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- Question
What will be the output of the program?
#include<stdio.h> int main() { printf("%d %d\n", 32<<1, 32<<0); printf("%d %d\n", 32<<-1, 32<<-0); printf("%d %d\n", 32>>1, 32>>0); printf("%d %d\n", 32>>-1, 32>>-0); return 0; }
Options- A. Garbage values
- B. 64 32
0 32
16 32
0 32 - C. All zeros
- D. 8 0
0 0
32 0
0 16 - Correct Answer
- 64 32
0 32
16 32
0 32
- 1. What will be the output of the program in 16 bit platform (Turbo C under DOS)?
#include<stdio.h> int main() { extern int i; i = 20; printf("%d\n", sizeof(i)); return 0; }
Options- A. 2
- B. 4
- C. vary from compiler
- D. Linker Error : Undefined symbol 'i' Discuss
- 2. 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
- 3. In the following program how long will the for loop get executed?
#include<stdio.h> int main() { int i=5; for(;scanf("%s", &i); printf("%d\n", i)); return 0; }
Options- A. The for loop would not get executed at all
- B. The for loop would get executed only once
- C. The for loop would get executed 5 times
- D. The for loop would get executed infinite times Discuss
- 4. 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
- 5. When we mention the prototype of a function?
Options- A. Defining
- B. Declaring
- C. Prototyping
- D. Calling Discuss
- 6. What will be the output of the program?
#include<stdio.h> #define PRINT(i) printf("%d,",i) int main() { int x=2, y=3, z=4; PRINT(x); PRINT(y); PRINT(z); return 0; }
Options- A. 2, 3, 4,
- B. 2, 2, 2,
- C. 3, 3, 3,
- D. 4, 4, 4, Discuss
- 7. What will be the output of the program
#include<stdio.h> void fun(int); int main(int argc) { printf("%d ", argc); fun(argc); return 0; } void fun(int i) { if(i!=4) main(++i); }
Options- A. 1 2 3
- B. 1 2 3 4
- C. 2 3 4
- D. 1 Discuss
- 8. Declare the following statement?
"A pointer to a function which receives an int pointer and returns float pointer".
Options- A.
float *(ptr)*int; - B.
float *(*ptr)(int) - C.
float *(*ptr)(int*) - D.
float (*ptr)(int)
Discuss
- 9. What do the following declaration signify?
char **argv;
Options- A. argv is a pointer to pointer.
- B. argv is a pointer to a char pointer.
- C. argv is a function pointer.
- D. argv is a member of function pointer. Discuss
- 10. Would the following typedef work?
typedef #include l;
Options- A. Yes
- B. No Discuss
More questions
Correct Answer: Linker Error : Undefined symbol 'i'
Explanation:
Linker Error : Undefined symbol
'i'
The statement extern int i specifies to the compiler that the memory for 'i' is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name 'i' is available in any other program with memory space allocated for it. Hence a linker error has occurred.
The statement extern int i specifies to the compiler that the memory for 'i' is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name 'i' is available in any other program with memory space allocated for it. Hence a linker error has occurred.
Correct Answer: Read and Write
Explanation:
r+ Open an existing file for update (reading and writing).
Correct Answer: The for loop would get executed infinite times
Explanation:
During the for loop execution scanf() ask input and then printf() prints that given input. This process will be continued repeatedly because, scanf() returns the number of input given, the condition is always true(user gives a input means it reurns '1').
Hence this for loop would get executed infinite times.
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"
Correct Answer: Declaring
Explanation:
A function prototype in C or C++ is a declaration of a function that omits the function body but does specify the function's name, argument types and return type.
While a function definition specifies what a function does, a function prototype can be thought of as specifying its interface.
Correct Answer: 2, 3, 4,
Explanation:
The macro PRINT(i) print("%d,", i); prints the given variable value in an integer format.
Step 1: int x=2, y=3, z=4; The variable x, y, z are declared as an integer type and initialized to 2, 3, 4 respectively.
Step 2: PRINT(x); becomes printf("%d,",x). Hence it prints '2'.
Step 3: PRINT(y); becomes printf("%d,",y). Hence it prints '3'.
Step 4: PRINT(z); becomes printf("%d,",z). Hence it prints '4'.
Hence the output of the program is 2, 3, 4.
Correct Answer: 1 2 3 4
Correct Answer:
float *(*ptr)(int*)Correct Answer: argv is a pointer to a char pointer.
Correct Answer: No
Explanation:
Because
typedef goes to work after preprocessing.
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