C++ Basics via @QuizBot
🎲 Quiz 'C++ Basics chapter 1 & 2' 🖊 10 questions · ⏱ 1 min
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3. Which of the following is NOT a characteristic of a well-defined algorithm?
a) Unambiguous instructions.
b) A finite number of steps.
c) The use of random numbers at every step.
d) A clear starting and ending point.
Answer C.
Use of random numbers at every step make it the algorithm unpredictable and unreliable.
a) Unambiguous instructions.
b) A finite number of steps.
c) The use of random numbers at every step.
d) A clear starting and ending point.
Answer C.
Use of random numbers at every step make it the algorithm unpredictable and unreliable.
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Question:
Given:
int x = 7;
int y = 3;
int z = x & y;
What is the value of z?
Explanation:
The question is asking us to determine the value of z after performing a bitwise AND operation between x and y.
Remember: z = x & y means z will store the result of a bitwise AND between x and y.
The bitwise AND operation works by comparing the corresponding bits of two numbers. It takes the bits one by one and the rules is:
• If both bits are 1, the result is 1.
• Otherwise, the result is 0.
First, we need to convert x (7) and y (3) into their binary representations:
• 7 = 0111 (Binary)
• 3 = 0011 (Binary)
Now, let's perform the bitwise AND operation:
0111 (x = 7)
& 0011 (y = 3)
0011 (z)
• Bit 0 (rightmost): 1 AND 1 = 1
• Bit 1: 1 AND 1 = 1
• Bit 2: 1 AND 0 = 0
• Bit 3 (leftmost): 0 AND 0 = 0
So, the result of the bitwise AND operation is 0011 in binary.
Finally, we convert the binary result (0011) back to decimal:
• 0011 (Binary) = 3 (Decimal)
Therefore, the value of z is 3.
Answer: z = 3👍5
QUIZ
Topic: Increment and Decrement
1. int a=5;
int x=++a + ++a;
cout<<x;
2. int a=5;
int b=6;
Cout<< ++a + b++;
Cout<<b;
3. int x, a, b, c;
a = 2;
b = 4;
c = 5;
x = a--+ b++ -++c;
cout<<x;
4. int x, a;
a = 2;
x = ++a *--a *++a;
cout<<x;
5. int x, a = 4;
x = ++a + ++a + ++a;
cout<<x;
Provide the output of each code snippets and justify your answer with a detailed explanation .
@ProgrammingWithFaith
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C++ Basics pinned «QUIZ Topic: Increment and Decrement 1. int a=5; int x=++a + ++a; cout<<x; 2. int a=5; int b=6; Cout<< ++a + b++; Cout<<b; 3. int x, a, b, c; a = 2; b = 4; c = 5; x = a--+ b++ -++c; cout<<x; 4. int x, a; a = 2; x = ++a *--a *++a; cout<<x;…»
C++ Basics
QUIZ Topic: Increment and Decrement 1. int a=5; int x=++a + ++a; cout<<x; 2. int a=5; int b=6; Cout<< ++a + b++; Cout<<b; 3. int x, a, b, c; a = 2; b = 4; c = 5; x = a--+ b++ -++c; cout<<x; 4. int x, a; a = 2; x = ++a *--a *++a; cout<<x;…
Hint:
One variable can't store two value at the same time.
C++ Basics
QUIZ Topic: Increment and Decrement 1. int a=5; int x=++a + ++a; cout<<x; 2. int a=5; int b=6; Cout<< ++a + b++; Cout<<b; 3. int x, a, b, c; a = 2; b = 4; c = 5; x = a--+ b++ -++c; cout<<x; 4. int x, a; a = 2; x = ++a *--a *++a; cout<<x;…
1.int a=5;
int x=++a + ++a;
(6+ ++a)
(7+7) coz one variable can't hold two different value , (6 and 7 ) at the same time so it become 7.
7+7=14
Answer 14
2.int a=5;
int b=6;
Cout<<++a +b++// post and pre increment .
6+6 // after this line b will be 7
Cout<<++a +b++; (6+6)
=12
Cout<<b;
=7
answer=12 and 7
3.int x,a,b,c;
a=2;
b=4;
c=5;
x=a-- + b++ - ++c;
(2+4)-6
=6-6=0
cout<<x;
Answer 0
4.int x,a;
a=2
x=++a *--a* ++a
(3*2) * ++a , but one variable can't hold two values at the same time.
So, (2*2)*++a
(4*3)=12
cout<<x;// 12
Answer 12
5.int x,a=4;
x=++a + ++a + ++a;
(++a + ++a) + ++a
(5+6) + ++a but we can't store two value at one variable at the same time.
So,
(6+6)+ ++a
(12)+ 7
Answer 19
int x=++a + ++a;
(6+ ++a)
(7+7) coz one variable can't hold two different value , (6 and 7 ) at the same time so it become 7.
7+7=14
Answer 14
2.int a=5;
int b=6;
Cout<<++a +b++// post and pre increment .
6+6 // after this line b will be 7
Cout<<++a +b++; (6+6)
=12
Cout<<b;
=7
answer=12 and 7
3.int x,a,b,c;
a=2;
b=4;
c=5;
x=a-- + b++ - ++c;
(2+4)-6
=6-6=0
cout<<x;
Answer 0
4.int x,a;
a=2
x=++a *--a* ++a
(3*2) * ++a , but one variable can't hold two values at the same time.
So, (2*2)*++a
(4*3)=12
cout<<x;// 12
Answer 12
5.int x,a=4;
x=++a + ++a + ++a;
(++a + ++a) + ++a
(5+6) + ++a but we can't store two value at one variable at the same time.
So,
(6+6)+ ++a
(12)+ 7
Answer 19
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C++ Basics
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