Performance of OR EQUAL vs EQUAL assignment in a for-loop
This question is mostly directed at compiled programming languages. It is purely out of curiosity because I believe the gain of performance in using one of the two operators would be so very small.
Consider a for-loop where when you meet a certain condition, you want to store true in a boolean:
b = false
for i in 1..N:
if someCondition(i):
b = true
moreThatNeedsToBeDone(i)
endfor
Now consider the same for loop with OR EQUAL instead
b = false
for i in 1..N:
if someCondition(i):
b |= true
moreThatNeedsToBeDone(i)
endfor
If the condition is met more than once, would the latter be faster in theory? Or atleast, would it do fewer operations? In general, the OR EQUAL evaluates the variable and if it is true, then it doesn't do anything hence there is no extra assignment compared to the EQUAL operator where it would STORE true multiple times. But writing this, I realize the OR EQUAL adds an extra operation anyways in order to evaluate/read the current value of the variable. So which would be faster or have fewer operations to do?
performance variable-assignment assignment-operator
asked Nov 25 '18 at 15:59
Fabrice DugasFabrice Dugas
149110
add a comment |
This question is mostly directed at compiled programming languages. It is purely out of curiosity because I believe the gain of performance in using one of the two operators would be so very small.
Consider a for-loop where when you meet a certain condition, you want to store true in a boolean:
b = false
for i in 1..N:
if someCondition(i):
b = true
moreThatNeedsToBeDone(i)
endfor
Now consider the same for loop with OR EQUAL instead
b = false
for i in 1..N:
if someCondition(i):
b |= true
moreThatNeedsToBeDone(i)
endfor
If the condition is met more than once, would the latter be faster in theory? Or atleast, would it do fewer operations? In general, the OR EQUAL evaluates the variable and if it is true, then it doesn't do anything hence there is no extra assignment compared to the EQUAL operator where it would STORE true multiple times. But writing this, I realize the OR EQUAL adds an extra operation anyways in order to evaluate/read the current value of the variable. So which would be faster or have fewer operations to do?
performance variable-assignment assignment-operator
asked Nov 25 '18 at 15:59
Fabrice DugasFabrice Dugas
149110
Depends on the...following the assignment. You can also break the loop once b is set to true is...is a no-op. The rest is marginal optimization that is so rarely relevant these days.
– TT.
Nov 25 '18 at 16:47
Yeah I thought about breaking once b is true but that's why I added the.... The...is something that needs to be done, in other words we can't break before the loop is over. I know it's really not relevant, just a question that popped in my head and wondered if anyone could provide an answer on SO.
– Fabrice Dugas
Nov 25 '18 at 23:15
add a comment |
This question is mostly directed at compiled programming languages. It is purely out of curiosity because I believe the gain of performance in using one of the two operators would be so very small.
Consider a for-loop where when you meet a certain condition, you want to store true in a boolean:
b = false
for i in 1..N:
if someCondition(i):
b = true
moreThatNeedsToBeDone(i)
endfor
Now consider the same for loop with OR EQUAL instead
b = false
for i in 1..N:
if someCondition(i):
b |= true
moreThatNeedsToBeDone(i)
endfor
If the condition is met more than once, would the latter be faster in theory? Or atleast, would it do fewer operations? In general, the OR EQUAL evaluates the variable and if it is true, then it doesn't do anything hence there is no extra assignment compared to the EQUAL operator where it would STORE true multiple times. But writing this, I realize the OR EQUAL adds an extra operation anyways in order to evaluate/read the current value of the variable. So which would be faster or have fewer operations to do?
performance variable-assignment assignment-operator
asked Nov 25 '18 at 15:59
Fabrice DugasFabrice Dugas
149110
This question is mostly directed at compiled programming languages. It is purely out of curiosity because I believe the gain of performance in using one of the two operators would be so very small.
