What do these 2 lines of code actually do?

-5

I don’t know Ruby but I need to understand how the input value is manipulated in the lines below.

I think that the input is converted to the sum of the values of its characters but the second line is confusing; does it take the final value of the sum and perform the bitwise operations or per iteration? Could you provide a simple explanation of the steps followed?

Thank you in advance!

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

asked Jan 1 at 13:48

LostIT

8318

1

Is it a checksum? A hash?

– matt
Jan 1 at 14:38

You need to tell us what input is. Not just its class (Integer?), but more generally what it represents. We also need to understand how the return value of your code is used (related to @matt's question, which you need to answer). Lastly, you need a third line of code containing just end. Please address these issues by editing your question rather than by elaborating in comments.

– Cary Swoveland
Jan 1 at 16:15

You have selected an answer that does nothing more than explain how inject works, with no reference to the particular calculation you wish to perform. There are thousands of explanations of inject/reduce within easy reach, many with greater detail than the explanation contained in the selected answer,. That show of laziness earns a downvote from me.

– Cary Swoveland
Jan 1 at 16:29

@CarySwoveland It does give me the information I needed under, "So in your case:".

– LostIT
Jan 2 at 0:15

That may be, but it's not at all clear from the question what you want. Judging from the voting I don't seem to be alone in my assessment of the quality of the question.

– Cary Swoveland
Jan 2 at 1:16

add a comment |

-5

I don’t know Ruby but I need to understand how the input value is manipulated in the lines below.

Thank you in advance!

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

asked Jan 1 at 13:48

LostIT

8318

1

Is it a checksum? A hash?

– matt
Jan 1 at 14:38

You need to tell us what input is. Not just its class (Integer?), but more generally what it represents. We also need to understand how the return value of your code is used (related to @matt's question, which you need to answer). Lastly, you need a third line of code containing just end. Please address these issues by editing your question rather than by elaborating in comments.

– Cary Swoveland
Jan 1 at 16:15

You have selected an answer that does nothing more than explain how inject works, with no reference to the particular calculation you wish to perform. There are thousands of explanations of inject/reduce within easy reach, many with greater detail than the explanation contained in the selected answer,. That show of laziness earns a downvote from me.

– Cary Swoveland
Jan 1 at 16:29

@CarySwoveland It does give me the information I needed under, "So in your case:".

– LostIT
Jan 2 at 0:15

That may be, but it's not at all clear from the question what you want. Judging from the voting I don't seem to be alone in my assessment of the quality of the question.

– Cary Swoveland
Jan 2 at 1:16

add a comment |

-5

I don’t know Ruby but I need to understand how the input value is manipulated in the lines below.

Thank you in advance!

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

asked Jan 1 at 13:48

LostIT

8318

I don’t know Ruby but I need to understand how the input value is manipulated in the lines below.

Thank you in advance!

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

ruby

asked Jan 1 at 13:48

LostIT

8318

asked Jan 1 at 13:48

LostIT

8318

asked Jan 1 at 13:48

LostIT

8318

asked Jan 1 at 13:48

LostIT

8318

asked Jan 1 at 13:48

LostIT

8318

1

Is it a checksum? A hash?

– matt
Jan 1 at 14:38

You need to tell us what input is. Not just its class (Integer?), but more generally what it represents. We also need to understand how the return value of your code is used (related to @matt's question, which you need to answer). Lastly, you need a third line of code containing just end. Please address these issues by editing your question rather than by elaborating in comments.

– Cary Swoveland
Jan 1 at 16:15

You have selected an answer that does nothing more than explain how inject works, with no reference to the particular calculation you wish to perform. There are thousands of explanations of inject/reduce within easy reach, many with greater detail than the explanation contained in the selected answer,. That show of laziness earns a downvote from me.

– Cary Swoveland
Jan 1 at 16:29

@CarySwoveland It does give me the information I needed under, "So in your case:".

– LostIT
Jan 2 at 0:15

That may be, but it's not at all clear from the question what you want. Judging from the voting I don't seem to be alone in my assessment of the quality of the question.

– Cary Swoveland
Jan 2 at 1:16

add a comment |

1

Is it a checksum? A hash?

– matt
Jan 1 at 14:38

You need to tell us what input is. Not just its class (Integer?), but more generally what it represents. We also need to understand how the return value of your code is used (related to @matt's question, which you need to answer). Lastly, you need a third line of code containing just end. Please address these issues by editing your question rather than by elaborating in comments.

– Cary Swoveland
Jan 1 at 16:15

You have selected an answer that does nothing more than explain how inject works, with no reference to the particular calculation you wish to perform. There are thousands of explanations of inject/reduce within easy reach, many with greater detail than the explanation contained in the selected answer,. That show of laziness earns a downvote from me.

– Cary Swoveland
Jan 1 at 16:29

@CarySwoveland It does give me the information I needed under, "So in your case:".

– LostIT
Jan 2 at 0:15

That may be, but it's not at all clear from the question what you want. Judging from the voting I don't seem to be alone in my assessment of the quality of the question.

– Cary Swoveland
Jan 2 at 1:16

Is it a checksum? A hash?

– matt
Jan 1 at 14:38

You need to tell us what input is. Not just its class (Integer?), but more generally what it represents. We also need to understand how the return value of your code is used (related to @matt's question, which you need to answer). Lastly, you need a third line of code containing just end. Please address these issues by editing your question rather than by elaborating in comments.

