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- Question
What will be the output of the program?
#include<stdio.h> int main() { const c = -11; const int d = 34; printf("%d, %d\n", c, d); return 0; }
Options- A. Error
- B. -11, 34
- C. 11, 34
- D. None of these
- Correct Answer
- -11, 34
ExplanationStep 1: const c = -11; The constant variable 'c' is declared and initialized to value "-11".Step 2: const int d = 34; The constant variable 'd' is declared as an integer and initialized to value '34'.
Step 3: printf("%d, %d\n", c, d); The value of the variable 'c' and 'd' are printed.
Hence the output of the program is -11, 34
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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