Binary addition which disapproves carrying
up vote
1
down vote
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I'm currently working with binary addition which returns False if there is carrying. My current code is :
def binaryadd(one, other):
str_1, str_2 = str(one), str(other)
for a,b in zip(str_1[::-1], str_2[::-1]):
if a == b == '1':
return False
return int(bin(rev_bin(one) + rev_bin(other))[2:])
so 10111 + 1000 would return 11111, 10110 + 1011 would return False. I think there would be more efficient codes ;such as to check overflows in addition, but I'm wondering which code could do it. Is there any better way to do it?
python binary addition
asked Nov 11 at 15:33
ILoveG11
325
add a comment |
up vote
1
down vote
favorite
I'm currently working with binary addition which returns False if there is carrying. My current code is :
def binaryadd(one, other):
str_1, str_2 = str(one), str(other)
for a,b in zip(str_1[::-1], str_2[::-1]):
if a == b == '1':
return False
return int(bin(rev_bin(one) + rev_bin(other))[2:])
so 10111 + 1000 would return 11111, 10110 + 1011 would return False. I think there would be more efficient codes ;such as to check overflows in addition, but I'm wondering which code could do it. Is there any better way to do it?
python binary addition
asked Nov 11 at 15:33
ILoveG11
325
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm currently working with binary addition which returns False if there is carrying. My current code is :
def binaryadd(one, other):
str_1, str_2 = str(one), str(other)
for a,b in zip(str_1[::-1], str_2[::-1]):
if a == b == '1':
return False
return int(bin(rev_bin(one) + rev_bin(other))[2:])
so 10111 + 1000 would return 11111, 10110 + 1011 would return False. I think there would be more efficient codes ;such as to check overflows in addition, but I'm wondering which code could do it. Is there any better way to do it?
python binary addition
asked Nov 11 at 15:33
ILoveG11
325
I'm currently working with binary addition which returns False if there is carrying. My current code is :
def binaryadd(one, other):
str_1, str_2 = str(one), str(other)
for a,b in zip(str_1[::-1], str_2[::-1]):
if a == b == '1':
return False
return int(bin(rev_bin(one) + rev_bin(other))[2:])
so 10111 + 1000 would return 11111, 10110 + 1011 would return False. I think there would be more efficient codes ;such as to check overflows in addition, but I'm wondering which code could do it. Is there any better way to do it?
python binary addition
python binary addition
asked Nov 11 at 15:33
ILoveG11
325
asked Nov 11 at 15:33
ILoveG11
325
asked Nov 11 at 15:33
ILoveG11
325
asked Nov 11 at 15:33
ILoveG11
325
asked Nov 11 at 15:33
ILoveG11
325
325
add a comment |
add a comment |
1 Answer
1
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0
down vote
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There is carry from the moment there is a "column" for which both operands have a 1, since in that case we generate a carry. A carry can also propagate, but only if there was already a carry generated.
We can check this with a bitwise and (&). In case there is no carry, we can just is a bitwise or (|):
def binaryadd(one, other):
if one & other:
return False
return one | otheror in a one-liner:
def binaryadd(one, other):
return not bool(one & other) and one | otheranswered Nov 11 at 15:36
Willem Van Onsem
142k16134225
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
There is carry from the moment there is a "column" for which both operands have a 1, since in that case we generate a carry. A carry can also propagate, but only if there was already a carry generated.
We can check this with a bitwise and (&). In case there is no carry, we can just is a bitwise or (|):
def binaryadd(one, other):
if one & other:
return False
return one | otheror in a one-liner:
def binaryadd(one, other):
return not bool(one & other) and one | otheranswered Nov 11 at 15:36
Willem Van Onsem
142k16134225
add a comment |
up vote
0
down vote
accepted
There is carry from the moment there is a "column" for which both operands have a 1, since in that case we generate a carry. A carry can also propagate, but only if there was already a carry generated.
We can check this with a bitwise and (&). In case there is no carry, we can just is a bitwise or (|):
def binaryadd(one, other):
if one & other:
return False
return one | otheror in a one-liner:
def binaryadd(one, other):
return not bool(one & other) and one | otheranswered Nov 11 at 15:36
Willem Van Onsem
142k16134225
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
There is carry from the moment there is a "column" for which both operands have a 1, since in that case we generate a carry. A carry can also propagate, but only if there was already a carry generated.
We can check this with a bitwise and (&). In case there is no carry, we can just is a bitwise or (|):
def binaryadd(one, other):
if one & other:
return False
return one | otheror in a one-liner:
def binaryadd(one, other):
return not bool(one & other) and one | otheranswered Nov 11 at 15:36
Willem Van Onsem
142k16134225
There is carry from the moment there is a "column" for which both operands have a 1, since in that case we generate a carry. A carry can also propagate, but only if there was already a carry generated.
We can check this with a bitwise and (&). In case there is no carry, we can just is a bitwise or (|):
def binaryadd(one, other):
if one & other:
return False
return one | otheror in a one-liner:
def binaryadd(one, other):
return not bool(one & other) and one | otheranswered Nov 11 at 15:36
Willem Van Onsem
142k16134225
edited Nov 11 at 15:41
answered Nov 11 at 15:36
Willem Van Onsem
142k16134225
answered Nov 11 at 15:36
Willem Van Onsem
142k16134225
answered Nov 11 at 15:36
Willem Van Onsem
142k16134225
142k16134225
add a comment |
add a comment |
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