Bitwise '&' operator

问题:

I am lacking some basic understanding in bitwise '&' operator.

5 = 101
4 = 100

So why the output of the below if condition is true cause and of bits 101 & 100 should be false:

#include <stdio.h>
main()
{
   if(5&4)
      printf("Yes\n");
}

回答1:

5 is 101

4 is 100

5 & 4 is not 0:

101 
100 &
↓↓↓
100

Problem solved ✓

---

Clarification:

In C, every non-zero value satisfies the if condition. Meaning, if you write:

if(-5) {
  if(100) {
     //Reachable code
  }
}

Whereas:

if(0) {
  destroyTheWorld(); //We are safe
}

回答2:

5 - 101
4 - 100
5&4 - 100

It is true.

回答3:

Because 0b100 & 0b101 equals 0b100 and the latter does not equal 0.

回答4:

Understanding bitwise operator truth tables is crucial. Consider the following, where A and B are inputs and Y is the output.

& (Bitwise And) When inputs A and B are true, output is true; otherwise output is false

A   B   Y
---------
0 | 0 | 0
0 | 1 | 0
1 | 0 | 0
1 | 1 | 1

| (Bitwise Or) When A or B or both inputs are true output is true; otherwise output is false

A   B   Y
---------
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 1

^ (Bitwise X-Or) When A and B are opposite states, output is true; otherwise output is false

A   B   Y
---------
0 | 0 | 0
0 | 1 | 1
1 | 0 | 1
1 | 1 | 0

! (Bitwise Not) Output is the opposite state of the input

A   Y
-----
0 | 1
1 | 0

Your Equation (5 & 4) == (0101 & 0100) == 0100 == 4 == true

  0101
& 0100
------
  0100

回答5:

0b101 & 0b100 = 0b100

or

5&4 = 4

and 4 is non-zero and prints Yes

回答6:

It enters the if condition. Because after the & operation it returns non-zero value. In C, for all non-zero value it's like returning true.