Here\'s the difference between IEEE 754 and IBM\'s HexadecimalFloating point Num
ID: 3612882 • Letter: H
Question
Here's the difference between IEEE 754 and IBM's HexadecimalFloating point Number 1. The 32 bit number is store in a binary form with 1 bitsign, 7 bit exponent and 24 bit mantissa. 2. The radix is 16 3. There's NO hidden bit i the mantissa 4. In the normalized form, the fist (leftmost ) digit of themantissa should be a nonzero hexadecimal digit. 5. The bias of the exponent is 64.For number -64, the number would need to be expressed as -0.25X 16^2. The normalized hexadeciaml radio floating pointrepresentation of the number Sign Exponent Mantissa 1 1000010 0100 0000 0000 0000 0000 0000
The arithmetic on this system is also similar to that ofIEEE754 in that last step of arithmetic operations is always apossible normalization of the exponent (E) and mantissa (M).
Question: What can really happen in the computation and what cannothappen in both operation???
For operation for Mantissa A. M << 1 B. M << 2 C. M << 3 D. M << 4 E. M << n, where n > 4 F. M >> 1 G. M >> 2 H. M >> 3 I. M >> 4 J. M >> n, where n > 4 K. Do nothing
For operation for Exponent a. E + 1 b. E + 2 c. E + 3 d. E + 4 e. E + n , where n > 4 f. E -1 g. E - 2 h. E - 3 i. E - 4 j. E - n, where n > 4 k. Do nothing
Can some experts tell me how to get a number in hexadecimalfloating point number as well??? 1. The 32 bit number is store in a binary form with 1 bitsign, 7 bit exponent and 24 bit mantissa. 2. The radix is 16 3. There's NO hidden bit i the mantissa 4. In the normalized form, the fist (leftmost ) digit of themantissa should be a nonzero hexadecimal digit. 5. The bias of the exponent is 64.
For number -64, the number would need to be expressed as -0.25X 16^2. The normalized hexadeciaml radio floating pointrepresentation of the number Sign Exponent Mantissa 1 1000010 0100 0000 0000 0000 0000 0000
The arithmetic on this system is also similar to that ofIEEE754 in that last step of arithmetic operations is always apossible normalization of the exponent (E) and mantissa (M).
Question: What can really happen in the computation and what cannothappen in both operation???
For operation for Mantissa A. M << 1 B. M << 2 C. M << 3 D. M << 4 E. M << n, where n > 4 F. M >> 1 G. M >> 2 H. M >> 3 I. M >> 4 J. M >> n, where n > 4 K. Do nothing
For operation for Exponent a. E + 1 b. E + 2 c. E + 3 d. E + 4 e. E + n , where n > 4 f. E -1 g. E - 2 h. E - 3 i. E - 4 j. E - n, where n > 4 k. Do nothing
Can some experts tell me how to get a number in hexadecimalfloating point number as well???
Explanation / Answer
please rate - thanks please remember cramster rule- 1 question per post. and this isreally 2 I don't understand part 1. Are you saying can you shift themantissa that # of bits, and can you add that amount to theexponent but here is how to do -64 in IBM hex floating point bit 0 (high order) is the sign bit, bits 1-7 are the exponentbias 64 and bits 8-31 are the mantissa the number is negative so the sign bit is1 64 in binary = 100 0000. to form the mantissa move the "decimal" point to the left in groupsof four, adding any 0's as filler as needed .01000000 1-zero needed as place holder, "decimal" moved 2 groups of 4 add 0's to the right until the mantissa has 24 digits mantissa = 0100 0000 0000 0000 00000000 we moved the "decimal" 2 places left, so the exponent will be 2.then bias it 64. 2+64=66 convert to binary 1000010 exponent = 100 0010 -64 in IBM hex floating point = 1100 0010 0100 0000 0000 0000 00000000 C 2 4 0 0 0 0 0