CODEWITHSUNDEEP
clogicbit-manipulationboolean-logicconditional-statements
amorimluc
amorimluc

Reputation: 1719

Why does this logical/bitwise operation return 1?

I don't understand why the following returns 1:

0x0102 && (0xff << 8)

From my understanding, bit-shifting 0xff by eight to the left results in 0x00; and anding that with anything would result in 0x00. Where am I wrong?

Upvotes: 1

Views: 291

Answers (5)

Theo20185
Theo20185

Reputation: 56

See In C Left shift (char) 0xFF by 8 and cast it to int

0xff << 8 yields -256. Since both sides of the AND logic gate are not 0, it will return true, or 1.

Upvotes: 0

Richard Schneider
Richard Schneider

Reputation: 35477

You are using logical AND, not bitwise AND. Replace && with &.

Try

0x0102 & (0xff << 8)

From my understanding, bit-shifting 0xff by eight to the left results in 0x00; and anding that with anything would result in 0x00. Where am I wrong?

0xFF is an int. Typically it is 16 or 32 bits depending on compiler settings. So shift left by 8 bits results in 0xFF00.

Upvotes: 3

sidyll
sidyll

Reputation: 59287

An integer constant is not composed by simply 8 bits, 0xFF << 8 is 0xFF00.

Upvotes: 1

Cornstalks
Cornstalks

Reputation: 38218

(0xff << 8) results in 0xff00. Bitwise ANDing that with 0x0102 will yield 0x0100 (which is true).

But, you aren't doing a bitwise AND. && is logical AND. & is bitwise AND. What you're basically doing is if 0x0102 is true AND if (0xff << 8) is true, which, since the arguments are non-zero, results in true (which gets converted to 1).

Upvotes: 3

Mitch Wheat
Mitch Wheat

Reputation: 300559

&& is logical AND. You want bitwise AND &, e.g.

0x0102 & (0xff << 8)

Also, bit-shifting 0xff by eight to the left results in 0xff00, since bitwise arithmetic is integer rather than byte.

Upvotes: 2

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