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- Question
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
- Correct Answer
- Error: unknown max in declaration/Constant expression required
ExplanationStep 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.
More questions
- 1. It is necessary that a header files should have a .h extension?
Options- A. Yes
- B. No Discuss
Correct Answer: No
Explanation:
No, the header files have any kind of extension.- 2. Bitwise | can be used to set multiple bits in number.
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
- 3. Bitwise & can be used to check if a bit in number is set or not.
Options- A. True
- B. False Discuss
Correct Answer: True
- 4. A pointer union CANNOT be created
Options- A. Yes
- B. No Discuss
Correct Answer: No
- 5. What will be the output of the program?
#include<stdio.h> #include<stdlib.h> int main() { union test { int i; float f; char c; }; union test *t; t = (union test *)malloc(sizeof(union test)); t->f = 10.10f; printf("%f", t->f); return 0; }
Options- A. 10
- B. Garbage value
- C. 10.100000
- D. Error Discuss
Correct Answer: 10.100000
- 6. What will be the output of the program?
#include<stdio.h> int main() { int x=4, y, z; y = --x; z = x--; printf("%d, %d, %d\n", x, y, z); return 0; }
Options- A. 4, 3, 3
- B. 4, 3, 2
- C. 3, 3, 2
- D. 2, 3, 3 Discuss
Correct Answer: 2, 3, 3
Explanation:
Step 1: int x=4, y, z; here variable x, y, z are declared as an integer type and variable x is initialized to 4.
Step 2: y = --x; becomes y = 3; because (--x) is pre-decrement operator.
Step 3: z = x--; becomes z = 3;. In the next step variable x becomes 2, because (x--) is post-decrement operator.
Step 4: printf("%d, %d, %d\n", x, y, z); Hence it prints "2, 3, 3".
- 7. Functions cannot return a floating point number
Options- A. Yes
- B. No Discuss
Correct Answer: No
Explanation:
A function can return floating point value.Example:
#include <stdio.h> float sub(float, float); /* Function prototype */ int main() { float a = 4.5, b = 3.2, c; c = sub(a, b); printf("c = %f\n", c); return 0; } float sub(float a, float b) { return (a - b); }Output:
c = 1.300000- 8. What will be the output of the program?
#include<stdio.h> int main() { int a=100, b=200, c; c = (a == 100 || b > 200); printf("c=%d\n", c); return 0; }
Options- A. c=100
- B. c=200
- C. c=1
- D. c=300 Discuss
Correct Answer: c=1
Explanation:
Step 1: int a=100, b=200, c;
Step 2: c = (a == 100 || b > 200);
becomes c = (100 == 100 || 200 > 200);
becomes c = (TRUE || FALSE);
becomes c = (TRUE);(ie. c = 1)
Step 3: printf("c=%d\n", c); It prints the value of variable i=1
Hence the output of the program is '1'(one).- 9. What will be the output of the program?
#include<stdio.h> int main() { int x=55; printf("%d, %d, %d\n", x<=55, x=40, x>=10); return 0; }
Options- A. 1, 40, 1
- B. 1, 55, 1
- C. 1, 55, 0
- D. 1, 1, 1 Discuss
Correct Answer: 1, 40, 1
Explanation:
Step 1: int x=55; here variable x is declared as an integer type and initialized to '55'.
Step 2: printf("%d, %d, %d\n", x<=55, x=40, x>=10);
In printf the execution of expressions is from Right to Left.
here x>=10 returns TRUE hence it prints '1'.
x=40 here x is assigned to 40 Hence it prints '40'.
x<=55 returns TRUE. hence it prints '1'.
Step 3: Hence the output is "1, 40, 1".- 10. What will be the output of the program?
#include<stdio.h> int main() { void fun(char*); char a[100]; a[0] = 'A'; a[1] = 'B'; a[2] = 'C'; a[3] = 'D'; fun(&a[0]); return 0; } void fun(char *a) { a++; printf("%c", *a); a++; printf("%c", *a); }
Options- A. AB
- B. BC
- C. CD
- D. No output Discuss
Correct Answer: BC
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