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.
Related
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.
I am currently reading about constants on the c++ tutorial from TutorialsPoint and, where it says:
Constants refer to fixed values that the program may not alter and they are called literals.
(Source)
I do not really get this. If constants are called literals and literals are data represented directly in the code, how can constants be considered as literals? I mean variables preceded with the const keyword are constants, but they are not literals, so how can you say that constants are literals?
Here:
const int MEANING = 42;
the value MEANING is a constant, 42 is a literal. There is no real relationship between the two terms, as can be seen here:
int n = 42;
where n is not a constant, but 42 is still a literal.
The major difference is that a constant may have an address in memory (if you write some code that needs such an address), whereas a literal never has an address.
I disagree with the claim "...There wasn't a thing called const in C originally so this was fine." const is actually one of the 32 C keywords. Google to see.
With that rested, I think the man missed something at TP. To be fair to them at Tutorials Point, they had an article that explained the difference thus (full quote, verbatim):
https://www.tutorialspoint.com/questions/category/Cplusplus
A literal is a value that is expressed as itself. For example, the number 25 or the string "Hello World" are both literals.
A constant is a data type that substitutes a literal. Constants are used when a specific, unchanging value is used various times during the program. For example, if you have a constant named PI that you'll be using at various places in your program to find the area, circumference, etc of a circle, this is a constant as you'll be reusing its value. But when you'll be declaring it as:
const float PI = 3.141;
The 3.141 is a literal that you're using. It doesn't have any memory address of its own and just sits in the source code.
Pls don't disparage those fellows doing what you call "random tutorials". Kids from poorer homes and less developed world can't afford your " good C++ textbooks " e.g. Scott Myers Effective C++ It is these online free tutorials they can have, and most of these tutorials do better explaining than the "good books".
By any means read them guys. Get confused some then come over here to StackOveflow or Quora to have your confusion cleared. Happy coding guys.
The author of the article is confused, and spreading that confusion to others (including you).
In C, literals are "constants". There wasn't a thing called const in C originally so this was fine.
C++ is a different language. In C++, literals are called "literals", and "constant" has a few meanings but generally is a const thing. The two concepts are different (although both kinds of things cannot be mutated after initial creation). We also have compile-time constants via constexpr which is yet another thing.
In general, read a good book rather than random tutorials written by randomers on the internet!
While the first part of the statement makes sense
Constants refer to fixed values that the program may not alter
the continuation
and they are called literals
is not really true.
Neil has already explained the semantical difference between the literal and the constant in his answer. But I would also like to add that the values of constant variables in C++ are not necessarily known at compile time.
// x might be obtained at runtime
// for instance, from the user input
void print_square(int x)
{
const int square = x*x;
std::cout << square << '\n';
}
Literals are values that are known at compile-time, which allows the compiler to put them to a separate read-only address space in the resulting binaries.
You can also enforce your variables to be known at compile-time by applying constexpr keyword (C++11).
constexpr int meaning = 42;
P.S. And I also do agree with a comment suggesting to use a good book instead of tutorialspoint.
If constants are called literals and literals are data represented directly in the code, how can constants be considered as literals?
The article from which you drew the quote is defining the word "constant" to be a synonym of "literal". The latter is the C++ standard's term for what it is describing. The former is what the C standard uses for the same concept.
I mean variables preceded with the const keyword are constants, but they are not literals, so how can you say that constants are literals?
And there you are providing an alternative definition for the term "constant", which, you are right, is inconsistent with the other. That's all. TP is using a different definition of the term than the one you are used to.
In truth, although the noun usage of "constant" appears in a couple of places in the C++ standard outside the defined term "null pointer constant", apparently with the meaning you propose here, I do not find an actual definition of that term, and especially not one matching yours. In truth, your definition is less plausible than TutorialPoint's, because an expression having const-qualified type can nevertheless designate an object that is modifiable (via a different expression).
Constant is simply a variable declared constant by keyword 'const' whose value after being declared shouldn't be altered during the course of the program (and if tried to alter it will result in an error).
On the other hand, literal is simply what is used and represented as it is typed in. For example, 25 when used in an expression (x+4*y+25) will be termed as literal.
Whenever we use String values or directly supply it in double quotes ("hello"), then that value in double quotes is called literal.
