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- Question
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.
- Correct Answer
- A structure which contains a char pointer which points to the first character of a file.
ExplanationThe fp is a structure which contains a char pointer which points to the first character of a file.
Input / Output problems
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- 1. Consider the following program and what will be content of t?
#include<stdio.h> int main() { FILE *fp; int t; fp = fopen("DUMMY.C", "w"); t = fileno(fp); printf("%d\n", t); return 0; }
Options- A. size of "DUMMY.C" file
- B. The handle associated with "DUMMY.C" file
- C. Garbage value
- D. Error in fileno() Discuss
Correct Answer: The handle associated with "DUMMY.C" file
Explanation:
fp = fopen("DUMMY.C", "w"); A file DUMMY.C is opened in write mode and returns the file pointer to fpt = fileno(fp); returns the handle for the fp stream and it stored in the variable t
printf("%d\n", t); It prints the handle number.
- 2. What is the purpose of "rb" in fopen() function used below in the code?
FILE *fp; fp = fopen("source.txt", "rb");
Options- A. open "source.txt" in binary mode for reading
- B. open "source.txt" in binary mode for reading and writing
- C. Create a new file "source.txt" for reading and writing
- D. None of above Discuss
Correct Answer: open "source.txt" in binary mode for reading
Explanation:
The file source.txt will be opened in the binary mode.- 3. Which files will get closed through the fclose() in the following program?
#include<stdio.h> int main() { FILE *fs, *ft, *fp; fp = fopen("A.C", "r"); fs = fopen("B.C", "r"); ft = fopen("C.C", "r"); fclose(fp, fs, ft); return 0; }
Options- A. "A.C" "B.C" "C.C"
- B. "B.C" "C.C"
- C. "A.C"
- D. Error in fclose() Discuss
Correct Answer: Error in fclose()
Explanation:
Extra parameter in call to fclose().- 4. In a file contains the line "I am a boy\r\n" then on reading this line into the array str using fgets(). What will str contain?
Options- A. "I am a boy\r\n\0"
- B. "I am a boy\r\0"
- C. "I am a boy\n\0"
- D. "I am a boy" Discuss
Correct Answer: "I am a boy\n\0"
Explanation:
Declaration: char *fgets(char *s, int n, FILE *stream);fgets reads characters from stream into the string s. It stops when it reads either n - 1 characters or a newline character, whichever comes first.
Therefore, the string str contain "I am a boy\n\0"
- 5. To print out a and b given below, which of the following printf() statement will you use?
#include<stdio.h> float a=3.14; double b=3.14;
Options- A. printf("%f %lf", a, b);
- B. printf("%Lf %f", a, b);
- C. printf("%Lf %Lf", a, b);
- D. printf("%f %Lf", a, b); Discuss
Correct Answer: printf("%f %lf", a, b);
Explanation:
To print a float value, %f is used as format specifier.To print a double value, %lf is used as format specifier.
Therefore, the answer is printf("%f %lf", a, b);
- 6. 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.
- 7. 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).- 8. 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 **'
- 9. 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
- 10. 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 Discuss
Correct Answer: 20
Explanation:
Step 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".
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- 1. Consider the following program and what will be content of t?