CLS
REM Declare Varibles
DIM MILES, YARDS, FEET, INCHES AS DOUBLE
DIM KM, METER, TINCH AS DOUBLE
DIM CM, TMETER AS DOUBLE
REM INPUT THE DATA
INPUT "ENTER THE DISTANCE IN MILES", MILES
INPUT "ENTER THE DISTANCE IN YARDS", YARDS
INPUT "ENTER THE DISTANCE IN FEET", FEET
INPUT "ENTER THE DISTANCE IN INCHES", INCHES
REM CONVERT INTO TOTAL INCHES
TINCH = 63360 * MILES + 36 * YARDS + 12 * FEET + INCHES
REM CONVERT INTO TOTAL METER
TMETER = TINCH / 39.37
REM CONVERT TO KILOMETER
KM = INT(TMETER / 1000)
REM CONVERT TO METER
METER = INT(TMETER - KM * 1000)
REM CONVERT TO CENTIMETER
CM = (TMETER - (KM * 1000) - METER) * 100
REM PRINT DETAILS
PRINT "KILOMETER", KM
PRINT "METER", METER
PRINT "CENTIMETER", CM;
Classic QBasic did not have the round function so you need to bring your own.
FUNCTION round# (num AS DOUBLE, dp AS INTEGER)
'WARNING: USE "#" at the end of constant values,
'or else you will get rounding errors:
' "num = .45" >> "num = .449999988079071
' "num = .45#" >> "num = .45"
DIM exp1 AS LONG, num2 AS LONG
exp1 = 10 ^ dp: num2 = num * exp1: round# = num2 / exp1
PRINT num
END FUNCTION
Borrowed this from, qbasicnews.com
If I remember correctly, you can use the PRINT USING statement.
cm_number = 73.53315
PRINT USING "##.##"; cm_number
OUTPUT: 73.53
"#" - represent digits,
"." - represents the decimal point position
Click here for a more detailed explanation.
Related
def call_result(label_result, n1, n2, n3):
num1 = (n1.get())
num2 = (n2.get())
num3 = (n3.get())
num4 = 100
num5 = 12
num6 = 2
main1 = float(int(num1))*float(int(num2)) / 12 / 2 / 100
main2 = float(int(num1)) / float(int(num3))
main3 = float(int(main2))+float(int(main1))
label_result.config(text="Result = %d" % float(main3))
return
I'm in trouble. Decimal numbers don't come when I do math
The answer is supposed to be 20.84 but this code produces 20.
The problem is due to the conversion between int and float
Assuming the three nums are strings you should just convert them to float directly and skip the intermediate step.
Moreover when you convert from float to int and then back to float you lost the decimal data because int does not store it.
For example
main2 = 23.34 # main2 is a float with value 23.34
main2 = int(main2) # main2 is now an int with value 23
# Remember int (short for integer) cannot store floating point values and will truncate the number to make an integer
main2 = float(main2) # main2 is now a float but the value is 23.0 because you lost the precision when it was converted to int
I recommend doing something like this
def call_result(label_result, n1, n2, n3):
num1 = float(n1.get())
num2 = float(n2.get())
num3 = float(n3.get())
num4 = 100
num5 = 12
num6 = 2
main1 = num1 * num2 / 12 / 2 / 100
main2 = num1 / num3
main3 = main2 + main1
label_result.config(text="Result = %.2f" % main3)
return
Another thing to take care of is the format specifier you use for the string. %d refers to an integer while %f is for floats. Even if main3 was a float you would have ended up with an integer in your string.
%.2f will truncate a float to a precision of 2 decimal places
"%.2f" % 123.45678 becomes 123.45
Take a look at the string formatting guide here
I'm writing a program that calculates the Factorial of 5 numbers and output the results in a Tabular form but I keep getting Zeros.
Factorial Formula:. n! = n×(n-1)!
I tried:
CLS
DIM arr(5) AS INTEGER
FOR x = 1 TO 5
INPUT "Enter Factors: ", n
NEXT x
f = 1
FOR i = 1 TO arr(n)
f = f * i
NEXT i
PRINT
PRINT "The factorial of input numbers are:";
PRINT
FOR x = 1 TO n
PRINT f(x)
NEXT x
END
and I'm expecting:
Numbers Factorrials
5 120
3 6
6 720
8 40320
4 24
You did some mistakes
FOR i = 1 TO arr(n)
where is n defined
you also never stored actual values into arr
PRINT f(x)
here you take from array f that is also not defined in your code
Possible solution to calculate arrays of factorials:
CLS
DIM arr(5) AS INTEGER
DIM ans(5) AS LONG
FOR x = 1 TO 5
INPUT "Enter Factors: ", arr(x)
f& = 1
FOR i = 1 TO arr(x)
f& = f& * i
NEXT i
ans(x) = f&
NEXT x
PRINT
PRINT "The factorial of input numbers are:";
PRINT
PRINT "Numbers", "Factorials"
FOR x = 1 TO 5
PRINT arr(x), ans(x)
NEXT x
END
I don't have a BASIC interpreter right in front of me, but I think this is what you're looking for:
CLS
DIM arr(5) AS INTEGER
DIM ans(5) AS LONG 'You need a separate array to store results in.
