# 5.4 Other clutter  (Page 2/2)

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``` c = a + b + d e = q + a + b```

becomes:

``` temp = a + b c = temp + de = q + temp```

Substituting for `a+b` eliminates some of the arithmetic. If the expression is reused many times, the savings can be significant. However, a compiler’s ability to recognize common subexpressions is limited, especially when there are multiple components, or their order is permuted. A compiler might not recognize that `a+b+c` and `c+b+a` are equivalent. And because of overflow and round-off errors in floating-point, in some situations they might not be equivalent. For important parts of the program, you might consider doing common subexpression elimination of complicated expressions by hand. This guarantees that it gets done. It compromises beauty somewhat, but there are some situations where it is worth it.

Here’s another example in which the function sin is called twice with the same argument:

``` x = r*sin(a)*cos(b); y = r*sin(a)*sin(b);z = r*cos(a);```

becomes:

``` temp = r*sin(a); x = temp*cos(b);y = temp*sin(b); z = r*cos(a);```

We have replaced one of the calls with a temporary variable. We agree, the savings for eliminating one transcendental function call out of five won’t win you a Nobel prize, but it does call attention to an important point: compilers typically do not perform common subexpression elimination over subroutine or function calls. The compiler can’t be sure that the subroutine call doesn’t change the state of the argument or some other variables that it can’t see.

The only time a compiler might eliminate common subexpressions containing function calls is when they are intrinsics, as in FORTRAN. This can be done because the compiler can assume some things about their side effects. You, on the other hand, can see into subroutines, which means you are better qualified than the compiler to group together common subexpressions involving subroutines or functions.

## Doing your own code motion

All of these optimizations have their biggest payback within loops because that’s where all of a program’s activity is concentrated. One of the best ways to cut down on runtime is to move unnecessary or repeated (invariant) instructions out of the main flow of the code and into the suburbs. For loops, it’s called hoisting instructions when they are pulled out from the top and sinking when they are pushed down below. Here’s an example:

``` DO I=1,N A(I) = A(I) / SQRT(X*X + Y*Y)ENDDO```

becomes:

``` TEMP = 1 / SQRT(X*X + Y*Y) DO I=1,NA(I) = A(I) * TEMP ENDDO```

We hoisted an expensive, invariant operation out of the loop and assigned the result to a temporary variable. Notice, too, that we made an algebraic simplification when we exchanged a division for multiplication by an inverse. The multiplication will execute much more quickly. Your compiler might be smart enough to make these transformations itself, assuming you have instructed the compiler that these are legal transformations; but without crawling through the assembly language, you can’t be positive. Of course, if you rearrange code by hand and the runtime for the loop suddenly goes down, you will know that the compiler has been sandbagging all along.

Sometimes you want to sink an operation below the loop. Usually, it’s some calculation performed each iteration but whose result is only needed for the last. To illustrate, here’s a sort of loop that is different from the ones we have been looking at. It searches for the final character in a character string:

``` while (*p != ’ ’) c = *p++;```

becomes:

``` while (*p++ != ’ ’); c = *(p-1);```

The new version of the loop moves the assignment of `c` beyond the last iteration. Admittedly, this transformation would be a reach for a compiler and the savings wouldn’t even be that great. But it illustrates the notion of sinking an operation very well.

Again, hoisting or sinking instructions to get them out of loops is something your compiler should be capable of doing. But often you can slightly restructure the calculations yourself when you move them to get an even greater benefit.

## Handling array elements in loops

Here’s another area where you would like to trust the compiler to do the right thing. When making repeated use of an array element within a loop, you want to be charged just once for loading it from memory. Take the following loop as an example. It reuses `X(I)` twice:

``` DO I=1,N XOLD(I) = X(I)X(I)= X(I) + XINC(I) ENDDO```

In reality, the steps that go into retrieving `X(I)` are just additional common subex- pressions: an address calculation (possibly) and a memory load operation. You can see that the operation is repeated by rewriting the loop slightly:

``` DO I=1,N TEMP= X(I)XOLD(I) = TEMP X(I)= TEMP + XINC(I)ENDDO```

FORTRAN compilers should recognize that the same `X(I)` is being used twice and that it only needs to be loaded once, but compilers aren’t always so smart. You sometimes have to create a temporary scalar variable to hold the value of an array element over the body of a loop. This is particularly true when there are subroutine calls or functions in the loop, or when some of the variables are `external` or `COMMON` . Make sure to match the types between the temporary variables and the other variables. You don’t want to incur type conversion overhead just because you are “helping” the compiler. For C compilers, the same kind of indexed expres- sions are an even greater challenge. Consider this code:

``` doinc(int xold[],int x[],int xinc[],int n){ for (i=0; i<n; i++) { xold[i]= x[i];x[i]= x[i]+ xinc[i];} }```

Unless the compiler can see the definitions of `x` , `xinc` , and `xold` , it has to assume that they are pointers leading back to the same storage, and repeat the loads and stores. In this case, introducing temporary variables to hold the values `x` , `xinc` , and `xold` is an optimization the compiler wasn’t free to make.

Interestingly, while putting scalar temporaries in the loop is useful for RISC and superscalar machines, it doesn’t help code that runs on parallel hardware. A parallel compiler looks for opportunities to eliminate the scalars or, at the very least, to replace them with temporary vectors. If you run your code on a parallel machine from time to time, you might want to be careful about introducing scalar temporary variables into a loop. A dubious performance gain in one instance could be a real performance loss in another.

find the 15th term of the geometric sequince whose first is 18 and last term of 387
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
how do they get the third part x = (32)5/4
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
kkk nice
algebra 2 Inequalities:If equation 2 = 0 it is an open set?
or infinite solutions?
Kim
The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
Al
y=10×
if |A| not equal to 0 and order of A is n prove that adj (adj A = |A|
rolling four fair dice and getting an even number an all four dice
Kristine 2*2*2=8
Differences Between Laspeyres and Paasche Indices
No. 7x -4y is simplified from 4x + (3y + 3x) -7y
how do you translate this in Algebraic Expressions
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials​ and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
what is system testing
what is the application of nanotechnology?
Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
Azam
anybody can imagine what will be happen after 100 years from now in nano tech world
Prasenjit
after 100 year this will be not nanotechnology maybe this technology name will be change . maybe aftet 100 year . we work on electron lable practically about its properties and behaviour by the different instruments
Azam
name doesn't matter , whatever it will be change... I'm taking about effect on circumstances of the microscopic world
Prasenjit
how hard could it be to apply nanotechnology against viral infections such HIV or Ebola?
Damian
silver nanoparticles could handle the job?
Damian
not now but maybe in future only AgNP maybe any other nanomaterials
Azam
Hello
Uday
I'm interested in Nanotube
Uday
this technology will not going on for the long time , so I'm thinking about femtotechnology 10^-15
Prasenjit
can nanotechnology change the direction of the face of the world
At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
the Beer law works very well for dilute solutions but fails for very high concentrations. why?
how did you get the value of 2000N.What calculations are needed to arrive at it
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