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- Question
What will be the output of the program?
#include<stdio.h> int main() { int x, y, z; x=y=z=1; z = ++x || ++y && ++z; printf("x=%d, y=%d, z=%d\n", x, y, z); return 0; }
Options- A. x=2, y=1, z=1
- B. x=2, y=2, z=1
- C. x=2, y=2, z=2
- D. x=1, y=2, z=1
- Correct Answer
- x=2, y=1, z=1
ExplanationStep 1: x=y=z=1; here the variables x ,y, z are initialized to value '1'.Step 2: z = ++x || ++y && ++z; becomes z = ( (++x) || (++y && ++z) ). Here ++x becomes 2. So there is no need to check the other side because ||(Logical OR) condition is satisfied.(z = (2 || ++y && ++z)). There is no need to process ++y && ++z. Hence it returns '1'. So the value of variable z is '1'
Step 3: printf("x=%d, y=%d, z=%d\n", x, y, z); It prints "x=2, y=1, z=1". here x is increemented in previous step. y and z are not increemented.
More questions
- 1. Point out the error in the following program.
#include<stdio.h> #include<stdarg.h> void varfun(int n, ...); int main() { varfun(3, 7, -11.2, 0.66); return 0; } void varfun(int n, ...) { float *ptr; int num; va_start(ptr, n); num = va_arg(ptr, int); printf("%d", num); }
Options- A. Error: too many parameters
- B. Error: invalid access to list member
- C. Error: ptr must be type of va_list
- D. No error Discuss
Correct Answer: Error: ptr must be type of va_list
- 2. Macros with arguments are allowed
Options- A. True
- B. False Discuss
Correct Answer: True
Explanation:
True, A macro may have arguments.Example: #define CUBE(X)(X*X*X)
- 3. 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
- 4. 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 Discuss
Correct Answer: Error
Explanation:
Step 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;
- 5. 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 **'
- 6. Point out the error in the program.
#include<stdio.h> int main() { const int x; x=128; printf("%d\n", x); return 0; }
Options- A. Error: unknown data type const int
- B. Error: const variable have been initialised when declared.
- C. Error: stack overflow in x
- D. No error Discuss
Correct Answer: Error: const variable have been initialised when declared.
Explanation:
A const variable has to be initialized when it is declared. later assigning the value to the const variable will result in an error "Cannot modify the const object".Hence Option B is correct
- 7. Point out the error in the program (in Turbo-C).
#include<stdio.h> #define MAX 128 int main() { const int max=128; char array[max]; char string[MAX]; array[0] = string[0] = 'A'; printf("%c %c\n", array[0], string[0]); return 0; }
Options- A. Error: unknown max in declaration/Constant expression required
- B. Error: invalid array string
- C. None of above
- D. No error. It prints A A Discuss
Correct Answer: Error: unknown max in declaration/Constant expression required
Explanation:
Step 1: A macro named MAX is defined with value 128Step 2: const int max=128; The constant variable max is declared as an integer data type and it is initialized with value 128.
Step 3: char array[max]; This statement reports an error "constant expression required". Because, we cannot use variable to define the size of array.
To avoid this error, we have to declare the size of an array as static. Eg. char array[10]; or use macro char array[MAX];
Note: The above program will print A A as output in Unix platform.
- 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 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.
- 10. Point out the error in 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", e1.name); e1.age=85; printf("%d", e1.age); printf("%f", e1.salary); return 0; }
Options- A. Error: RValue required
- B. Error: cannot modify const object
- C. Error: LValue required in strcpy
- D. No error Discuss
Correct Answer: Error: cannot modify const object
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- 1. Point out the error in the following program.