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- Question
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
- Correct Answer
- Error: cannot convert parameter 1 from 'const int **' to 'int **'
- 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 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
- 7. 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
- 8. 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
- 9. 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
- 10. 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
Constants problems
Search Results
Correct Answer: Read and Write
Explanation:
Correct Answer: Trh
Explanation:
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.
Correct Answer: A structure which contains a char pointer which points to the first character of a file.
Explanation:
Correct Answer: The handle associated with "DUMMY.C" file
Explanation:
t = 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.
Correct Answer: open "source.txt" in binary mode for reading
Explanation:
Correct Answer: Before modification arr[3] = 4
After modification arr[3] = 10
Explanation:
arr[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
Correct Answer: 20
Explanation:
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".
Correct Answer: 128
Explanation:
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"
Correct Answer: Error: ++needs a value
Explanation:
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.
Correct Answer: i= FFE4 ptr=10 j=FFE2 ptr=20
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