how to add 896 using bitwise operation?
Question: how to express b = a + 896 ... (1)
, and a = b - 896 ... (2)
with bitwise operation instead of addition/subtraction?
Motivation (not related to this question):
To convert between IEEE754 32-bit and 64-bit floating point's exponent value, I need to add/subtract 896, because of the equations:
127 + exponent_value = fp32_biased_exponent // bit 30-23
1023 + exponent_value = fp64_biased_exponent // bit 62-52
Therefore b = a + 896 ... (1)
, and a = b - 896 ... (2)
.
In program it looks like this
uint64_t b = ((uint64_t)f32.biased_exponent + 896) & 0x7ff;
uint64_t a = (uint32_t)(((uint64_t)fp64.biased_exponent - 896) & 0xff;
floating-point bit-manipulation ieee-754
add a comment |
Question: how to express b = a + 896 ... (1)
, and a = b - 896 ... (2)
with bitwise operation instead of addition/subtraction?
Motivation (not related to this question):
To convert between IEEE754 32-bit and 64-bit floating point's exponent value, I need to add/subtract 896, because of the equations:
127 + exponent_value = fp32_biased_exponent // bit 30-23
1023 + exponent_value = fp64_biased_exponent // bit 62-52
Therefore b = a + 896 ... (1)
, and a = b - 896 ... (2)
.
In program it looks like this
uint64_t b = ((uint64_t)f32.biased_exponent + 896) & 0x7ff;
uint64_t a = (uint32_t)(((uint64_t)fp64.biased_exponent - 896) & 0xff;
floating-point bit-manipulation ieee-754
Try to have a look at stackoverflow.com/questions/15327169/…
– Giovanni Cerretani
Nov 13 '18 at 8:38
add a comment |
Question: how to express b = a + 896 ... (1)
, and a = b - 896 ... (2)
with bitwise operation instead of addition/subtraction?
Motivation (not related to this question):
To convert between IEEE754 32-bit and 64-bit floating point's exponent value, I need to add/subtract 896, because of the equations:
127 + exponent_value = fp32_biased_exponent // bit 30-23
1023 + exponent_value = fp64_biased_exponent // bit 62-52
Therefore b = a + 896 ... (1)
, and a = b - 896 ... (2)
.
In program it looks like this
uint64_t b = ((uint64_t)f32.biased_exponent + 896) & 0x7ff;
uint64_t a = (uint32_t)(((uint64_t)fp64.biased_exponent - 896) & 0xff;
floating-point bit-manipulation ieee-754
Question: how to express b = a + 896 ... (1)
, and a = b - 896 ... (2)
with bitwise operation instead of addition/subtraction?
Motivation (not related to this question):
To convert between IEEE754 32-bit and 64-bit floating point's exponent value, I need to add/subtract 896, because of the equations:
127 + exponent_value = fp32_biased_exponent // bit 30-23
1023 + exponent_value = fp64_biased_exponent // bit 62-52
Therefore b = a + 896 ... (1)
, and a = b - 896 ... (2)
.
In program it looks like this
uint64_t b = ((uint64_t)f32.biased_exponent + 896) & 0x7ff;
uint64_t a = (uint32_t)(((uint64_t)fp64.biased_exponent - 896) & 0xff;
floating-point bit-manipulation ieee-754
floating-point bit-manipulation ieee-754
edited Nov 13 '18 at 4:54
Leedehai
asked Nov 13 '18 at 4:49
LeedehaiLeedehai
1,016619
1,016619
Try to have a look at stackoverflow.com/questions/15327169/…
– Giovanni Cerretani
Nov 13 '18 at 8:38
add a comment |
Try to have a look at stackoverflow.com/questions/15327169/…
– Giovanni Cerretani
Nov 13 '18 at 8:38
Try to have a look at stackoverflow.com/questions/15327169/…
– Giovanni Cerretani
Nov 13 '18 at 8:38
Try to have a look at stackoverflow.com/questions/15327169/…
– Giovanni Cerretani
Nov 13 '18 at 8:38
add a comment |
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Try to have a look at stackoverflow.com/questions/15327169/…
– Giovanni Cerretani
Nov 13 '18 at 8:38