FOR x = 1 TO 5
INPUT "Enter Factors: ", arr(x)
NEXT x
FOR x = 1 to 5
f& = 1
FOR i = 1 TO arr(x)
f& = f& * i
NEXT i
ans(x) = f&
NEXT x
PRINT
PRINT "The factorial of input numbers are:";
PRINT
PRINT "Numbers", "Factorials"
FOR x = 1 TO 5
PRINT STR$(arr(x)), ans(x)
NEXT x
END
Just a comment though: In programming, you should avoid reusing variables unless you are short on memory. It can be done right, but it creates many opportunities for hard to find bugs in larger programs.
Possible solution to calculate arrays of factorials and square roots:
CLS
PRINT "Number of values";: INPUT n
DIM arr(n) AS INTEGER
DIM ans(n) AS LONG
FOR x = 1 TO n
PRINT "Enter value"; x;: INPUT arr(x)
f& = 1
FOR i = 1 TO arr(x)
f& = f& * i
NEXT i
ans(x) = f&
NEXT x
PRINT
PRINT "The factorial/square root of input numbers are:";
PRINT
PRINT "Number", "Factorial", "Squareroot"
FOR x = 1 TO n
PRINT arr(x), ans(x), SQR(arr(x))
NEXT x
END
I am trying to convert some data with the indicated measurements package, but I'm not succeeding on it.
My data:
Long Lat
62ᵒ36.080 58ᵒ52.940
61ᵒ28.020 54ᵒ59.940
62ᵒ07.571 56ᵒ48.873
62ᵒ04.929 57ᵒ33.605
63ᵒ01.419 60ᵒ30.349
63ᵒ09.555 61ᵒ29.199
63ᵒ43.499 61ᵒ23.590
64ᵒ34.175 62ᵒ30.304
63ᵒ16.342 59ᵒ16.437
60ᵒ55.090 54ᵒ49.269
61ᵒ28.013 54ᵒ59.928
62ᵒ07.868 56ᵒ48.040
62ᵒ04.719 57ᵒ32.120
62ᵒ36.083 58ᵒ51.766
63ᵒ01.644 60ᵒ30.714
64ᵒ33.897 62ᵒ30.772
63ᵒ43.604 61ᵒ23.426
63ᵒ09.288 61ᵒ29.888
63ᵒ16.722 59ᵒ16.204
What I'm trying:
library(measurements)
library(readxl)
coord = read.table('coord_converter.txt', header = T, stringsAsFactors = F)
# change the degree symbol to a space
lat = gsub('°','', coord$Lat)
long = gsub('°','', coord$Long)
# convert from decimal minutes to decimal degrees
lat = measurements::conv_unit(lat, from = 'deg_dec_min', to = 'dec_deg')
long = measurements::conv_unit(long, from = 'deg_dec_min', to = 'dec_deg')
What I'm getting with this penultimate line:
Warning messages:
In split(as.numeric(unlist(strsplit(x, " "))) * c(3600, 60), f = rep(1:length(x), : NAs introduced by coercion
In as.numeric(unlist(strsplit(x, " "))) * c(3600, 60) : longer object length is not a multiple of shorter object length
In split.default(as.numeric(unlist(strsplit(x, " "))) * c(3600, : data length is not a multiple of split variable
Can someone point my mistake or make a suggestion of how to proceed?
Thank you!
I think the issue here was that after gsub call, degrees and minutes were not space delimited, as required by measurements::conv_unit.
