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Thus, if both bits in the compared position are 1, the bit in the resulting binary representation is 1 (1 × 1 = 1); otherwise, the result is 0 (1 × 0 = 0 and 0 × 0 = 0). For example: 0101 (decimal 5) AND 0011 (decimal 3) = 0001 (decimal 1) The operation may be used to determine whether a particular bit is set (1) or cleared (0). For example ...
For example, a string of three bits can represent up to eight distinct values as illustrated in Table 1. As the number of bits composing a string increases, the number of possible 0 and 1 combinations increases exponentially. A single bit allows only two value-combinations, two bits combined can make four separate values, three bits for eight ...
Bit indexing correlates to the positional notation of the value in base 2. For this reason, bit index is not affected by how the value is stored on the device, such as the value's byte order. Rather, it is a property of the numeric value in binary itself. This is often utilized in programming via bit shifting: A value of 1 << n corresponds to ...
Sign bit: 1 bit; Exponent width: 19 bits; Significand precision: 237 bits (236 explicitly stored) The format is written with an implicit lead bit with value 1 unless the exponent is all zeros. Thus only 236 bits of the significand appear in the memory format, but the total precision is 237 bits (approximately 71 decimal digits: log 10 (2 237 ...
Thus only 23 fraction bits of the significand appear in the memory format, but the total precision is 24 bits (equivalent to log 10 (2 24) ≈ 7.225 decimal digits) for normal values; subnormals have gracefully degrading precision down to 1 bit for the smallest non-zero value. The bits are laid out as follows: The real value assumed by a given ...
The format is written with the significand having an implicit integer bit of value 1 (except for special data, see the exponent encoding below). With the 52 bits of the fraction (F) significand appearing in the memory format, the total precision is therefore 53 bits (approximately 16 decimal digits, 53 log 10 (2) ≈ 15.955). The bits are laid ...
→ -1 load the int value −1 onto the stack iconst_0 03 0000 0011 → 0 load the int value 0 onto the stack iconst_1 04 0000 0100 → 1 load the int value 1 onto the stack iconst_2 05 0000 0101 → 2 load the int value 2 onto the stack iconst_3 06 0000 0110 → 3 load the int value 3 onto the stack iconst_4 07 0000 0111 → 4
The 8 decimal values whose digits are all 8s or 9s have four codings each. The bits marked x in the table above are ignored on input, but will always be 0 in computed results. (The 8 × 3 = 24 non-standard encodings fill in the gap from 10 3 = 1000 and 2 10 - 1 = 1023.)