[Solved] Why Do Computer Counts from Zero.
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Actually, it doesn't simplify the calculation of the address. It just aids the ability to interchangeably address the whole array and the address of its first element (e.g. in C,
&arrayand&array[0]both give the same address). In languages like FORTRAN (from 1957 to current day) which start indices from 1, the formula iselement(n) = (address_of_array - size_of_the_element) + n * size_of_the_element
which is just as simple as starting from 0 as
(address_of_array - size_of_the_element)is a compile time constant value. Considering the millions of errors over the decades that starting from zero has perpetuated (e.g. forgetting to add one to the bounds when declaring the array or getting the end point wrong in loops or wasting space by deliberately skipping element 0), doing the compile time calculation would have been a small price to pay.jsc42 wrote:
element(n) = (address_of_array - size_of_the_element) + n * size_of_the_element
which is just as simple as starting from 0 as
(address_of_array - size_of_the_element)is a compile time constant value.(address_of_array - size_of_the_element) is not a compile time constant value in any but the simplest of cases, i.e. where the array is at a fixed memory location every time the program is run, which is almost never.
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That may have made sense in the 1960's, but not really anymore. A simple addition and possibly a shift takes very time relative to all the other processing happening in a computer.
You don't know what else the computer is doing so don't assume that there are plenty of cycles. If every app wastes 10% of the "extra" cycles it doesn't take long to bog the whole thing down.
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Oh yes. Little tiny wasted instructions add up fast when you do them as often as array accesses. And remember, when this was decided, even HUGE computers ran at much, much, lower speeds - an IBM 360/195 could execute 16,000,000 instructions per second (at best), compared to the around 6,000,000,000 that your PC is (probably) capable of executing - it's difficult to be sure, due to multiple cores, pipelines, and caches, all of which were a lot smaller or nonexistent in those days. And remember, a 360/195 cost about £3,000,000 back in 1982, when beer was £0.30 per pint...so you didn't want to waste anything! :laugh:
The universe is composed of electrons, neutrons, protons and......morons. (ThePhantomUpvoter)
OriginalGriff wrote:
wasted instructions add up fast
Which is why I detest animated widgets. X|
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jsc42 wrote:
element(n) = (address_of_array - size_of_the_element) + n * size_of_the_element
which is just as simple as starting from 0 as
(address_of_array - size_of_the_element)is a compile time constant value.(address_of_array - size_of_the_element) is not a compile time constant value in any but the simplest of cases, i.e. where the array is at a fixed memory location every time the program is run, which is almost never.
You are correct! What I meant (and I admit that the language that I used was far from rigorous) was that the relative offset is unvarying. If, for example, the offset of the start of the memory allocated for the array is 100 from a location (relative addressing / stack frame relative / absolute / whatever) and each element takes up 4 address locations (e.g. bytes), then all that the compiler does for 1-based indices is to use 96 + n * 4 (because 100 - 4 = 96) instead of 100 +(n - 1) * 4 in all of its address calculations for the array; this is as simple as using 100 + n * 4 for 0-based indices. This can be extended for arrays with specifiable lower bounds e.g. in this example if lwb is the lower bound, the 0-based index offset would be 100 + (n - lwb) * 4 [or more optimally (100 - lwb * 4) + n * 4] or for 1-based index offset would be (100 - (lwb + 1) * 4) + n * 4 = (96 - lwb * 4) + n * 4.
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Link to the article For More detail explanation(PDF) Counting arrays from 0 simplifies the computation of the memory address of each element. If an array is stored at a given position in memory (it’s called the address) the position of each element can be computed as
element(n) = address + n * size_of_the_element
//If you consider the first element the first, the computation becomeselement(n) = address + (n-1) * size_of_the_element
//Not a huge difference but it adds an unnecessary subtraction for each access. -
You are correct! What I meant (and I admit that the language that I used was far from rigorous) was that the relative offset is unvarying. If, for example, the offset of the start of the memory allocated for the array is 100 from a location (relative addressing / stack frame relative / absolute / whatever) and each element takes up 4 address locations (e.g. bytes), then all that the compiler does for 1-based indices is to use 96 + n * 4 (because 100 - 4 = 96) instead of 100 +(n - 1) * 4 in all of its address calculations for the array; this is as simple as using 100 + n * 4 for 0-based indices. This can be extended for arrays with specifiable lower bounds e.g. in this example if lwb is the lower bound, the 0-based index offset would be 100 + (n - lwb) * 4 [or more optimally (100 - lwb * 4) + n * 4] or for 1-based index offset would be (100 - (lwb + 1) * 4) + n * 4 = (96 - lwb * 4) + n * 4.
Introduction of the lower bound would definitely negate any advantage of 0 based indexing. With no specified lower bound the cost at the instruction level would depend on the instruction set and processor, but with the instruction sets I am familiar with, 0 based indexing was definitely less expensive. Very simple case using x86 32 bit instructions and assuming that the array base in ESI and array index in EBX, accessing a 32 bit array value can be done in a single instruction in both cases: mov eax, [esi + 4*ebx] ; 0 based indexing mov eax, [esi + 4*ebx + 4] ; 1 based indexing On even the most modern x86 processor, the instruction with an offset takes more memory than the one without. On older x86 processors it would also cost CPU cycles, and on more primitive instruction sets you wouldn't even be able to encode the address calculations into a single instruction at all. One interpretation of your argument for same cost might be that in the one-based case that you just start with esi pointing to 4 bytes before, which would be true but you still have to count the cost of adjusting esi, though in some cases that calculation could be amortized across many array accesses.
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You don't know what else the computer is doing so don't assume that there are plenty of cycles. If every app wastes 10% of the "extra" cycles it doesn't take long to bog the whole thing down.
That is assuming that the developers of system code have not already optimized the code where it can be of the most benefit. Also, it assumes that the compiler is not smart enough to do some optimizations.
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That is assuming that the developers of system code have not already optimized the code where it can be of the most benefit. Also, it assumes that the compiler is not smart enough to do some optimizations.
Not at all. I'm saying that I could have a gazillion apps running.
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Not at all. I'm saying that I could have a gazillion apps running.
Relative to system calls, and library calls, the code in the application usually takes up little of the resources unless you have an application that is heavily processor bound.
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Relative to system calls, and library calls, the code in the application usually takes up little of the resources unless you have an application that is heavily processor bound.
And let's keep it that way.
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Link to the article For More detail explanation(PDF) Counting arrays from 0 simplifies the computation of the memory address of each element. If an array is stored at a given position in memory (it’s called the address) the position of each element can be computed as
element(n) = address + n * size_of_the_element
//If you consider the first element the first, the computation becomeselement(n) = address + (n-1) * size_of_the_element
//Not a huge difference but it adds an unnecessary subtraction for each access.It was also handy back in the days when only single array indexes existed and you wanted dual indexes because everything mathematically matched up really handily without taking extra steps. for (i = 0; i < 10; i++)// Not at all like the format of older languages! { for (j = 0; j < 10: j++) { ind= i + j * 10; val[ind] = some value } } Or the other way around, find the two indexes from the current index: for (ind = 0; ind < 100; ind++) { j = ind/10; i = ind - (j * 10); }