– Cary Swoveland
Jan 1 at 16:15

You have selected an answer that does nothing more than explain how inject works, with no reference to the particular calculation you wish to perform. There are thousands of explanations of inject/reduce within easy reach, many with greater detail than the explanation contained in the selected answer,. That show of laziness earns a downvote from me.

– Cary Swoveland
Jan 1 at 16:29

@CarySwoveland It does give me the information I needed under, "So in your case:".

– LostIT
Jan 2 at 0:15

That may be, but it's not at all clear from the question what you want. Judging from the voting I don't seem to be alone in my assessment of the quality of the question.

– Cary Swoveland
Jan 2 at 1:16

add a comment |

1 Answer
1

active

oldest

votes

inject is the same as reduce, and is similar to reduce in many other languages.

There are a number of different ways to call it but the way shown in the question is this:

inject(initial) { |memo, obj| block } → obj

If you specify a block, then for each element in enum the block is
passed an accumulator value (memo) and the element. If you specify a
symbol instead, then each element in the collection will be passed to
the named method of memo. In either case, the result becomes the new
value for memo. At the end of the iteration, the final value of memo
is the return value for the method.

If you do not explicitly specify an initial value for memo, then the
first element of collection is used as the initial value of memo.

So in your case:

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

The initial value of sum is 0, which is used for the first iteration, but for each subsequent iteration the result of the block is used for sum in the next iteration.

For example the following should produce the same value without using inject/reduce.

sum = 0

input.to_s.each_char do |ch|

    sum = (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

answered Jan 1 at 14:38

nPn

8,27362245

add a comment |

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1 Answer
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active

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1 Answer
1

active

oldest

votes

inject is the same as reduce, and is similar to reduce in many other languages.

There are a number of different ways to call it but the way shown in the question is this:

inject(initial) { |memo, obj| block } → obj

If you specify a block, then for each element in enum the block is
passed an accumulator value (memo) and the element. If you specify a
symbol instead, then each element in the collection will be passed to
the named method of memo. In either case, the result becomes the new
value for memo. At the end of the iteration, the final value of memo
is the return value for the method.

If you do not explicitly specify an initial value for memo, then the
first element of collection is used as the initial value of memo.

So in your case:

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

The initial value of sum is 0, which is used for the first iteration, but for each subsequent iteration the result of the block is used for sum in the next iteration.

For example the following should produce the same value without using inject/reduce.

sum = 0

input.to_s.each_char do |ch|

    sum = (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

answered Jan 1 at 14:38

nPn

8,27362245

add a comment |

inject is the same as reduce, and is similar to reduce in many other languages.

There are a number of different ways to call it but the way shown in the question is this:

inject(initial) { |memo, obj| block } → obj

If you specify a block, then for each element in enum the block is
passed an accumulator value (memo) and the element. If you specify a
symbol instead, then each element in the collection will be passed to
the named method of memo. In either case, the result becomes the new
value for memo. At the end of the iteration, the final value of memo
is the return value for the method.

If you do not explicitly specify an initial value for memo, then the
first element of collection is used as the initial value of memo.

So in your case:

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

The initial value of sum is 0, which is used for the first iteration, but for each subsequent iteration the result of the block is used for sum in the next iteration.

For example the following should produce the same value without using inject/reduce.

sum = 0

input.to_s.each_char do |ch|

    sum = (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

answered Jan 1 at 14:38

nPn

8,27362245

add a comment |

inject is the same as reduce, and is similar to reduce in many other languages.

There are a number of different ways to call it but the way shown in the question is this:

inject(initial) { |memo, obj| block } → obj

If you specify a block, then for each element in enum the block is
passed an accumulator value (memo) and the element. If you specify a
symbol instead, then each element in the collection will be passed to
the named method of memo. In either case, the result becomes the new
value for memo. At the end of the iteration, the final value of memo
is the return value for the method.

If you do not explicitly specify an initial value for memo, then the
first element of collection is used as the initial value of memo.

So in your case:

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

The initial value of sum is 0, which is used for the first iteration, but for each subsequent iteration the result of the block is used for sum in the next iteration.

For example the following should produce the same value without using inject/reduce.

sum = 0

input.to_s.each_char do |ch|

    sum = (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

answered Jan 1 at 14:38

nPn

8,27362245

inject is the same as reduce, and is similar to reduce in many other languages.

There are a number of different ways to call it but the way shown in the question is this:

inject(initial) { |memo, obj| block } → obj

If you specify a block, then for each element in enum the block is
passed an accumulator value (memo) and the element. If you specify a
symbol instead, then each element in the collection will be passed to
the named method of memo. In either case, the result becomes the new
value for memo. At the end of the iteration, the final value of memo
is the return value for the method.

If you do not explicitly specify an initial value for memo, then the
first element of collection is used as the initial value of memo.

So in your case:

input.to_s.each_char.inject(0) do |sum, ch|

    (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

The initial value of sum is 0, which is used for the first iteration, but for each subsequent iteration the result of the block is used for sum in the next iteration.

For example the following should produce the same value without using inject/reduce.

sum = 0

input.to_s.each_char do |ch|

    sum = (sum << 8) ^ (ch.ord) ^ (sum >> 4)

end

answered Jan 1 at 14:38

nPn

8,27362245

answered Jan 1 at 14:38

nPn

8,27362245

answered Jan 1 at 14:38

nPn

8,27362245

answered Jan 1 at 14:38

nPn

8,27362245

add a comment |

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