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- Question
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
- Correct Answer
- CURIOUSTAB
Explanationprintf(5+"CuriousTab\n"); In the printf statement, it skips the first 5 characters and it prints "CURIOUSTAB"
Strings problems
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- 1. 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".
- 2. What will be the output of the program?
#include<stdio.h> int main() { char str1[] = "Hello"; char str2[10]; char *t, *s; s = str1; t = str2; while(*t=*s) *t++ = *s++; printf("%s\n", str2); return 0; }
Options- A. Hello
- B. HelloHello
- C. No output
- D. ello Discuss
Correct Answer: Hello
- 3. What will be the output of the program?
#include<stdio.h> int main() { printf(5+"Good Morning\n"); return 0; }
Options- A. Good Morning
- B. Good
- C. M
- D. Morning Discuss
Correct Answer: Morning
Explanation:
printf(5+"Good Morning\n"); It skips the 5 characters and prints the given string.Hence the output is "Morning"
- 4. 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'.
- 5. 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"
- 6. 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
- 7. 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
- 8. 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".
- 9. 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 Discuss
Correct Answer: 11
Explanation:
The 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.- 10. 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".
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- 1. What will be the output of the program?