I am trying to access a data block, the way it is define is as follows
DATA NAME /'X1','X2','X3','X4','X5','X6','X7','X8','X9','10','11',00028650
1'12','13','14','15','16','17','18','19','20','21','22','23','24'/ 00028660
The code is on paper. Note this is an old code, the only thing i am trying to do is understand how the array is being indexed. I am not trying to compile it.
The way it is accessed is as follows
I = 0
Loop
I = I + 1
write (06,77) (NAME(J,I),J=1,4) //this is inside a write statement.
end loop //77 is a format statement.
Not sure how it is being indexed, if you guys can shed some light that would be great.
The syntax (expr, intvar=int1,int2[,int3]) widely refers to an implied DO loop. There are several places where such a thing may occur, and an input/output statement is one such place.
An implied DO loop evaluates the expression expr with the do control integer variable intvar sequentially taking the values initially int1 in steps of int3 until the value int2 is reached/passed. This loop control is exactly as one would find in a do loop statement.
In the case of the question, the expression is name(j,i), the integer variable j is the loop variable, taking values between the bounds 1 and 4. [The step size int3 is not provided so is treated as 1.] The output statement is therefore exactly like
write(6,77) name(1,i), name(2,i), name(3,i), name(4,i)
as we should note that elements of the implied loop are expanded in order. i itself comes from the loop containing this output statement.
name here may refer to a function, but given the presence of a data statement initializing it, it must somehow be declared as a rank-2 (character) array. The initialization is not otherwise important.
Related
I have seen some very weird for loops when reading other people's code. I have been trying to search for a full syntax explanation for the for loop in C but it is very hard because the word "for" appears in unrelated sentences making the search almost impossible to Google effectively.
This question came to my mind after reading this thread which made me curious again.
The for here:
for(p=0;p+=(a&1)*b,a!=1;a>>=1,b<<=1);
In the middle condition there is a comma separating the two pieces of code, what does this comma do? The comma on the right side I understand as it makes both a>>=1 and b<<=1.
But within a loop exit condition, what happens? Does it exit when p==0, when a==1 or when both happen?
It would be great if anyone could help me understand this and maybe point me in the direction of a full for loop syntax description.
The comma is not exclusive of for loops; it is the comma operator.
x = (a, b);
will do first a, then b, then set x to the value of b.
The for syntax is:
for (init; condition; increment)
...
Which is somewhat (ignoring continue and break for now) equivalent to:
init;
while (condition) {
...
increment;
}
So your for loop example is (again ignoring continue and break) equivalent to
p=0;
while (p+=(a&1)*b,a!=1) {
...
a>>=1,b<<=1;
}
Which acts as if it were (again ignoring continue and break):
p=0;
while (true) {
p+=(a&1)*b;
if (a == 1) break;
...
a>>=1;
b<<=1;
}
Two extra details of the for loop which were not in the simplified conversion to a while loop above:
If the condition is omitted, it is always true (resulting in an infinite loop unless a break, goto, or something else breaks the loop).
A continue acts as if it were a goto to a label just before the increment, unlike a continue in the while loop which would skip the increment.
Also, an important detail about the comma operator: it is a sequence point, like && and || (which is why I can split it in separate statements and keep its meaning intact).
Changes in C99
The C99 standard introduces a couple of nuances not mentioned earlier in this explanation (which is very good for C89/C90).
First, all loops are blocks in their own right. Effectively,
for (...) { ... }
is itself wrapped in a pair of braces
{
for (...) { ... }
}
The standard sayeth:
ISO/IEC 9899:1999 §6.8.5 Iteration statements
¶5 An iteration statement is a block whose scope is a strict subset of the scope of its
enclosing block. The loop body is also a block whose scope is a strict subset of the scope
of the iteration statement.
This is also described in the Rationale in terms of the extra set of braces.
Secondly, the init portion in C99 can be a (single) declaration, as in
for (int i = 0; i < sizeof(something); i++) { ... }
Now the 'block wrapped around the loop' comes into its own; it explains why the variable i cannot be accessed outside the loop. You can declare more than one variable, but they must all be of the same type:
for (int i = 0, j = sizeof(something); i < j; i++, j--) { ... }
The standard sayeth:
ISO/IEC 9899:1999 §6.8.5.3 The for statement
The statement
for ( clause-1 ; expression-2 ; expression-3 ) statement
behaves as follows: The expression expression-2 is the controlling expression that is
evaluated before each execution of the loop body. The expression expression-3 is
evaluated as a void expression after each execution of the loop body. If clause-1 is a
declaration, the scope of any variables it declares is the remainder of the declaration and
the entire loop, including the other two expressions; it is reached in the order of execution
before the first evaluation of the controlling expression. If clause-1 is an expression, it is
evaluated as a void expression before the first evaluation of the controlling expression.133)
Both clause-1 and expression-3 can be omitted. An omitted expression-2 is replaced by a
nonzero constant.
