x = ++2<<4%3*3
So i assumed that by precedence :
List item
++2
4%3
what is get in (2) *3.
but then i wondered doesn't ++2 mean increase x by 2? so the answer would be in terms of x :S ?
PLEASE NOTE: this is not a code im trying to compile its, a question i have had problems in at a written coding excercise where im supposed to calculate x.
then i wondered doesn't ++2 mean increase x by 2?
No. ++2 is illegal. The pre-increment can be used only with an expression that evaluates to an lvalue.
but then i wondered doesn't ++2 mean increase x by 2?
No; it attempts to increment the literal 2 (which is not allowed). x += 2 increases x by 2.
Perhaps you (or the person you got the code from) meant to write
x += 2 << 4 % 3 * 3; // whitespace is your friend, use it
in which case, you're adding 16 to x:
x += ( 2 << ( ( 4 % 3 ) * 3 ) )
x += ( 2 << ( 1 * 3 ) )
x += ( 2 << 3 )
x += 16
Related
I'm trying to find a efficient algorithm in C, which provides me all options of a given equation.
I have equation AX + BY = M, where A, B and M i got on input (scanf).
For example lets have: 5X + 10Y = 45
1st option: 5 * 9 + 10 * 0
2nd option: 5 * 7 + 10 * 1
n-th option: 5 * 1 +
10 * 4
And also I need to count how many possible options exist?
Some tips, hints?
I forgot to say that X and Y are in Z and >= 0, so there is no infinite options.
The question makes sense if you restrict to non-negative unknowns.
Rewrite the equation as
AX = M - BY.
There can be positive solutions as long as the RHS is positive, i.e.
BY ≤ M,
or
Y ≤ M/B.
Then for a given Y, there is a solution iff
A|(M - BY)
You can code this in Python as
for Y in range(M / B + 1):
if (M - B * Y) % A == 0:
X= (M - B * Y) / A
The solutions are
9 0
7 1
5 2
3 3
1 4
The number of iterations equals M / B. If A > B, it is better to swap X and Y.
you can calcule every solution if you put some limit in your input value, for example: use X and Y in a value included from 0 to 9... in this way you can use for to calculate every solution.
The number of solution is infinite:
find a first solution like: X=9, Y=0.
you can create another solution by using:
X' = X+2*p
Y' = Y-p
For any p in Z.
This proves your program will never terminate.
I just started learning prolog recently and am confused about how the recursion works. I have written this code:
test(X,B):- B is X + 2, B < 22, test(B,_).
test(X,Y).
I want it to return 20 or 21, but instead, if I call, say, test(4,X) it returns 6 over and over again (7 times to be exact), like this:
X = 6 ;
X = 6 ;
X = 6 ;
...
etc. So I was wondering what I was doing wrong. Thanks in advance for the help! Really appreciate it.
You misinterpreted the result you got, look:
?- test(4,X).
X = 6
; X = 6
; X = 6
; X = 6
; X = 6
; X = 6
; X = 6
; X = 6
; true. % <====
Note the true at the end, it means: Yes, this is true for any X!
Here is what you probably meant to write - trying to simulate your exact way of expression:
mtest(X0, X) :-
X1 is X0+2,
X1 < 22,
mtest(X1, X).
mtest(X0, X) :-
X is X0,
X+2 >= 22.
Some things are remarkable:
Note variable X in the first clause: It "hands back" the result.
The second clause needs to express the opposite of the very condition you gave in the first clause, because clauses are read independently of each other. In your original program the fact test(X,Y). stated that the relation is true for everything! (You got a warning for that, didn't you?) For example, also test(7,1000). succeeded.
The X is X0 is here to ensure that the second argument will be only a number but not an expression.
In any case, as a beginner, try to use library(clpfd), clpfd instead of (is)/2. Or even better, start with successor-arithmetics.
I want to shift an array within stateflow by one element.
In matlab, I would use circshift or this code:
>> x = [1:5]
x =
1 2 3 4 5
>> x(2:end) = x(1:end-1)
x =
1 1 2 3 4
>> x(1) = 0 % New Value
x =
0 1 2 3 4
How can I implement this in stateflow action language.
Embedded matlab-function is not possible because of realtime-target.
I tried this:
{x[2:end] = x[1:end-1];
x[1] = 0;}
but thats a syntax error.
For loop should be possible as well, but thats strange to me in matlab :-)
Thanks in advance
Both answers in comments helped a lot: Matlab Function works also with my target and circshift works fine.
This seems to be the #1 thing that is asked when dealing with Remainder/Mod, and I'm kind of hitting a wall with it. I'm teaching myself to program with a textbook and a chuck of C code.
Seeing as I don't really have an instructor to say, "No, no. It actually works like this", I thought I'd try my hand here. I haven't found a conclusive answer to the mathematical part of this, though.
So... I'm under the impression that this is a pretty rare occurrence, but I'd still like to know what it is that happens underneath the shiny compiling. Plus, this textbook would like for me to supply all values that are possible when using negative remainders, per the C89 Standard. Would it be much to ask if someone could check to see if this math is sound?