For example, this works fine (for this reproducible example I also changed "ᵒ" to "°"):
library(measurements)
#read your data
txt <-
"Long Lat
62°36.080 58°52.940
61°28.020 54°59.940
62°07.571 56°48.873
62°04.929 57°33.605
63°01.419 60°30.349
63°09.555 61°29.199
63°43.499 61°23.590
64°34.175 62°30.304
63°16.342 59°16.437
60°55.090 54°49.269
61°28.013 54°59.928
62°07.868 56°48.040
62°04.719 57°32.120
62°36.083 58°51.766
63°01.644 60°30.714
64°33.897 62°30.772
63°43.604 61°23.426
63°09.288 61°29.888
63°16.722 59°16.204"
coord <- read.table(text = foo, header = TRUE, stringsAsFactors = F)
# change the degree symbol to a space
lat = gsub('°',' ', coord$Lat)
long = gsub('°',' ', coord$Long)
# convert from decimal minutes to decimal degrees
lat = measurements::conv_unit(lat, from = 'deg_dec_min', to = 'dec_deg')
long = measurements::conv_unit(long, from = 'deg_dec_min', to = 'dec_deg')
yields...
> cbind(long, lat)
long lat
[1,] "62.6013333333333" "58.8823333333333"
[2,] "61.467" "54.999"
[3,] "62.1261833333333" "56.81455"
[4,] "62.08215" "57.5600833333333"
[5,] "63.02365" "60.5058166666667"
[6,] "63.15925" "61.48665"
[7,] "63.7249833333333" "61.3931666666667"
[8,] "64.5695833333333" "62.5050666666667"
[9,] "63.2723666666667" "59.27395"
[10,] "60.9181666666667" "54.82115"
[11,] "61.4668833333333" "54.9988"
[12,] "62.1311333333333" "56.8006666666667"
[13,] "62.07865" "57.5353333333333"
[14,] "62.6013833333333" "58.8627666666667"
[15,] "63.0274" "60.5119"
[16,] "64.56495" "62.5128666666667"
[17,] "63.7267333333333" "61.3904333333333"
[18,] "63.1548" "61.4981333333333"
[19,] "63.2787" "59.2700666666667"
I have converted a decimal number to binary using STR$() in QBASIC. But I need a way to convert decimal number to binary without using string functions. Thanks.
My Code :
CLS
INPUT N
WHILE N <> 0
E = N MOD 2
B$ = STR$(E)
N = FIX(N / 2)
C$ = B$ + C$
WEND
PRINT "Output "; C$
END
This code sample converts a numeric value to a binary string in Basic.
PRINT "Enter value";
INPUT Temp#
Out3$ = ""
IF Temp# >= False THEN
Digits = False
DO
IF 2 ^ (Digits + 1) > Temp# THEN
EXIT DO
END IF
Digits = Digits + 1
LOOP
FOR Power = Digits TO 0 STEP -1
IF Temp# - 2 ^ Power >= False THEN
Temp# = Temp# - 2 ^ Power
Out3$ = Out3$ + "1"
ELSE
Out3$ = Out3$ + "0"
END IF
NEXT
END IF
PRINT Out3$
END
When you want to display an integer value as binary, it seems logical to me to store it in a string variable, because it's only for display. So I'm not really sure what you are trying to do here.
Maybe you were looking for LTRIM$ so you would get outputs like 11010 instead of 1 1 0 1 0 ?
You could store it in an integer value like in the code below. But, although the integer value will look the same as the string variable, it will in fact be a completely different value.
CLS
INPUT "Type a decimal number:", N
S$ = ""
I = 0
P = 1
WHILE (N <> 0)
' get right most bit and shift right
E = N AND 1
N = INT(N / 2) ' bit shift right
' format for dsplay
S$ = LTRIM$(STR$(E)) + S$
I = I + (E * P)
P = P * 10
WEND
PRINT "Binary as string="; S$
PRINT "Binary as int="; I
END
How do I scale a number up to the nearest ten, hundred, thousand, etc...
Ex.
num = 11 round up to 20
num = 15 round up to 20
num = 115 round up to 200
num = 4334 round up to 5000
I guess this formula might work? Unless you have more examples to show.
power = floor(log10(n))
result = (floor(n/(10^power)) + 1) * 10^power
import math
exp = math.log10(num)
exp = math.floor(exp)
out = math.ceil(num/10**exp)
out = out * 10**exp
Convert the number to a decimal (i.e. 11 goes to 1.1, 115 goes to 1.15), then take the ceiling of the number, then multiply it back. Example:
public static int roundByScale(int toRound) {
int scale = (int)Math.pow(10.0, Math.floor(Math.log10(toRound)));
double dec = toRound / scale;
int roundDec = (int)Math.ceil(dec);
return roundDec * scale;
}
In this case, if you input 15, it will be divided by 10 to become 1.5, then rounded up to 2, then the method will return 2 * 10 which is 20.
public static int ceilingHighestPlaceValue(int toCeil)
{
int placeValue = Math.Pow(10,toCeil.ToString().Length()-1);
double temp = toCeil / placeValue;
return= ceil(temp) * placeValue;
}