133) Thus, clause-1 specifies initialization for the loop, possibly declaring one or more variables for use in
the loop; the controlling expression, expression-2, specifies an evaluation made before each iteration,
such that execution of the loop continues until the expression compares equal to 0; and expression-3
specifies an operation (such as incrementing) that is performed after each iteration.
The comma simply separates two expressions and is valid anywhere in C where a normal expression is allowed. These are executed in order from left to right. The value of the rightmost expression is the value of the overall expression.
for loops consist of three parts, any of which may also be empty; one (the first) is executed at the beginning, and one (the third) at the end of each iteration. These parts usually initialize and increment a counter, respectively; but they may do anything.
The second part is a test that is executed at the beginning of each execution. If the test yields false, the loop is aborted. That's all there is to it.
The C style for loop consists of three expressions:
for (initializer; condition; counter) statement_or_statement_block;
The initializer runs once, when the loop starts.
The condition is checked before each iteration. The loop runs as long it evaluates to true.
The counter runs once after each iteration.
Each of these parts can be an expression valid in the language you write the loop in. That means they can be used more creatively. Anything you want to do beforehand can go into the initializer, anything you want to do in between can go into the condition or the counter, up to the point where the loop has no body anymore.
To achieve that, the comma operator comes in very handy. It allows you to chain expressions together to form a single new expression. Most of the time it is used that way in a for loop, the other implications of the comma operator (e.g. value assignment considerations) play a minor role.
Even though you can do clever things by using syntax creatively - I would stay clear of it until I find a really good reason to do so. Playing code golf with for loops makes code harder to read and understand (and maintain).
The wikipedia has a nice article on the for loop as well.
Everything is optional in a for loop. We can initialize more than one variable, we can check for more than one condition, we can iterate more than one variable using the comma operator.
The following for loop will take you into an infinite loop. Be careful by checking the condition.
for(;;)
Konrad mentioned the key point that I'd like to repeat: The value of the rightmost expression is the value of the overall expression.
A Gnu compiler stated this warning when I put two tests in the "condition" section of the for loop
warning: left-hand operand of comma expression has no effect
What I really intended for the "condition" was two tests with an "&&" between. Per Konrad's statement, only the test on to the right of the comma would affect the condition.
the for loop is execution for particular time for(;;)
the syntex for for loop
for(;;)
OR
for (initializer; condition; counter)
e.g (rmv=1;rmv<=15;rmv++)
execution to 15 times in for block
1.first initializ the value because start the value
(e.g)rmv=1 or rmv=2
2.second statement is test the condition is true or false ,the condition true no.of time execution the for loop and the condition is false terminate for block,
e.g i=5;i<=10 the condition is true
i=10;i<10 the condition is false terminate for block,
3.third is increment or decrement
(e.g)rmv++ or ++rmv
I am reading data structures and I am following book "Fundamentals of data structures in C++" - E. Horowitz, S. Sahni & D. Mehta.
While reading rules for step counts in time complexity, I am stuck at the following statement.
Can anyone please explain me the following paragraph especially bold literals.
The assignment statement variable = expression has a step count equal to that expression unless the size of variable is a function of instance characteristics.
The non-bold part is straight-forward: the code of an assignment is at a minimum the cost of computing what is to be assigned. The bold part is simply saying that if, once you have finished that calculation, the amount of work to assign that value is not a constant (that is, it is a function of the problem size), then you have to take that into account as well.
For example, how long it takes to assign a value to an integer variable would be a constant, while copying a string won't be if its length is determined by the size of the problem.
Hi I am new here and want to solve this problem:
do k=1,31
Data H(1,k)/0/
End do
do l=1,21
Data H(l,1)/0.5*(l-1)/
End do
do m=31,41
Data H(17,m)/0/
End do
do n=17,21
Data H(n,41)/0.5*(n-17)/
End do
I get error for l and n saying that it is a syntax error in DATA statement. Anyone know how to solve this problem?
You have three problems here, and not just with the "l" and "n" loops.
The first problem is that the values in a data statement cannot be arbitrary expressions. In particular, they must be constants; 0.5*(l-1) is not a constant.
The second problem is that the bounds in the object lists must also be constant (expressions); l is not a constant expression.
For the first, it's also worth noting that * in a data value list has a special meaning, and it isn't the multiplication operator. * gives a repeat count, and a repeat count of 0.5 is not valid.
You can fix the second point quite simply, by using such constructions as
data H(1,1:31) /31*0./ ! Note the repeat count specifier
outside a loop, or using an implied loop
data (H(1,k),k=1,31) /31*0./
To do something for the "l" loop is more tedious
data H(1:21,1) /0., 0.5, 1., 1.5, ... /
and we have to be very careful about the number of values specified. This cannot be dynamic.
