how to display exponent value after calulation to textbox in iphone sdk.For example say 6.4516e-10.i am not getting answer for it in my textbox after calculating 10 * 6.4516e-10.please tell me solution..
Use StringWithFormat and exponent formations:
From the man page on printf:
eE The argument is printed in the style e `[-d.ddd+-dd]' where is one digit before the decimal point and the number after is equal to the precision specification for the argument; when the precision is missing, 6 digits are produced.
So you would want a format something like: %.4e
float n = 6.4516e-10;
n = n * 10;
NSLog(#"n: %.4e", n);
2011-08-29 07:36:38.158 Test[39477:707] n: 6.4516e-09
Related
I don't understand how floating point numbers are represented in hex notation in Swift. Apple's documentation shows that 0xC.3p0 is equal to 12.1875 in decimal. Can someone walk me through how to do that conversion? I understand that before the decimal hex value 0xC = 12. The 3p0 after the decimal is where I am stumped.
From the documentation:
Floating-Point Literals
...
Hexadecimal floating-point literals consist of a 0x prefix, followed
by an optional hexadecimal fraction, followed by a hexadecimal
exponent. The hexadecimal fraction consists of a decimal point
followed by a sequence of hexadecimal digits. The exponent consists of
an upper- or lowercase p prefix followed by a sequence of decimal
digits that indicates what power of 2 the value preceding the p is
multiplied by. For example, 0xFp2 represents 15 × 22, which evaluates
to 60. Similarly, 0xFp-2 represents 15 × 2-2, which evaluates to 3.75.
In your case
0xC.3p0 = (12 + 3/16) * 2^0 = 12.1875
Another example:
0xAB.CDp4 = (10*16 + 11 + 12/16 + 13/16^2) * 2^4 = 2748.8125
This format is very similar to the %a printf-format (see for example
http://pubs.opengroup.org/onlinepubs/009695399/functions/fprintf.html).
It can be used to specify a floating point number directly in its
binary IEEE 754 representation, see Why does Swift use base 2 for the exponent of hexadecimal floating point values?
for more information.
Interpret 0xC.3p0 using the place value system:
C (or 12) is in the 16^0 place
3 is in the 16^-1 place (and 3/16 == 0.1875)
p says the exponent follows (like the e in 6.022e23 in base 10)
0 is the exponent (in base 10) that is the power of 2 (2^0 == 1)
So putting it all together
0xC.3p0 = (12 + (3/16)) * 2^0 = 12.1875
In order to sum up what I've read, you can see those representations as follow:
0xC.3p0 = (12*16^0 + 3*16^-1) * 2^0 = 12.1875
From Martin R's example above :
0xAB.CDp4 = (10*16^1 + 11*16^0 + 12*16^-1 + 13*16^-2) * 2^4 = 2748.8125
The 0xC is 12, as you said. The decimal part is ((1/16)*3)*10^0.
So you need to take the decimal part and divide it by 16. Then you need to multiply it by 2 raised to the power of the number after the p
Hexadecimal -(0-9,A=10,B=11,C=12,D=13,E=14,F=15) and p0 means 2^0
ex: - 0xC = 12 (0x prefix represents hexadecimal)
After the decimal part as in 0xC.3p0 we divide the numbers with the power of 16
So here its 3/16 = 0.1875
so 0xC.3p0 = (12 + (3/16) ) 2^0
If it was 0xC.43p0 then for the 4 we would use 4/(16), for 3 we would use 3/(16 ^2) and similarly if the decimal part increases.
ex: 0xC.231p1 = (12 + 2/16 + 3/(256) + 1/(16^3)) 2^1 = 24.27392578125
I'm working through the first basic playground in https://github.com/nettlep/learn-swift using XCode
What exactly is happening with this expression?
0xC.3p0 == 12.1875
I've learned about hexadecimal literals and the special "p" notation that indicates a power of 2.
0xF == 15
0xFp0 == 15 // 15 * 2^0
If I try 0xC.3 I get the error: Hexadecimal floating point literal must end with an exponent.
I found this nice overview of numeric literals and another deep explanation, but I didn't see something that explains what .3p0 does.