Consider a for-loop where when you meet a certain condition, you want to store true in a boolean:
b = false
for i in 1..N:
if someCondition(i):
b = true
moreThatNeedsToBeDone(i)
endfor
Now consider the same for loop with OR EQUAL instead
b = false
for i in 1..N:
if someCondition(i):
b |= true
moreThatNeedsToBeDone(i)
endfor
If the condition is met more than once, would the latter be faster in theory? Or atleast, would it do fewer operations? In general, the OR EQUAL evaluates the variable and if it is true, then it doesn't do anything hence there is no extra assignment compared to the EQUAL operator where it would STORE true multiple times. But writing this, I realize the OR EQUAL adds an extra operation anyways in order to evaluate/read the current value of the variable. So which would be faster or have fewer operations to do?
performance variable-assignment assignment-operator
performance variable-assignment assignment-operator
asked Nov 25 '18 at 15:59
Fabrice DugasFabrice Dugas
149110
asked Nov 25 '18 at 15:59
Fabrice DugasFabrice Dugas
149110
edited Nov 25 '18 at 23:18
Fabrice Dugas
asked Nov 25 '18 at 15:59
Fabrice DugasFabrice Dugas
149110
asked Nov 25 '18 at 15:59
Fabrice DugasFabrice Dugas
149110
asked Nov 25 '18 at 15:59
Fabrice DugasFabrice Dugas
149110
149110
Depends on the...following the assignment. You can also break the loop once b is set to true is...is a no-op. The rest is marginal optimization that is so rarely relevant these days.
– TT.
Nov 25 '18 at 16:47
Yeah I thought about breaking once b is true but that's why I added the.... The...is something that needs to be done, in other words we can't break before the loop is over. I know it's really not relevant, just a question that popped in my head and wondered if anyone could provide an answer on SO.
– Fabrice Dugas
Nov 25 '18 at 23:15
add a comment |
Depends on the...following the assignment. You can also break the loop once b is set to true is...is a no-op. The rest is marginal optimization that is so rarely relevant these days.
– TT.
Nov 25 '18 at 16:47
Yeah I thought about breaking once b is true but that's why I added the.... The...is something that needs to be done, in other words we can't break before the loop is over. I know it's really not relevant, just a question that popped in my head and wondered if anyone could provide an answer on SO.
– Fabrice Dugas
Nov 25 '18 at 23:15
Depends on the
... following the assignment. You can also break the loop once b is set to true is ... is a no-op. The rest is marginal optimization that is so rarely relevant these days.– TT.
Nov 25 '18 at 16:47
Depends on the
... following the assignment. You can also break the loop once b is set to true is ... is a no-op. The rest is marginal optimization that is so rarely relevant these days.– TT.
Nov 25 '18 at 16:47
Yeah I thought about breaking once b is true but that's why I added the
.... The ... is something that needs to be done, in other words we can't break before the loop is over. I know it's really not relevant, just a question that popped in my head and wondered if anyone could provide an answer on SO.– Fabrice Dugas
Nov 25 '18 at 23:15
Yeah I thought about breaking once b is true but that's why I added the
.... The ... is something that needs to be done, in other words we can't break before the loop is over. I know it's really not relevant, just a question that popped in my head and wondered if anyone could provide an answer on SO.– Fabrice Dugas
Nov 25 '18 at 23:15
add a comment |
1 Answer
1
active
oldest
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Most current compilers optimize to the best run time, meaning the resulting assembler or machine code would probably be identical. You could out of interest take a peek at the assembly generated for the two versions. I bet they are identical.
If the compiler would translate literally, the |= operation will not involve evaluating the variable to know if OR'ing the value should be done or not. The compiler will simply emit an or instruction as that would be equivalent and faster than first checking the variable (which could end up clearing the instruction pipeline).
Without optimization, the assembly generated could be (ax is a processor register):
or ax,1 ; for b|=true
mov ax,1 ; for b=true
I doubt on current processors this makes any difference in execution speed at all. Even if it would make a difference, we're talking about such a marginal optimization versus load times from memory to cache to processor/registers, or versus thread switches/process switches on CPU's, or versus branch misprediction in pipelined microprocessors and such.
answered Nov 26 '18 at 3:47
TT.TT.
12.5k63464
add a comment |
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Most current compilers optimize to the best run time, meaning the resulting assembler or machine code would probably be identical. You could out of interest take a peek at the assembly generated for the two versions. I bet they are identical.
