2=t; }

Correct answer: Hello IndiaBIXswap receives copies of the two pointers and exchanges the copies, which vanish when the function returns. The array in main is untouched, so the strings print in their original order. To swap the caller’s pointers the function would have to take char ** and be called as swap(&pstr[0], &pstr[1]).

51.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    char *names[] = { "Suresh", "Siva", "Sona", "Baiju", "Ritu"};
    int i;
    char *t;
    t = names[3];
    names[3] = names[4];
    names[4] = t;
    for(i=0; i<=4; i++)
        printf("%s,", names[i]);
    return 0;
}

A.Suresh, Siva, Sona, Baiju, Ritu
B.Suresh, Siva, Sona, Ritu, Baiju
C.Suresh, Siva, Baiju, Sona, Ritu
D.Suresh, Siva, Ritu, Sona, Baiju

Answer: B

Correct answer: Suresh, Siva, Sona, Ritu, Baiju

The array holds pointers, and the code exchanges two of them through the temporary t - the strings themselves are never touched or copied. So elements 3 and 4 swap places, giving Suresh, Siva, Sona, Ritu, Baiju. Swapping pointers rather than character arrays is exactly why an array of pointers is preferred for sorting strings.

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52.

The C library function _________ breaks string s1 into a series of tokens using the delimiter s2.

A.char *strtok(char *s1,const char *s2);
B.char *strtok(char *s2,const char *s1);
C.char *strstr(char *s1,const char *s2);
D.char *strstr(char *s2,const char *s1);

Answer: A

Correct answer: char *strtok(char *s1,const char *s2);

strtok splits s1 into tokens separated by any of the characters in s2. The first call passes the string; every later call passes NULL to continue where it left off. Note that it modifies the string, writing a null over each delimiter, and keeps state between calls, which makes it unsuitable for reentrant or threaded code.

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53.

The mem functions are meant for _______

A.returning a pointer to the token
B.manipulating objects as character arrays
C.returning a pointer for implemented-defined string
D.returning a pointer to first occurrence of string in another string

Answer: B

Correct answer: manipulating objects as character arrays

The mem functions treat their arguments as blocks of bytes of a stated length, with no special meaning for the null character. That is what lets them copy, compare, search and fill arbitrary data - structures, integers, binary buffers - and not just text.

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54.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    static char s[25] = "The cocaine man";
    int i=0;
    char ch;
    ch = s[++i];
    printf("%c", ch);
    ch = s[i++];
    printf("%c", ch);
    ch = i++[s];
    printf("%c", ch);
    ch = ++i[s];
    printf("%c", ch);
    return 0;
}

A.hhe!
B.he c
C.The c
D.Hhec

Answer: A

Correct answer: hhe!

Take the four statements in turn, with s = "The cocaine man" and i = 0.

s[++i] - i becomes 1 first, so this is s[1] = h.
s[i++] - the old i, still 1, so h again; i becomes 2.
i++[s] - the same as s[i++], so s[2] = e; i becomes 3.
++i[s] - subscript binds tighter than the prefix ++, so this increments the character s[3], a space (32), to 33 = !.

Output: hhe!

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55.

Output of following program

#include <stdio.h> 
int fun(char *p) 
{ 
    if (p == NULL || *p == '\0') return 0; 
    int current = 1, i = 1; 
    while (*(p+current)) 
    { 
        if (p[current] != p[current-1]) 
        { 
            p[i] = p[current]; 
            i++; 
        } 
        current++; 
    } 
    *(p+i)='\0'; 
    return i; 
} 
  
int main() 
{ 
    char str[] = "geeksskeeg"; 
    fun(str); 
    puts(str); 
    return 0; 
}

A.gekskeg
B.geeksskeeg
C.geeks
D.Garbage Values

Answer: A

Correct answer: gekskeg

The function removes adjacent duplicates: current scans the string while i marks where to write, and a character is kept only when it differs from the one before it. On geeksskeeg the second e, the second s and the second e of ee are dropped, leaving gekskeg. The result is written back in place and terminated with a null.

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56.

What does the function void *memmove(void *s1, const void *s2, size_t n); return?

A.returns the value of s1
B.returns the value of s2
C.doesn’t return any value
D.returns the value of s1 and s2

Answer: A

Correct answer: returns the value of s1

memmove returns its destination pointer, s1 - as memcpy, strcpy and strcat all do. Returning the destination lets a call be used directly as an argument to another function. Its distinctive property is that it works correctly even when the two regions overlap, which memcpy does not guarantee.

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57.

What is the return value of strxfrm()?

A.length of the transformed string, not including the terminating null-character
B.length of the transformed string, including the terminating null-character
C.display the transformed string, not including the terminating null character
D.display the transformed string, not including the terminating null-character

Answer: A

Correct answer: length of the transformed string, not including the terminating null-character

strxfrm transforms a string so that comparing two transformed strings with strcmp gives the same order as comparing the originals with strcoll. It returns the length of the transformed string, excluding the null - so a caller can size the buffer by calling it once with a length of 0.

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58.

Which among the given options compares at most n characters of string ch to string s?

A.int strncmp(ch, s, n)
B.int strcmp(ch, s)
C.int strncmp(s, ch, n)
D.int strcmp(s, ch)

Answer: A

Correct answer: int strncmp(ch, s, n)

strncmp is the bounded form of strcmp: it compares at most n characters, stopping early at a difference or at a terminating null. The order of the first two arguments matters only for the sign of the result, but the question asks for ch compared to s, so ch comes first.

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59.

What will be the output of the program ?

#include <stdio.h>

int main()
{
    char str = "IndiaBIX";
    printf("%s\n", str);
    return 0;
}

A.Error
B.IndiaBIX
C.Base address of str
D.No output

Answer: A

Correct answer: Error

char str declares a single character, and a string literal cannot be assigned to one - the initialiser is a pointer to an array of characters. The declaration would have to be char str[] = "IndiaBIX"; or char *str = "IndiaBIX";.

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60.

What will be the output of the program ?

#include <stdio.h>
void swap(char *, char *);

int main()
{
    char *pstr[2] = {"Hello", "IndiaBIX"};
    swap(pstr[0], pstr[1]);
    printf("%s\n%s", pstr[0], pstr[1]);
    return 0;
}
void swap(char *t1, char *t2)
{
    char *t;
    t=t1;
    t1=t2;
    t2=t;
}

A.IndiaBIXHello
B.Address of "Hello" and "IndiaBIX"
C.HelloIndiaBIX
D.IelloHndiaBIX

Answer: C

Correct answer: Hello IndiaBIX

swap receives copies of the two pointers and exchanges the copies, which vanish when the function returns. The array in main is untouched, so the strings print in their original order. To swap the caller’s pointers the function would have to take char ** and be called as swap(&pstr[0], &pstr[1]).

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