The third problem is that you cannot specify explicit initialization for an element more than once. Look at your first two loops: if this worked you'd be initializing H(1,1) twice. Even though the same value is given, this is still invalid.
Well, actually you have four problems. The fourth is related to the point about dynamic number of values. You probably don't want to be doing explicit initialization. Whilst it's possible to do what it looks like you want to do, just use assignment where these restrictions don't apply.
do l=1,21
H(l,1) = 0.5*(l-1)
End do
Yes, there are times when complicated explicit initialization is a desirable thing, but in this case, in what I assume is new code, keeping things simple is good. An "initialization" portion of your code which does the assignments is far more "modern".
Apologies for such an entirely uninformed question, but I don't know any SAS and just need to know what one line of code does, so I hope someone can help.
I have a loop over an array of variables, and an if clause that is based on a comparison to .Z, but this variable is defined nowhere, so I'm guessing this is some sort of SAS syntax trick. Here's the loop:
ARRAY PTYPE{*} X4216 X4316 X4416 X4816 X4916 X5016;
DO I=1 TO DIM(PTYPE);
IF (PTYPE{I}<=.Z) THEN PUT &ID= PTYPE{I}=;
END;
So on the first iteration, the loop would check whether the value in X4216 is smaller than .Z, and then...? ID is another varuable in the dataset, but I have no idea what's happening on the right hand side of that if clause. I've briefly consulted the SAS documentation to figure out that ampersands refer to macros, but my knowledge of SAS is to limited to understand what's happening.
Can anyone enlighten me?
.Z is a special missing value. In SAS a missing value (what you might call a NULL value) is indicated by a period. There are also 27 other special missing values that are indicated by a period followed by a letter or an underscore. The missing values are distinct and are all considered smaller than any actual number. .Z is the "largest". So PTYPE{I}<=.Z is basically testing if the value is missing. You could instead use MISSING(PTYPE{I}) to make the same test. The right hand side is writing out the name and the value of the variable in the array with a missing value and also the name and value of the variable named in the macro variable ID.
I've got experience in a lot of other programming languages, but I'm having a lot of difficulty with Stata syntax. I've got a statement that evaluates with no problem if I put in values, but I can't figure out why it's not evaluating variables like I expect it to.
gen j=5
forvalues i = 1(1)5 {
replace TrustBusiness_local=`i' if TrustBusiness_local2==`j'
replace j=`j'-1
}
If I replace i and j with 1 and 5 respectively, like I'm expecting to happen from the code above, then it works fine, but I get an if not found error otherwise, which hasn't produced meaningful results when Googled. Does anyone see what I don't see? I hate to brute-force something that could so simply be done with a loop.
Easy to understand once you approach it the right way!
Problem 1. You never defined local macro j. That in itself is not an error, but it often leads to errors. Macros that don't exist are equivalent to empty strings, so Stata sees in this example the code
if TrustBusiness_local2==`j'
as
if TrustBusiness_local2==
which is illegal; hence the error message.
Problem 2. There is no connection of principle between a variable you called j and a local macro called j but referenced using single quotes. A variable in Stata is a variable (namely, column) in your dataset; that doesn't mean a variable otherwise in the sense of any programming language. Variables meaning single values can be held in Stata within scalars or within macros. Putting a constant into a variable, Stata sense, is legal, but usually bad style. If you have millions of observations, for example, you now have a column j with millions of values of 5 within it.
Problem 3. You could, legally, go
local j "j"
so that now the local macro j contains the text "j", which depending on how you use it could be interpreted as a variable name. It's hard to see why you would want to do that here, but it would be legal.
Problem 4. Your whole example doesn't even need a loop as it appears to mean
replace TrustBusiness_local= 6 - TrustBusiness_local2 if inlist(TrustBusiness_local2, 1,2,3,4,5)
and, depending on your data, the if qualifier could be redundant. Flipping 5(1)1 to 1(1)5 is just a matter of subtracting from 6.
Problem 5. Your example written as a loop in Stata style could be
local j = 5
forvalues i = 1/5 {
replace TrustBusiness_local=`i' if TrustBusiness_local2==`j'
local j=`j'-1
}
and it could be made more concise, but given Problem 4 that no loop is needed, I will leave it there.
Problem 6. What you talking about are, incidentally, not if statements so far as Stata is concerned, as the if qualifier used in your examples is not the same as the if command.
The problem of translating one language's jargon into another can be challenging. See my comments at http://www.stata.com/statalist/archive/2008-08/msg01258.html After experience in other languages, the macro manipulations of Stata seemed at first strange to me too; they are perhaps best understood as equivalent to shell programming.
I wouldn't try to learn Stata by Googling. Read [U] from beginning to end. (A similar point was made in the reply to your previous question at use value label in if command in Stata but you don't want to believe it!)