Strings Questions and Answers

Q: Consider the following C statements:char *str1 = "Hello; /* Statement S1 */char *str2 = "Hello;"; /* Statement S2 */int *str3 = "Hello"; /* Statement S3 */Which of the following options is/are correct?

The correct option is S1 has a lexical error and S3 has a semantic error.In S1 the string literal "Hello; is never closed before the end of the line. The lexical analyser cannot form a valid string token, so this is a lexical error.S2 is valid: the semicolon inside the quotes is part of the string, and the statement ends with its own semicolon.S3 is well formed syntactically, but it initialises an int * with a string literal, whose type is a char array that decays to char *. That type mismatch is detected during semantic analysis, so it is a semantic error.Calling S1's error syntactic, or saying S2 has any error, misplaces the phase that catches each fault.

Q: Consider the following C program:#include \(\lt\)stdio.h\(\gt\)void stringcopy(char *, char *);int main(){char a[30] = "@#Hello World!";stringcopy(a, a + 2);printf("%s\n", a);return 0;}void stringcopy(char *s, char *t) {while(*t)*s++ = *t++;}Which ONE of the following will be the output of the program?

The program prints Hello World!d!.The array holds "@#Hello World!": 14 characters in a[0] to a[13], then the terminator in a[14]. The call stringcopy(a, a + 2) copies from the third character onwards into the start of the array.The loop while(*t) *s++ = *t++; stops when *t is the terminator, without copying it. So it copies the 12 characters 'Hello World!' into a[0] to a[11], and a[12] and a[13] keep their old values, 'd' and '!'. The original terminator at a[14] is untouched.printf then prints a[0] to a[13]: Hello World!d!. 'Hello World!' alone would be printed only if the terminator were copied too. The original string and 'ello World!' would need different copy offsets.

Q: What is the output of following program? # include <stdio.h> int main() { char str1[] = "GeeksQuiz"; char str2[] = {'G', 'e', 'e', 'k', 's', 'Q', 'u', 'i', 'z'}; int n1 = sizeof(str1)/sizeof(str1[0]); int n2 = sizeof(str2)/sizeof(str2[0]); printf("n1 = %d, n2 = %d", n1, n2); return 0; }

Correct answer: n1 = 10, n2 = 9str1 is initialised from a string literal, so the compiler adds a terminating null: 9 letters plus '\0' is 10 bytes. str2 is initialised from an explicit list of nine characters, with no null added, so it is 9 bytes - and, lacking a terminator, it is an array of characters rather than a usable C string.

Q: What is the output of following program? #include <stdio.h> void swap(char *str1, char *str2) { char *temp = str1; str1 = str2; str2 = temp; } int main() { char *str1 = "Geeks"; char *str2 = "Quiz"; swap(str1, str2); printf("str1 is %s, str2 is %s", str1, str2); return 0; }

Correct answer: str1 is Geeks, str2 is Quizswap receives copies of the two pointers and exchanges the copies; the originals in main are untouched. To swap the caller’s pointers the function would have to take char ** and be called as swap(&str1, &str2).

Q: Predict the output? #include <stdio.h> int fun(char *str1) { char *str2 = str1; while(*++str1); return (str1-str2); } int main() { char *str = "GeeksQuiz"; printf("%d", fun(str)); return 0; }

Correct answer: 9while(*++str1); pre-increments before testing, so the character at index 0 is never examined and the pointer stops on the terminating null at index 9. The difference str1 - str2 is therefore 9 - which happens to be the length here, but only because the string is non-empty; on an empty string this code would run past the end.

Q: What does the following fragment of C-program print? char c[] = "GATE2011"; char *p =c; printf("%s", p + p[3] - p[1]) ;

Correct answer: 2011The subscripts give character codes: p[3] is 'E' = 69 and p[1] is 'A' = 65, so the expression is\[ p + 69 - 65 = p + 4 \]Printing from index 4 of "GATE2011" gives 2011.

Q: #include <stdio.h> int main() { char str[] = "GeeksQuiz"; printf("%s %s %s\n", &str[5], &5[str], str+5); printf("%c %c %c\n", *(str+6), str[6], 6[str]); return 0; }

Correct answer: Quiz Quiz Quiz u u uSubscripting is defined as a[i] == *(a + i), and since addition commutes, i[a] means the same thing. So &str[5], &5[str] and str+5 are all the address of the sixth character - each prints Quiz. Likewise *(str+6), str[6] and 6[str] are all the character u.

