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- Question
What will be the output of the program (Turbo C in 16 bit platform DOS)?
#include<stdio.h> #include<string.h> int main() { char *str1 = "India"; char *str2 = "CURIOUSTAB"; char *str3; str3 = strcat(str1, str2); printf("%s %s\n", str3, str1); return 0; }
Options- A. CuriousTab India
- B. CuriousTab CuriousTab
- C. India India
- D. Error
- Correct Answer
- CuriousTab CuriousTab
ExplanationIt prints 'CuriousTab CuriousTab' in TurboC (in 16 bit platform).It may cause a 'segmentation fault error' in GCC (32 bit platform).
Strings problems
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- 1. What will be the output of the program?
#include<stdio.h> int main() { char str[25] = "CuriousTab"; printf("%s\n", &str+2); return 0; }
Options- A. Garbage value
- B. Error
- C. No output
- D. diaCURIOUSTAB Discuss
Correct Answer: Garbage value
Explanation:
Step 1: char str[25] = "CuriousTab"; The variable str is declared as an array of characteres and initialized with a string "CuriousTab".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 "CuriousTab". Therefore &str+2 is another memory location.
Hence it prints the Garbage value.
- 2. What will be the output of the program?
#include<stdio.h> #include<string.h> int main() { printf("%c\n", "abcdefgh"[4]); return 0; }
Options- A. Error
- B. d
- C. e
- D. abcdefgh Discuss
Correct Answer: e
Explanation:
printf("%c\n", "abcdefgh"[4]); It prints the 5 character of the string "abcdefgh".Hence the output is 'e'.
- 3. What will be the output of the program?
#include<stdio.h> #include<string.h> int main() { char sentence[80]; int i; printf("Enter a line of text\n"); gets(sentence); for(i=strlen(sentence)-1; i >=0; i--) putchar(sentence[i]); return 0; }
Options- A. The sentence will get printed in same order as it entered
- B. The sentence will get printed in reverse order
- C. Half of the sentence will get printed
- D. None of above Discuss
Correct Answer: The sentence will get printed in reverse order
- 4. What will be the output of the program If characters 'a', 'b' and 'c' enter are supplied as input?
#include<stdio.h> int main() { void fun(); fun(); printf("\n"); return 0; } void fun() { char c; if((c = getchar())!= '\n') fun(); printf("%c", c); }
Options- A. abc abc
- B. bca
- C. Infinite loop
- D. cba Discuss
Correct Answer: cba
Explanation:
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
- 5. 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; }
Options- A. hhe!
- B. he c
- C. The c
- D. Hhec Discuss
Correct Answer: hhe!
- 6. What will be the output of the program?
#include<stdio.h> int main() { int i; char a[] = "\0"; if(printf("%s", a)) printf("The string is not empty\n"); else printf("The string is empty\n"); return 0; }
Options- A. The string is not empty
- B. The string is empty
- C. No output
- D. 0 Discuss
Correct Answer: The string is empty
Explanation:
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 empty".
- 7. What will be the output of the following program in 16 bit platform assuming that 1022 is memory address of the string "Hello1" (in Turbo C under DOS)?
#include<stdio.h> int main() { printf("%u %s\n", &"Hello1", &"Hello2"); return 0; }
Options- A. 1022 Hello2
- B. Hello1 1022
- C. Hello1 Hello2
- D. 1022 1022
- E. Error Discuss
Correct Answer: 1022 Hello2
Explanation:
In printf("%u %s\n", &"Hello", &"Hello");.The %u format specifier tells the compiler to print the memory address of the "Hello1".
The %s format specifier tells the compiler to print the string "Hello2".
Hence the output of the program is "1022 Hello2".
- 8. If the size of pointer is 4 bytes then What will be the output of the program?
#include<stdio.h> int main() { char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"}; printf("%d, %d", sizeof(str), strlen(str[0])); return 0; }
Options- A. 22, 4
- B. 25, 5
- C. 24, 5
- D. 20, 2 Discuss
Correct Answer: 24, 5
Explanation:
Step 1: char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"}; The variable str is declared as an pointer to the array of 6 strings.Step 2: printf("%d, %d", sizeof(str), strlen(str[0]));
sizeof(str) denotes 6 * 4 bytes = 24 bytes. Hence it prints '24'
strlen(str[0])); becomes strlen(Frogs)). Hence it prints '5';
Hence the output of the program is 24, 5
Hint: If you run the above code in 16 bit platform (Turbo C under DOS) the output will be 12, 5. Because the pointer occupies only 2 bytes. If you run the above code in Linux (32 bit platform), the output will be 24, 5 (because the size of pointer is 4 bytes).
- 9. What will be the output of the program?
#include<stdio.h> int main() { char str[] = "Nagpur"; str[0]='K'; printf("%s, ", str); str = "Kanpur"; printf("%s", str+1); return 0; }
Options- A. Kagpur, Kanpur
- B. Nagpur, Kanpur
- C. Kagpur, anpur
- D. Error Discuss
Correct Answer: Error
Explanation:
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"
- 10. What will be the output of the program?
#include<stdio.h> int main() { char p[] = "%d\n"; p[1] = 'c'; printf(p, 65); return 0; }
Options- A. A
- B. a
- C. c
- D. 65 Discuss
Correct Answer: A
Explanation:
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'.
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- 1. What will be the output of the program?