I've forked the code and upgraded this lesson to XCode 7 / Swift 2 -- here's the specific line.
This is Hexadecimal exponential notation.
By convention, the letter P (or p, for "power") represents times two
raised to the power of ... The number after the P is decimal and
represents the binary exponent.
...
Example: 1.3DEp42 represents hex(1.3DE) × dec(2^42).
For your example, we get:
0xC.3p0 represents 0xC.3 * 2^0 = 0xC.3 * 1 = hex(C.3) = 12.1875
where hex(C.3) = dec(12.{3/16}) = dec(12.1875)
As an example, you can try 0xC.3p1 (equals hex(C.3) * dec(2^1)), which yields double the value, i.e., 24.375.
You can also study the binary exponent growth in a playground for hex-value 1:
// ...
print(0x1p-3) // 1/8 (0.125)
print(0x1p-2) // 1/4 (0.25)
print(0x1p-1) // 1/2 (0.5)
print(0x1p1) // 2.0
print(0x1p2) // 4.0
print(0x1p3) // 8.0
// ...
Finally, this is also explained in Apple`s Language Reference - Lexical Types: Floating-Point Literals:
Hexadecimal floating-point literals consist of a 0x prefix, followed
by an optional hexadecimal fraction, followed by a hexadecimal
exponent. The hexadecimal fraction consists of a decimal point
followed by a sequence of hexadecimal digits. The exponent consists
of an upper- or lowercase p prefix followed by a sequence of decimal
digits that indicates what power of 2 the value preceding the p is
multiplied by. For example, 0xFp2 represents 15 x 2^2, which
evaluates to 60. Similarly, 0xFp-2 represents 15 x 2^-2, which
evaluates to 3.75.
I don't understand how floating point numbers are represented in hex notation in Swift. Apple's documentation shows that 0xC.3p0 is equal to 12.1875 in decimal. Can someone walk me through how to do that conversion? I understand that before the decimal hex value 0xC = 12. The 3p0 after the decimal is where I am stumped.
From the documentation:
Floating-Point Literals
...
Hexadecimal floating-point literals consist of a 0x prefix, followed
by an optional hexadecimal fraction, followed by a hexadecimal
exponent. The hexadecimal fraction consists of a decimal point
followed by a sequence of hexadecimal digits. The exponent consists of
an upper- or lowercase p prefix followed by a sequence of decimal
digits that indicates what power of 2 the value preceding the p is
multiplied by. For example, 0xFp2 represents 15 × 22, which evaluates
to 60. Similarly, 0xFp-2 represents 15 × 2-2, which evaluates to 3.75.
In your case
0xC.3p0 = (12 + 3/16) * 2^0 = 12.1875
Another example:
0xAB.CDp4 = (10*16 + 11 + 12/16 + 13/16^2) * 2^4 = 2748.8125
This format is very similar to the %a printf-format (see for example
http://pubs.opengroup.org/onlinepubs/009695399/functions/fprintf.html).
It can be used to specify a floating point number directly in its
binary IEEE 754 representation, see Why does Swift use base 2 for the exponent of hexadecimal floating point values?
for more information.
Interpret 0xC.3p0 using the place value system:
C (or 12) is in the 16^0 place
3 is in the 16^-1 place (and 3/16 == 0.1875)
p says the exponent follows (like the e in 6.022e23 in base 10)
0 is the exponent (in base 10) that is the power of 2 (2^0 == 1)
So putting it all together
0xC.3p0 = (12 + (3/16)) * 2^0 = 12.1875
In order to sum up what I've read, you can see those representations as follow:
0xC.3p0 = (12*16^0 + 3*16^-1) * 2^0 = 12.1875
From Martin R's example above :
0xAB.CDp4 = (10*16^1 + 11*16^0 + 12*16^-1 + 13*16^-2) * 2^4 = 2748.8125
The 0xC is 12, as you said. The decimal part is ((1/16)*3)*10^0.