For example, printf("This is literal");
And if you are assigning a string value to a variable then thereafter you will refer to the variable (which could be declared constant if desired) and not exclusively to the value you have stored in it, i.e., only till the point you are supplying a value (string type of any other type) to the variable, the value is referred to as literal value, after that the variable is talked about whenever referring that value.
Once again, the value(25) in expression : x+4*y+25 is literal.
The value(4) in the term 4*y is also a literal (since it is exactly as we see it and is known to compiler beforehand).
--> The value(4) in the term 4*y is called numerical coefficient in algebraic terms and y is called literal coefficient in algebraic terms.
Hence,
All the above explanation I have given is in computer terms only. The meaning of literals and constants in Algebra are somewhat different than used in computer terms.
"Constants refer to fixed values that the program may not alter and they are called literals. (Source)"
The sentence construction is weird which is leading to the confusion.
Here, the the "they" that are referring to are the the fixed values and not constants. I would phrase it as "Constants refer to fixed values, that the program may not alter, called literals." which is less confusing I hope.
Constants are variables that can't vary, whereas Literals are literally numbers/letters that indicate the value of a variable or constant.
I can explain it this way.
Basically, constants are variables whose value cannot change.
Literals are notations that represent fixed values. These values can be Strings numbers etc
Literals can be assigned to variables
Code :
var a = 10;
var name = "Simba";
const pi = 3.14;
Here a and name are variables. pi is a constant. ( Constants are those variables whose value doesn't change. )
Here 10, "Simba" and 3.14 are literals.
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".
Prior understanding
According to The ArraySlice structure as reported by standardml.org (I don't have an SML 97 manual to check, only an SML 90 PDF manual), ArraySlice.copyVec gets an Array.array parameter as the destination, and not an ArraySlice.slice as one (or at least me) would intuitively expects. Of course, one can use ArraySlice.base to get an array and an index for respectively the dst and di parameters to copyVec. Surprisingly, copyVec from ArraySlice, does not even have a single parameter of type ArraySlice.slice. Fortunately, its src parameter is of type VectorSlice.slice, as intuitively expected.
The question
What's the rationale for ArraySlice.copyVec? Why doesn't it get an ArraySlice.slice as dst?
Presumably, because that would require the additional constraint that both slices have the same length (or at least dst is larger than src). The current API gets by without such an extra side condition. Also, it would probably be somewhat more cumbersome to use.
I was reading Hacker News and this article came up. It contains a raytracer that the code is written on the back of a business card. I decided it would be a good academic challenge to translate the c++ to python, but there's a few concepts I'm stuck on.
First, this function comes up: i T(v o,v d,f& t,v& n){...} Which is translated to int Tracer(vector o, vector d, float& t, vector& n){...} What does the float& mean? I know that in other places & is used as a == is that the case here? Can you do that in c++?
Second, I noticed these three lines:
for(i k=19;k--;) //For each columns of objects
for(i j=9;j--;) //For each line on that columns
if(G[j]&1<<k){
I know the << is a the bit shift, and I assume the & is ==. Are the for loops just like one for loop in an other?
Finally, this line: v p(13,13,13); I am not quite sure what it does. Does it create a class labeled by p that extends v (vector) with the defaults of 13,13,13?
These are probably dumb questions, but I want to see if I can understand this and my searching didn't come up with anything. Thank you in advance!
What does the float& mean?
Here, & means "reference", so the argument is passed by reference.
I know that in other places & is used as a == is that the case here?
& means various things in various contexts, but it never means ==. In this case, it's not an operator either; it's part of a type specification, meaning that it's a reference type.
I know the << is a the bit shift, and I assume the & is ==
No, it's a bitwise and operator. The result has its bits set where a bit is set in both operands. Here, with 1<<k as one operand, the result is the kth bit of G[j]; so this tests whether that bit is set.
Are the for loops just like one for loop in an other?
Yes. If you don't use braces around a for-loop's body, then the body is a single statement. So in this case, the body of the first loop is the second loop. To make this clear, I would recommend indenting the body of the loop, and using braces whether or not they are strictly necessary. But of course, I don't write (deliberately) obfuscated code.
Finally, this line: v p(13,13,13);
v is a class with a constructor taking three arguments. This declares an variable called p, of type v, initialised using that constructor; i.e. the three co-ordinates are initialised to 13.
When you seeVector& n it is referencing the vector passed into the function. This means that you can change n inside of this function without having to copy it to another Vector or without returning the Vector. This previous answer should be helpful to you.