Expressions problems
- 1. 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.
- 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'
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=-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".
- 4. 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 Discuss
Correct Answer: 0, 0, 0
Explanation:
Step 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".
- 5. 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.
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.
- 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".
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. 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
- 8. 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".
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".
- 9. 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).
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).
- 10. 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, 3, 1, 1
- B. 2, 3, 1, 2
- C. 1, 2, 3, 1
- D. 3, 3, 1, 2 Discuss
Correct Answer: -2, 3, 1, 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 && 1;
becomes m = TRUE && 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,k are increemented by '1'(one).
Hence the output is "-2, 3, 1, 1".
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