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- Question
What will be the output of the program?
#include<stdio.h> int get(); int main() { const int x = get(); printf("%d", x); return 0; } int get() { return 20; }
Options- A. Garbage value
- B. Error
- C. 20
- D. 0
- Correct Answer
- 20
ExplanationStep 1: int get(); This is the function prototype for the funtion get(), it tells the compiler returns an integer value and accept no parameters.Step 2: const int x = get(); The constant variable x is declared as an integer data type and initialized with the value "20".
The function get() returns the value "20".
Step 3: printf("%d", x); It prints the value of the variable x.
Hence the output of the program is "20".
Constants problems
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- 1. What will be the output of the program (in Turbo C)?
#include<stdio.h> int fun(int *f) { *f = 10; return 0; } int main() { const int arr[5] = {1, 2, 3, 4, 5}; printf("Before modification arr[3] = %d", arr[3]); fun(&arr[3]); printf("\nAfter modification arr[3] = %d", arr[3]); return 0; }
Options- A. Before modification arr[3] = 4
After modification arr[3] = 10 - B. Error: cannot convert parameter 1 from const int * to int *
- C. Error: Invalid parameter
- D. Before modification arr[3] = 4
After modification arr[3] = 4
Discuss
Correct Answer: Before modification arr[3] = 4
After modification arr[3] = 10
Explanation:
Step 1: const int arr[5] = {1, 2, 3, 4, 5}; The constant variable arr is declared as an integer array and initialized toarr[0] = 1, arr[1] = 2, arr[2] = 3, arr[3] = 4, arr[4] = 5
Step 2: printf("Before modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 4).
Step 3: fun(&arr[3]); The memory location of the arr[3] is passed to fun() and arr[3] value is modified to 10.
A const variable can be indirectly modified by a pointer.
Step 4: printf("After modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 10).
Hence the output of the program is
Before modification arr[3] = 4
After modification arr[3] = 10
- 2. What will be the output of the program?
#include<stdio.h> int fun(int **ptr); int main() { int i=10; const int *ptr = &i; fun(&ptr); return 0; } int fun(int **ptr) { int j = 223; int *temp = &j; printf("Before changing ptr = %5x\n", *ptr); const *ptr = temp; printf("After changing ptr = %5x\n", *ptr); return 0; }
Options- A. Address of i
Address of j - B. 10
223 - C. Error: cannot convert parameter 1 from 'const int **' to 'int **'
- D. Garbage value Discuss
Correct Answer: Error: cannot convert parameter 1 from 'const int **' to 'int **'
- 3. Which of the following operations can be performed on the file "NOTES.TXT" using the below code?
FILE *fp; fp = fopen("NOTES.TXT", "r+");
Options- A. Reading
- B. Writing
- C. Appending
- D. Read and Write Discuss
Correct Answer: Read and Write
Explanation:
r+ Open an existing file for update (reading and writing).- 4. On executing the below program what will be the contents of 'target.txt' file if the source file contains a line "To err is human"?
#include<stdio.h> int main() { int i, fss; char ch, source[20] = "source.txt", target[20]="target.txt", t; FILE *fs, *ft; fs = fopen(source, "r"); ft = fopen(target, "w"); while(1) { ch=getc(fs); if(ch==EOF) break; else { fseek(fs, 4L, SEEK_CUR); fputc(ch, ft); } } return 0; }
Options- A. r n
- B. Trh
- C. err
- D. None of above Discuss
Correct Answer: Trh
Explanation:
The file source.txt is opened in read mode and target.txt is opened in write mode. The file source.txt contains "To err is human".Inside the while loop,
ch=getc(fs); The first character('T') of the source.txt is stored in variable ch and it's checked for EOF.
if(ch==EOF) If EOF(End of file) is true, the loop breaks and program execution stops.
If not EOF encountered, fseek(fs, 4L, SEEK_CUR); the file pointer advances 4 character from the current position. Hence the file pointer is in 5th character of file source.txt.
fputc(ch, ft); It writes the character 'T' stored in variable ch to target.txt.
The while loop runs three times and it write the character 1st and 5th and 11th characters ("Trh") in the target.txt file.
- 5. What does fp point to in the program?
#include<stdio.h> int main() { FILE *fp; fp=fopen("trial", "r"); return 0; }
Options- A. The first character in the file
- B. A structure which contains a char pointer which points to the first character of a file.
- C. The name of the file.
- D. The last character in the file. Discuss
Correct Answer: A structure which contains a char pointer which points to the first character of a file.
Explanation:
The fp is a structure which contains a char pointer which points to the first character of a file.- 6. 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"
- 7. 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.
- 8. 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
- 9. 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".
- 10. 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
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- 1. What will be the output of the program (in Turbo C)?