So you need to take the decimal part and divide it by 16. Then you need to multiply it by 2 raised to the power of the number after the p
Hexadecimal -(0-9,A=10,B=11,C=12,D=13,E=14,F=15) and p0 means 2^0
ex: - 0xC = 12 (0x prefix represents hexadecimal)
After the decimal part as in 0xC.3p0 we divide the numbers with the power of 16
So here its 3/16 = 0.1875
so 0xC.3p0 = (12 + (3/16) ) 2^0
If it was 0xC.43p0 then for the 4 we would use 4/(16), for 3 we would use 3/(16 ^2) and similarly if the decimal part increases.
ex: 0xC.231p1 = (12 + 2/16 + 3/(256) + 1/(16^3)) 2^1 = 24.27392578125
I have a, very long, integer. The integer is represented by a array of unsigned chars.
Example: the integer 1234 with base 10 is represented in the array as [4,3,2,1], [2,2,3,2] (base 8) and [2,13,4] (base 16)
Now I want to convert my integer with base n to another integer with base m. In my persued for a answer I came accross Wallar's algorithm, originally from here.
from math import *
def baseExpansion(n,c,b):
j = 0
base10 = sum([pow(c,len(n)-k-1)*n[k] for k in range(0,len(n))])
while floor(base10/pow(b,j)) != 0: j = j+1
return [floor(base10/pow(b,j-p)) % b for p in range(1,j+1)]
At first I thought this was my answer but unfortunately it is not. The problem I have is that the algorithm computes the sum. In my case this is a problem because the variable base10 is of type unsigned integer of 32 bits. Therefore when my integer, represented as a array, has more then 10 digits it can not convert the number anymore. Anyone has a solution?
Here's the school-book algorithm for doing what you're trying. You start with a representation for zero and call it a running total. Then, for each digit of the number to be converted, starting with the most significant and going to the least significant, 1) multiply the running total by the base of the source number and 2) add the digit to the running total. Now all you need is algorithms to do the multiplication and addition (and you can actually do both at once). Here's how to do that: 1) set the current digit to a variable, call it "carry", 2) for each digit in your new number, starting with the least significant and going to the most significant: 2a) set carry to the current digit in the new number times the output base plus carry, 2b) set the current digit to carry mod the output base, 2c) set carry to carry divided by the output base. And that should do it. There is an implementation of what you are trying to do somewhere here: http://www.cis.ksu.edu/~howell/calculator/comparison.html
I want to convert the decimal number 27 into binary such a way that , first the digit 2 is converted and its binary value is placed in an array and then the digit 7 is converted and its binary number is placed in that array. what should I do?
thanks in advance
That's called binary-coded decimal. It's easiest to work right-to-left. Take the value modulo 10 (% operator in C/C++/ObjC) and put it in the array. Then integer-divide the value by 10 (/ operator in C/C++/ObjC). Continue until your value is zero. Then reverse the array if you need most-significant digit first.
If I understand your question correctly, you want to go from 27 to an array that looks like {0010, 0111}.
If you understand how base systems work (specifically the decimal system), this should be simple.
First, you find the remainder of your number when divided by 10. Your number 27 in this case would result with 7.
Then you integer divide your number by 10 and store it back in that variable. Your number 27 would result in 2.
How many times do you do this?
You do this until you have no more digits.
How many digits can you have?
Well, if you think about the number 100, it has 3 digits because the number needs to remember that one 10^2 exists in the number. On the other hand, 99 does not.
The answer to the previous question is 1 + floor of Log base 10 of the input number.
Log of 100 is 2, plus 1 is 3, which equals number of digits.
Log of 99 is a little less than 2, but flooring it is 1, plus 1 is 2.
In java it is like this:
int input = 27;
int number = 0;
int numDigits = Math.floor(Log(10, input)) + 1;
int[] digitArray = new int [numDigits];
for (int i = 0; i < numDigits; i++) {
number = input % 10;
digitArray[numDigits - i - 1] = number;
input = input / 10;
}
return digitArray;
Java doesn't have a Log function that is portable for any base (it has it for base e), but it is trivial to make a function for it.
double Log( double base, double value ) {
return Math.log(value)/Math.log(base);
}
Good luck.