CODEWITHSUNDEEP
c#
Abanoub
Abanoub

Reputation: 3809

Get the leftmost 8 bits

how do I get the leftmost 8bits of array?

byte[] buf = new byte[] { 0x83, 0x00, 0xEE, 0x03, 0x26, 0x6D, 0x14, 0x00, 0xF1, 0x65, 0x27, 0x00, 0x19, 0x02, 0xD8, 0x0F, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xDB, 0xD7, 0x0F, 0x08, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x2B, 0x04, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x07, 0x00, 0x41, 0x00, 0x64, 0x00, 0xE4, 0x00, 0x00, 0x00, 0xDD, 0x0A, 0x18, 0x19, 0x00, 0x00, 0x79, 0x91, 0x87, 0x00, 0x00, 0x01, 0x00, 0xA8, 0x02, 0x00, 0x00, 0x64, 0x00, 0x00, 0x00, 0x34, 0x00, 0x6A, 0x18, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0xC2, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x00, 0x03, 0x0A, 0x7E, 0x42, 0x6C, 0x75, 0x65, 0x57, 0x61, 0x76, 0x65, 0x7E, 0x00, 0x09, 0x42, 0x6C, 0x61, 0x63, 0x6B, 0x44, 0x75, 0x73, 0x74 };

are the leftmost 8bits are buf[0] and the rightmost 8bits are buf[buf.Length]? or how do i get it? thanks.

Upvotes: 0

Views: 209

Answers (2)

KomodoDave
KomodoDave

Reputation: 7319

1 byte = 8 bits.

In the usual case, an array is a linear, contiguous block of memory. So strictly speaking, an array typically points to the first block in allocated memory space.

Leftmost (smallest index) element in array is leftmost element in memory, i.e. order is the same.

Answer: buf[0]

Upvotes: 1

thumbmunkeys
thumbmunkeys

Reputation: 20764

no, rightmost are

     buf[buf.Length-1]

leftmost are

     buf[0]

Upvotes: 2

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