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What will be the output of the program?
#include<stdio.h> int main() { static int a[20]; int i = 0; a[i] = i ; printf("%d, %d, %d\n", a[0], a[1], i); return 0; }
Options- A. 1, 0, 1
- B. 1, 1, 1
- C. 0, 0, 0
- D. 0, 1, 0
- Correct Answer
- 0, 0, 0
ExplanationStep 1: static int a[20]; here variable a is declared as an integer type and static. If a variable is declared as static and it will be automatically initialized to value '0'(zero).Step 2: int i = 0; here vaiable i is declared as an integer type and initialized to '0'(zero).
Step 3: a[i] = i ; becomes a[0] = 0;
Step 4: printf("%d, %d, %d\n", a[0], a[1], i);
Here a[0] = 0, a[1] = 0(because all staic variables are initialized to '0') and i = 0.
Step 4: Hence the output is "0, 0, 0".
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- 1. What will be the output of the program?
#include<stdio.h> int main() { int i=-3, j=2, k=0, m; m = ++i || ++j && ++k; printf("%d, %d, %d, %d\n", i, j, k, m); return 0; }
Options- A. 2, 2, 0, 1
- B. 1, 2, 1, 0
- C. -2, 2, 0, 0
- D. -2, 2, 0, 1 Discuss
Correct Answer: -2, 2, 0, 1
Explanation:
Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.Step 2: m = ++i || ++j && ++k; here (++j && ++k;) this code will not get executed because ++i has non-zero value.
becomes m = -2 || ++j && ++k;
becomes m = TRUE || ++j && ++k; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of variable 'i' only increemented by '1'(one). The variable j,k are not increemented.
Hence the output is "-2, 2, 0, 1".
- 2. What will be the output of the program?
#include<stdio.h> int main() { int k, num=30; k = (num>5? (num <=10? 100 : 200): 500); printf("%d\n", num); return 0; }
Options- A. 200
- B. 30
- C. 100
- D. 500 Discuss
Correct Answer: 30
Explanation:
Step 1: int k, num=30; here variable k and num are declared as an integer type and variable num is initialized to '30'.
Step 2: k = (num>5 ? (num <=10 ? 100 : 200): 500); This statement does not affect the output of the program. Because we are going to print the variable num in the next statement. So, we skip this statement.
Step 3: printf("%d\n", num); It prints the value of variable num '30'
Step 3: Hence the output of the program is '30'
- 3. What will be the output of the program?
#include<stdio.h> int main() { int i=2; int j = i + (1, 2, 3, 4, 5); printf("%d\n", j); return 0; }
Options- A. 4
- B. 7
- C. 6
- D. 5 Discuss
Correct Answer: 7
Explanation:
Because, comma operator used in the expression i (1, 2, 3, 4, 5). The comma operator has left-right associativity. The left operand is always evaluated first, and the result of evaluation is discarded before the right operand is evaluated. In this expression 5 is the right most operand, hence after evaluating expression (1, 2, 3, 4, 5) the result is 5, which on adding to i results into 7.- 4. What will be the output of the program?
#include<stdio.h> int main() { int i=-3, j=2, k=0, m; m = ++i && ++j || ++k; printf("%d, %d, %d, %d\n", i, j, k, m); return 0; }
Options- A. 1, 2, 0, 1
- B. -3, 2, 0, 1
- C. -2, 3, 0, 1
- D. 2, 3, 1, 1 Discuss
Correct Answer: -2, 3, 0, 1
Explanation:
Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j are increemented by '1'(one).
Hence the output is "-2, 3, 0, 1".
- 5. What will be the output of the program?
#include<stdio.h> int main() { int x=12, y=7, z; z = x!=4 || y == 2; printf("z=%d\n", z); return 0; }
Options- A. z=0
- B. z=1
- C. z=4
- D. z=2 Discuss
Correct Answer: z=1
Explanation:
Step 1: int x=12, y=7, z; here variable x, y and z are declared as an integer and variable x and y are initialized to 12, 7 respectively.Step 2: z = x!=4 || y == 2;
becomes z = 12!=4 || 7 == 2;
then z = (condition true) || (condition false); Hence it returns 1. So the value of z=1.Step 3: printf("z=%d\n", z); Hence the output of the program is "z=1".
- 6. Assuming, integer is 2 byte, What will be the output of the program?
#include<stdio.h> int main() { printf("%x\n", -2<<2); return 0; }
Options- A. ffff
- B. 0
- C. fff8
- D. Error Discuss
Correct Answer: fff8
Explanation:
The integer value 2 is represented as 00000000 00000010 in binary system.
Negative numbers are represented in 2's complement method.
1's complement of 00000000 00000010 is 11111111 11111101 (Change all 0s to 1 and 1s to 0).
2's complement of 00000000 00000010 is 11111111 11111110 (Add 1 to 1's complement to obtain the 2's complement value).
Therefore, in binary we represent -2 as: 11111111 11111110.
After left shifting it by 2 bits we obtain: 11111111 11111000, and it is equal to "fff8" in hexadecimal system.- 7. 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".
- 8. What will be the output of the program?
#include<stdio.h> int main() { int i=3; i = i++; printf("%d\n", i); return 0; }
Options- A. 3
- B. 4
- C. 5
- D. 6 Discuss
Correct Answer: 4
- 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() { 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).
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- 1. What will be the output of the program?