If the compiler would translate literally, the |= operation will not involve evaluating the variable to know if OR'ing the value should be done or not. The compiler will simply emit an or instruction as that would be equivalent and faster than first checking the variable (which could end up clearing the instruction pipeline).
Without optimization, the assembly generated could be (ax is a processor register):
or ax,1 ; for b|=true
mov ax,1 ; for b=true
I doubt on current processors this makes any difference in execution speed at all. Even if it would make a difference, we're talking about such a marginal optimization versus load times from memory to cache to processor/registers, or versus thread switches/process switches on CPU's, or versus branch misprediction in pipelined microprocessors and such.
answered Nov 26 '18 at 3:47
TT.TT.
12.5k63464
add a comment |
Most current compilers optimize to the best run time, meaning the resulting assembler or machine code would probably be identical. You could out of interest take a peek at the assembly generated for the two versions. I bet they are identical.
If the compiler would translate literally, the |= operation will not involve evaluating the variable to know if OR'ing the value should be done or not. The compiler will simply emit an or instruction as that would be equivalent and faster than first checking the variable (which could end up clearing the instruction pipeline).
Without optimization, the assembly generated could be (ax is a processor register):
or ax,1 ; for b|=true
mov ax,1 ; for b=true
I doubt on current processors this makes any difference in execution speed at all. Even if it would make a difference, we're talking about such a marginal optimization versus load times from memory to cache to processor/registers, or versus thread switches/process switches on CPU's, or versus branch misprediction in pipelined microprocessors and such.
answered Nov 26 '18 at 3:47
TT.TT.
12.5k63464
add a comment |
Most current compilers optimize to the best run time, meaning the resulting assembler or machine code would probably be identical. You could out of interest take a peek at the assembly generated for the two versions. I bet they are identical.
If the compiler would translate literally, the |= operation will not involve evaluating the variable to know if OR'ing the value should be done or not. The compiler will simply emit an or instruction as that would be equivalent and faster than first checking the variable (which could end up clearing the instruction pipeline).
Without optimization, the assembly generated could be (ax is a processor register):
or ax,1 ; for b|=true
mov ax,1 ; for b=true
I doubt on current processors this makes any difference in execution speed at all. Even if it would make a difference, we're talking about such a marginal optimization versus load times from memory to cache to processor/registers, or versus thread switches/process switches on CPU's, or versus branch misprediction in pipelined microprocessors and such.
answered Nov 26 '18 at 3:47
TT.TT.
12.5k63464
Most current compilers optimize to the best run time, meaning the resulting assembler or machine code would probably be identical. You could out of interest take a peek at the assembly generated for the two versions. I bet they are identical.
If the compiler would translate literally, the |= operation will not involve evaluating the variable to know if OR'ing the value should be done or not. The compiler will simply emit an or instruction as that would be equivalent and faster than first checking the variable (which could end up clearing the instruction pipeline).
Without optimization, the assembly generated could be (ax is a processor register):
or ax,1 ; for b|=true
mov ax,1 ; for b=true
I doubt on current processors this makes any difference in execution speed at all. Even if it would make a difference, we're talking about such a marginal optimization versus load times from memory to cache to processor/registers, or versus thread switches/process switches on CPU's, or versus branch misprediction in pipelined microprocessors and such.
answered Nov 26 '18 at 3:47
TT.TT.
12.5k63464
answered Nov 26 '18 at 3:47
TT.TT.
12.5k63464
answered Nov 26 '18 at 3:47
TT.TT.
12.5k63464
answered Nov 26 '18 at 3:47
TT.TT.
12.5k63464
12.5k63464
add a comment |
add a comment |
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Depends on the
...following the assignment. You can also break the loop once b is set to true is...is a no-op. The rest is marginal optimization that is so rarely relevant these days.– TT.
Nov 25 '18 at 16:47
Yeah I thought about breaking once b is true but that's why I added the
.... The...is something that needs to be done, in other words we can't break before the loop is over. I know it's really not relevant, just a question that popped in my head and wondered if anyone could provide an answer on SO.– Fabrice Dugas
Nov 25 '18 at 23:15