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- Question
What will be the output of the program?
#include<stdio.h> int main() { const int x=5; const int *ptrx; ptrx = &x; *ptrx = 10; printf("%d\n", x); return 0; }
Options- A. 5
- B. 10
- C. Error
- D. Garbage value
- Correct Answer
- Error
ExplanationStep 1: const int x=5; The constant variable x is declared as an integer data type and initialized with value '5'.Step 2: const int *ptrx; The constant variable ptrx is declared as an integer pointer.
Step 3: ptrx = &x; The address of the constant variable x is assigned to integer pointer variable ptrx.
Step 4: *ptrx = 10; Here we are indirectly trying to change the value of the constant vaiable x. This will result in an error.
To change the value of const variable x we have to use *(int *)&x = 10;
Constants problems
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- 1. What will be the output of the program?
#include<stdio.h> #include<stdlib.h> union employee { char name[15]; int age; float salary; }; const union employee e1; int main() { strcpy(e1.name, "K"); printf("%s %d %f", e1.name, e1.age, e1.salary); return 0; }
Options- A. Error: RValue required
- B. Error: cannot convert from 'const int *' to 'int *const'
- C. Error: LValue required in strcpy
- D. No error Discuss
Correct Answer: No error
Explanation:
The output will be (in 16-bit platform DOS):
K 75 0.000000- 2. What will be the output of the program?
#include<stdio.h> int main() { const char *s = ""; char str[] = "Hello"; s = str; while(*s) printf("%c", *s++); return 0; }
Options- A. Error
- B. H
- C. Hello
- D. Hel Discuss
Correct Answer: Hello
Explanation:
Step 1: const char *s = ""; The constant variable s is declared as an pointer to an array of characters type and initialized with an empty string.Step 2: char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello".
Step 3: s = str; The value of the variable str is assigned to the variable s. Therefore str contains the text "Hello".
Step 4: while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s.
Hence the output of the program is "Hello".
- 3. What will be the output of the program in TurboC?
#include<stdio.h> int fun(int **ptr); int main() { int i=10, j=20; const int *ptr = &i; printf(" i = %5X", ptr); printf(" ptr = %d", *ptr); ptr = &j; printf(" j = %5X", ptr); printf(" ptr = %d", *ptr); return 0; }
Options- A. i= FFE2 ptr=12 j=FFE4 ptr=24
- B. i= FFE4 ptr=10 j=FFE2 ptr=20
- C. i= FFE0 ptr=20 j=FFE1 ptr=30
- D. Garbage value Discuss
Correct Answer: i= FFE4 ptr=10 j=FFE2 ptr=20
- 4. What will be the output of the program?
#include<stdio.h> int main() { const int i=0; printf("%d\n", i++); return 0; }
Options- A. 10
- B. 11
- C. No output
- D. Error: ++needs a value Discuss
Correct Answer: Error: ++needs a value
Explanation:
This program will show an error "Cannot modify a const object".Step 1: const int i=0; The constant variable 'i' is declared as an integer and initialized with value of '0'(zero).
Step 2: printf("%d\n", i++); Here the variable 'i' is increemented by 1(one). This will create an error "Cannot modify a const object".
Because, we cannot modify a const variable.
- 5. What will be the output of the program?
#include<stdio.h> int main() { int y=128; const int x=y; printf("%d\n", x); return 0; }
Options- A. 128
- B. Garbage value
- C. Error
- D. 0 Discuss
Correct Answer: 128
Explanation:
Step 1: int y=128; The variable 'y' is declared as an integer type and initialized to value "128".Step 2: const int x=y; The constant variable 'x' is declared as an integer and it is initialized with the variable 'y' value.
Step 3: printf("%d\n", x); It prints the value of variable 'x'.
Hence the output of the program is "128"
- 6. What will be the output of the program?
#include<stdio.h> int main() { const c = -11; const int d = 34; printf("%d, %d\n", c, d); return 0; }
Options- A. Error
- B. -11, 34
- C. 11, 34
- D. None of these Discuss
Correct Answer: -11, 34
Explanation:
Step 1: const c = -11; The constant variable 'c' is declared and initialized to value "-11".Step 2: const int d = 34; The constant variable 'd' is declared as an integer and initialized to value '34'.
Step 3: printf("%d, %d\n", c, d); The value of the variable 'c' and 'd' are printed.
Hence the output of the program is -11, 34
- 7. What will the function rewind() do?
Options- A. Reposition the file pointer to a character reverse.
- B. Reposition the file pointer stream to end of file.
- C. Reposition the file pointer to begining of that line.
- D. Reposition the file pointer to begining of file. Discuss
Correct Answer: Reposition the file pointer to begining of file.
Explanation:
rewind() takes the file pointer to the beginning of the file. so that the next I/O operation will take place at the beginning of the file.
Example: rewind(FilePointer);- 8. Which standard library function will you use to find the last occurance of a character in a string in C?
Options- A. strnchar()
- B. strchar()
- C. strrchar()
- D. strrchr() Discuss
Correct Answer: strrchr()
Explanation:
strrchr() returns a pointer to the last occurrence of character in a string.Example:
#include <stdio.h> #include <string.h> int main() { char str[30] = "12345678910111213"; printf("The last position of '2' is %d.\n", strrchr(str, '2') - str); return 0; }Output: The last position of '2' is 14.
- 9. Input/output function prototypes and macros are defined in which header file?
Options- A. conio.h
- B. stdlib.h
- C. stdio.h
- D. dos.h Discuss
Correct Answer: stdio.h
Explanation:
stdio.h, which stands for "standard input/output header", is the header in the C standard library that contains macro definitions, constants, and declarations of functions and types used for various standard input and output operations.- 10. Can you use the fprintf() to display the output on the screen?
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
Explanation:
Do like this fprintf(stdout, "%s %d %f", str, i, a);
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- 1. What will be the output of the program?