1) 9%4
9 - (2) * 4 = 1 //this is a value based on x - (x/y) * y
(2) * 4 + (1) = 9 //this is a check based on (x/y) * y + (x%y) = x
2) -9%4
9 - (2) * 4 = 1; 9 - (3) * 4 = -3 //these are the possible values
(2) * 4 + (1) = 9; (3) * 4 + (-3) = 9 //these are the checks
3) 9%-4
Same values as #2??
I tried computing with negatives in the expressions, and came up with ridiculous things such as 17 and -33. Are they 1 and -3 for #3 as well??
4) -9%-4
Same as #1??
In algebraic division, negative signs "cancel". Do they do the same here, or is there something else going on?
I think the thing that gets me confused the most is the negatives. The way I learned algebra in school (5-6 years ago), they are "attached" to their numbers. In programming, since they are unary operators, is that not so? Example: When filling in the value for x on #2, x = 9 instead of x = -9.
I sincerely appreciate any help.
Here you need the mathematical definition on remainder.
Given two integer numbers m, d, we say that r is the remainder of the division of m and d if r satisfies two conditions:
There exists another integer k such that m == k * d + r , and
0 <= r < d.
For positive numbers, in C, we have m % d == r and m / d == k, just by following the definition above.
From the definition, it can be obtainded that 3 % 2 == 1 and 3 / 2 == 1.
Other examples:
4 / 3 == 1 and 5 / 3 == 1, in despite of 5.0/3.0 == 1.6666 (which
would round to 2.0).
4 % 3 == 1 and 5 % 3 == 2.
You can trust also in the formula r = m - k * d, which in C is written as:
m % d == m - (m / d) * d
However, in the standard C, the integer division follows the rule: round to 0.
Thus, with negative operands C offer different results that the mathematical ones.
We would have:
(-4) / 3 == -1, (-4) % 3 == -1 (in C), but in plain maths: (-4) / 3 = -2, (-4) % 3 = 2.
In plain maths, the remainder is always nonnegative, and less than the abs(d).
In standard C, the remainder always has the sign of the first operand.
+-----------------------+
| m | d | / | % |
+-----+-----+-----+-----+
| 4 | 3 | 1 | 1 |
+-----+-----+-----+-----+
| -4 | 3 | -1 | -1 |
+-----+-----+-----+-----+
| 4 | -3 | -1 | 1 |
+-----+-----+-----+-----+
| -4 | -3 | 1 | -1 |
+-----------------------+
Remark: This description (in the negative case) is for standard C99/C11 only. You must be carefull with your compiler version, and do some tests.
Like Barmar's linked answer says modulus in a mathematical sense means that numbers are the same class for a ring (my algebra theory is a bit rusty so sorry the terms might be a bit loosely used:)).
So modulus 5 means that you have a ring of size 5. i.e. 0,1,2,3,4 when you add 1 to 4 you are back at zero. so -9,-4,1,6,11,16 are all the same modulo 5 because they are all equivalent. This is actually very important for various algebra theorems but for normal programmers it's pretty much useless.
Basically the standards were unspecified so the modulus returned for negative numbers just has to be one of those equivalent classes of numbers. It's not a remainder. Your best bet in situations like this is to operate on absolute values when doing modulo operators if you want basic integer division. If you are using more advanced techniques (like public key encryption) you'll probably need to brush up on your math a little more.
For now I'd say still with positive ints in this case and have fun programming something interesting.
I am trying to understand the MARIE assembly language. I don't quite understand skipcond for
doing things like <, or >, or multiply or divide.
I am taking this simple program:
x = 1
while x < 10 do
x = x +1
endwhile;
What I don't understand is how to use certain skip conditions:
Skipcond 800 if AC > 0,
Skipcond 400 if AC = 0,
Skipcond 000 if AC < 0
Now, I know I would subtract x from 10 and test using skipcond.
I am not sure which one and why. I guess if I knew how they really work maybe it would be easier to understand. Why is it used to compare to zero?
This is what I have:
100 load one
101 store x
102 subt ten
103 skipcond400 if x-10 = 0? // or skpcond000 x -10 < 0??
while x < 10 do
x = x + 1
will jump out of the loop as soon as x equals 10. If you subtract 10 from x, you'll get a negative value until x equals 10 (and the value is 0). So using skpcond000 would be wrong as it would jump out too soon. So skpcond400 is correct.
Perhaps it is easier to understand if you change the C code so it will be closer to the assembly code:
Original: while (x < 10) do
Subtract 10: while ((x - 10) < 0) do
Use != instead of <: while ((x - 10) != 0) do
Also note that you have to increase x after the condition to reproduce identical behaviour to the while loop.
This may help. There are many ways to write this but I think this is the easiest way to understand what is happening in the loop. Note: usually variables are placed at the bottom of the program.
while x<10
x = x+1
Org 100
Load One / loads accumulator = 1 from a decimal constant
Store X / initialize the var x = 1
loop, Load X / loads x into the accumulator
Subt Ten / compares x to 10
Skipcond 000 / if ac < 0 i.e. if x < 10 run rest of loop body
JUMP Endloop / if ac => 10 terminate loop
Load X / begin the Loop
ADD One / add 1 to x
Store X / store new value in X
JUMP loop / continue loop
Endloop Halt / ends loop
One = DEC 1 Constant
Ten = DEC 10 Constant
X = 0 Variable