Q: In below program, what would you put in place of “?” to print “Quiz”? #include <stdio.h> int main() { char arr[] = "GeeksQuiz"; printf("%s", ?); return 0; }

Correct answer: (arr+5)"GeeksQuiz" has Q at index 5, counting from 0, so arr+5 points at the start of Quiz and %s prints from there to the terminating null. arr+4 would give sQuiz.

Q: Output? int main() { char a[2][3][3] = {'g','e','e','k','s','q','u','i','z'}; printf("%s ", **a); return 0; }

Correct answer: geeksquizThe array is 2×3×3 = 18 bytes in one contiguous block. The nine initialisers fill the first nine, and the remaining bytes are zero-filled, as a partly initialised array always is. **a is the address of the first byte, and %s prints from there to the first null - which is byte 10. So the whole run of nine characters comes out, with no garbage after it.

Q: #include <stdio.h> void my_toUpper(char* str, int index) { *(str + index) &= ~32; } int main() { char* arr = "geeksquiz"; my_toUpper(arr, 0); my_toUpper(arr, 5); printf("%s", arr); return 0; }

Correct answer: Compiler dependentarr points to a string literal, and modifying one is undefined behaviour. Most compilers place literals in read-only memory, so the write faults at run time; others allow it, and the program prints GeeksQuiz. The fix is char arr[] = "geeksquiz";, which gives a modifiable copy. (The &= ~32 trick works because upper and lower case differ by exactly bit 5.)

Consider the following C statements:

char *str1 = "Hello; /* Statement S1 */
char *str2 = "Hello;"; /* Statement S2 */
int *str3 = "Hello"; /* Statement S3 */

Which of the following options is/are correct?

A.S1 and S2 have syntactic errors
B.S2 has a lexical error and S3 has a syntactic error
C.S1 has a lexical error and S3 has a semantic error
D.S1 has a syntactic error and S3 has a semantic error

Exam Year(s)

GATE CS - 2026,

Answer: C

The correct option is S1 has a lexical error and S3 has a semantic error.

In S1 the string literal "Hello; is never closed before the end of the line. The lexical analyser cannot form a valid string token, so this is a lexical error.

S2 is valid: the semicolon inside the quotes is part of the string, and the statement ends with its own semicolon.

S3 is well formed syntactically, but it initialises an int * with a string literal, whose type is a char array that decays to char *. That type mismatch is detected during semantic analysis, so it is a semantic error.

Calling S1's error syntactic, or saying S2 has any error, misplaces the phase that catches each fault.

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Consider the following C program:

#include \(\lt\)stdio.h\(\gt\) void stringcopy(char *, char *); int main(){ char a[30] = "@#Hello World!"; stringcopy(a, a + 2); printf("%s\n", a); return 0; } void stringcopy(char *s, char *t) { while(*t) *s++ = *t++; }

Which ONE of the following will be the output of the program?

A.@#Hello World!
B.Hello World!
C.ello World!
D.Hello World!d!

Exam Year(s)

GATE CS - 2025,

Answer: D

The program prints Hello World!d!.

The array holds "@#Hello World!": 14 characters in a[0] to a[13], then the terminator in a[14]. The call stringcopy(a, a + 2) copies from the third character onwards into the start of the array.

The loop while(*t) *s++ = *t++; stops when *t is the terminator, without copying it. So it copies the 12 characters 'Hello World!' into a[0] to a[11], and a[12] and a[13] keep their old values, 'd' and '!'. The original terminator at a[14] is untouched.

printf then prints a[0] to a[13]: Hello World!d!. 'Hello World!' alone would be printed only if the terminator were copied too. The original string and 'ello World!' would need different copy offsets.

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What is the output of following program?

# include <stdio.h>
  
int main() 
{ 
   char str1[] = "GeeksQuiz"; 
   char str2[] = {'G', 'e', 'e', 'k', 's', 'Q', 'u', 'i', 'z'}; 
   int n1 = sizeof(str1)/sizeof(str1[0]); 
   int n2 = sizeof(str2)/sizeof(str2[0]); 
   printf("n1 = %d, n2 = %d", n1, n2); 
   return 0; 
}

A.n1 = 10, n2 = 9
B.n1 = 10, n2 = 10
C.n1 = 9, n2 = 9
D.n1 = 9, n2 = 10

Answer: A

Correct answer: n1 = 10, n2 = 9

str1 is initialised from a string literal, so the compiler adds a terminating null: 9 letters plus '\0' is 10 bytes. str2 is initialised from an explicit list of nine characters, with no null added, so it is 9 bytes - and, lacking a terminator, it is an array of characters rather than a usable C string.

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What is the output of following program?

