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- Question
What will be the output of the program?
#include<stdio.h> int main() { char str[] = "India\0CURIOUSTAB\0"; printf("%d\n", sizeof(str)); return 0; }
Options- A. 10
- B. 6
- C. 5
- D. 11
- Correct Answer
- 11
ExplanationThe following examples may help you understand this problem:1. sizeof("") returns 1 (1*).
2. sizeof("India") returns 6 (5 + 1*).
3. sizeof("CURIOUSTAB") returns 4 (3 + 1*).
4. sizeof("India\0CURIOUSTAB") returns 10 (5 + 1 + 3 + 1*).
Here '\0' is considered as 1 char by sizeof() function.5. sizeof("India\0CURIOUSTAB\0") returns 11 (5 + 1 + 3 + 1 + 1*).
Here '\0' is considered as 1 char by sizeof() function.
Strings problems
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- 1. What will be the output of the program?
#include<stdio.h> int main() { char t; char *p1 = "India", *p2; p2=p1; p1 = "CURIOUSTAB"; printf("%s %s\n", p1, p2); return 0; }
Options- A. India CURIOUSTAB
- B. CURIOUSTAB India
- C. India India
- D. CURIOUSTAB CURIOUSTAB Discuss
Correct Answer: CURIOUSTAB India
Explanation:
Step 1: char *p1 = "India", *p2; The variable p1 and p2 is declared as an pointer to a character value and p1 is assigned with a value "India".Step 2: p2=p1; The value of p1 is assigned to variable p2. So p2 contains "India".
Step 3: p1 = "CURIOUSTAB"; The p1 is assigned with a string "CURIOUSTAB"
Step 4: printf("%s %s\n", p1, p2); It prints the value of p1 and p2.
Hence the output of the program is "CURIOUSTAB India".
- 2. If char=1, int=4, and float=4 bytes size, What will be the output of the program?
#include<stdio.h> int main() { char ch = 'A'; printf("%d, %d, %d", sizeof(ch), sizeof('A'), sizeof(3.14f)); return 0; }
Options- A. 1, 2, 4
- B. 1, 4, 4
- C. 2, 2, 4
- D. 2, 4, 8 Discuss
Correct Answer: 1, 4, 4
Explanation:
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
- 3. What will be the output of the program?
#include<stdio.h> #include<string.h> int main() { static char str1[] = "dills"; static char str2[20]; static char str3[] = "Daffo"; int i; i = strcmp(strcat(str3, strcpy(str2, str1)), "Daffodills"); printf("%d\n", i); return 0; }
Options- A. 0
- B. 1
- C. 2
- D. 4 Discuss
Correct Answer: 0
- 4. What will be the output of the program?
#include<stdio.h> int main() { printf(5+"CuriousTab\n"); return 0; }
Options- A. Error
- B. CuriousTab
- C. CURIOUSTAB
- D. None of above Discuss
Correct Answer: CURIOUSTAB
Explanation:
printf(5+"CuriousTab\n"); In the printf statement, it skips the first 5 characters and it prints "CURIOUSTAB"- 5. What will be the output of the program?
#include<stdio.h> int main() { char str = "CuriousTab"; printf("%s\n", str); return 0; }
Options- A. Error
- B. CuriousTab
- C. Base address of str
- D. No output Discuss
Correct Answer: Error
Explanation:
The line char str = "CuriousTab"; generates "Non portable pointer conversion" error.To eliminate the error, we have to change the above line to
char *str = "CuriousTab"; (or) char str[] = "CuriousTab";
Then it prints "CuriousTab".
- 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 empty\n"); else printf("The string is not empty\n"); return 0; }
Options- A. The string is empty
- B. The string is not empty
- C. No output
- D. 0 Discuss
Correct Answer: The string is not 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 not empty".
- 7. What will be the output of the program?
#include<stdio.h> #include<string.h> int main() { printf("%d\n", strlen("123456")); return 0; }
Options- A. 6
- B. 12
- C. 7
- D. 2 Discuss
Correct Answer: 6
Explanation:
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".
- 8. What will be the output of the program in Turbo C?
#include<stdio.h> int main() { char str[10] = "India"; str[6] = "CURIOUSTAB"; printf("%s\n", str); return 0; }
Options- A. India CURIOUSTAB
- B. CURIOUSTAB
- C. India
- D. Error Discuss
Correct Answer: Error
Explanation:
str[6] = "CURIOUSTAB"; - Nonportable pointer conversion.- 9. What will be the output of the program?
#include<stdio.h> void swap(char *, char *); int main() { char *pstr[2] = {"Hello", "CuriousTab"}; 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; }
Options- A. CuriousTab
Hello - B. Address of "Hello" and "CuriousTab"
- C. Hello
CuriousTab - D. Iello
HndiaCURIOUSTAB
Discuss
Correct Answer: Hello
CuriousTab
Explanation:
Step 1: void swap(char *, char *); This prototype tells the compiler that the function swap accept two strings as arguments and it does not return anything.Step 2: char *pstr[2] = {"Hello", "CuriousTab"}; The variable pstr is declared as an pointer to the array of strings. It is initialized to
pstr[0] = "Hello", pstr[1] = "CuriousTab"
Step 3: swap(pstr[0], pstr[1]); The swap function is called by "call by value". Hence it does not affect the output of the program.
If the swap function is "called by reference" it will affect the variable pstr.
Step 4: printf("%s\n%s", pstr[0], pstr[1]); It prints the value of pstr[0] and pstr[1].
Hence the output of the program is
Hello
CuriousTab- 10. What will be the output of the program?
#include<stdio.h> int main() { static char mess[6][30] = {"Don't walk in front of me...", "I may not follow;", "Don't walk behind me...", "Just walk beside me...", "And be my friend." }; printf("%c, %c\n", *(mess[2]+9), *(*(mess+2)+9)); return 0; }
Options- A. t, t
- B. k, k
- C. n, k
- D. m, f Discuss
Correct Answer: k, k
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- 1. What will be the output of the program?