Confusion in evaluating == and &

Question

Why does the below statement:

int a = 10;
a&8 == 8;

return false (0) ?

I know that == has more precedence over &, but still it should check whether 8 == 8 and it should have evaluated true, but its evaluating as false.

Can anyone please help me through reasoning.

phoxis · Accepted Answer · 2013-08-06T08:52:58.087

23

a & 8 == 8 is a & (8 == 8) because == has higher precedence than & (here)

So 8 == 8 is 1 and a = 10 which is 1010 in binary

So

  00000001
& 00001010
  --------
  00000000

Interesting fact: Although false is a zero and true is any non-zero value in C Language. But, if the equality operator evaluates to be true it guarantees to evaluate to the value 1 and 0 otherwise. Referrer C99 Section 6.5.9 Paragraph 3

edited Aug 06 '13 at 08:52

answered Aug 06 '13 at 08:42

phoxis

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jxh · Answer 2 · 2013-08-06T09:09:31.913

The result of 8 == 8 when converted to int is 1.

a & 8 == 8 thus becomes a & 1, which is true if the least significant bit of a is set. This would only be true of a is odd.

Since a is 10, then a & 1 is 0, since 10 is even.

The & in this case is a bitwise-and and not a logical-and. A logical-and would have resulted in the expression a && 8 == 8 being true, since both sides of the && operator would have been non-zero.

However, the bitwise-and produces a result in which "each bit in the result is set if and only if each of the corresponding bits in the converted operands is set" (C.11 §6.5.10 ¶4). For an even number, the 1s bit is 0, so the result of bitwise-and with 1 is 0.

Since the result of the bitwise-and is 0, the expression is treated as false.

score 5 · Answer 3 · answered Aug 06 '13 at 08:42

5

Because 10 is 1010 in binary and result of 8 == 8 is 1. so 0001 & 1010 is false.

answered Aug 06 '13 at 08:42

Jeyaram

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score 3 · Answer 4 · answered Aug 06 '13 at 08:44

3

a&8 == 8,

= a&1 // because 8 == 8, is true

= 0 // 1010 & 1 = 0

answered Aug 06 '13 at 08:44

mohit

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score 2 · Answer 5 · answered Aug 06 '13 at 08:43

2

Because 8 == 8 is true, which in C equals 1, 1 & 10 equals 0, which in C is false.

answered Aug 06 '13 at 08:43

jbr

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Joni · Answer 6 · 2013-08-06T10:57:35.067

1

a&8 == 8 is parsed as a & (8==8), which is the same as a & 1. Since a=10 is even its binary representation has a 0 in the units position, so the result of a & 1 is 0.

edited Aug 06 '13 at 10:57

answered Aug 06 '13 at 08:42

Joni

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score 1 · Answer 7 · answered Aug 06 '13 at 08:47

1

As you are performing bitwise and(&) operation, a&1(8==8) is true,i.e 10 & 1 results to all 0's. O represents false.

00000000 00001010 & 00000000 00001000 = 00000000 00000000 = False (assumed int is 16 bit)

answered Aug 06 '13 at 08:47

Raju

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Confusion in evaluating == and &

7 Answers7