#include <stdio.h> 
void swap(char *str1, char *str2) 
{ 
  char *temp = str1; 
  str1 = str2; 
  str2 = temp; 
}   
    
int main() 
{ 
  char *str1 = "Geeks"; 
  char *str2 = "Quiz"; 
  swap(str1, str2); 
  printf("str1 is %s, str2 is %s", str1, str2); 
  return 0; 
}

A.str1 is Quiz, str2 is Geeks
B.str1 is Geeks, str2 is Quiz
C.str1 is Geeks, str2 is Geeks
D.str1 is Quiz, str2 is Quiz

Answer: B

Correct answer: str1 is Geeks, str2 is Quiz

swap receives copies of the two pointers and exchanges the copies; the originals in main are untouched. To swap the caller’s pointers the function would have to take char ** and be called as swap(&str1, &str2).

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Predict the output?

#include <stdio.h> 
int fun(char *str1) 
{ 
  char *str2 = str1; 
  while(*++str1); 
  return (str1-str2); 
} 
  
int main() 
{ 
  char *str = "GeeksQuiz"; 
  printf("%d", fun(str)); 
  return 0; 
}

A.10
B.9
C.8
D.Random Number

Answer: B

Correct answer: 9

while(*++str1); pre-increments before testing, so the character at index 0 is never examined and the pointer stops on the terminating null at index 9. The difference str1 - str2 is therefore 9 - which happens to be the length here, but only because the string is non-empty; on an empty string this code would run past the end.

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What does the following fragment of C-program print?

char c[] = "GATE2011";  
char *p =c;  
printf("%s", p + p[3] - p[1]) ;

A.GATE2011
B.E2011
C.2011
D.011

Answer: C

Correct answer: 2011

The subscripts give character codes: p[3] is 'E' = 69 and p[1] is 'A' = 65, so the expression is

\[ p + 69 - 65 = p + 4 \]

Printing from index 4 of "GATE2011" gives 2011.

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#include <stdio.h> 
int main() 
{ 
    char str[] = "GeeksQuiz"; 
    printf("%s %s %s\n", &str[5], &5[str], str+5); 
    printf("%c %c %c\n", *(str+6), str[6], 6[str]); 
    return 0; 
}

A.Runtime Error
B.Compiler Error
C.uiz uiz uizu u u
D.Quiz Quiz Quizu u u

Answer: D

Correct answer: Quiz Quiz Quiz u u u

Subscripting is defined as a[i] == *(a + i), and since addition commutes, i[a] means the same thing. So &str[5], &5[str] and str+5 are all the address of the sixth character - each prints Quiz. Likewise *(str+6), str[6] and 6[str] are all the character u.

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In below program, what would you put in place of “?” to print “Quiz”?

#include <stdio.h> 
int main()  
{  
  char arr[] = "GeeksQuiz";  
  printf("%s", ?);  
  return 0;  
}

A.arr
B.(arr+5)
C.(arr+4)
D.Not possible

Answer: B

Correct answer: (arr+5)

"GeeksQuiz" has Q at index 5, counting from 0, so arr+5 points at the start of Quiz and %s prints from there to the terminating null. arr+4 would give sQuiz.

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Output?

int main() 
{ 
    char a[2][3][3] = {'g','e','e','k','s','q','u','i','z'}; 
    printf("%s ", **a); 
    return 0; 
}

A.Compiler Error
B.geeksquiz followed by garbage characters
C.geeksquiz
D.Runtime Error

Answer: C

Correct answer: geeksquiz

The array is 2×3×3 = 18 bytes in one contiguous block. The nine initialisers fill the first nine, and the remaining bytes are zero-filled, as a partly initialised array always is. **a is the address of the first byte, and %s prints from there to the first null - which is byte 10. So the whole run of nine characters comes out, with no garbage after it.

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10.

#include <stdio.h> 
   
void my_toUpper(char* str, int index) 
{ 
    *(str + index) &= ~32; 
} 
   
int main() 
{ 
    char* arr = "geeksquiz"; 
    my_toUpper(arr, 0); 
    my_toUpper(arr, 5); 
    printf("%s", arr); 
    return 0; 
}

A.GeeksQuiz
B.geeksquiz
C.Compiler dependent

Answer: C

Correct answer: Compiler dependent

arr points to a string literal, and modifying one is undefined behaviour. Most compilers place literals in read-only memory, so the write faults at run time; others allow it, and the program prints GeeksQuiz. The fix is char arr[] = "geeksquiz";, which gives a modifiable copy. (The &= ~32 trick works because upper and lower case differ by exactly bit 5.)

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