I would like to convert $14.83984375$ to Binary using IEEE 754 with 32 bits.
The true solution is: $01000001011011010111000000000000$.
But I do not get it.
My solution is:
1. the sign is positive: 0
2. 14 in binary equals $1110$ $0.83984375$ equals $110101100...0$
3. So the exponent is 3 and we get: $3+127 = 10000010$ in bits
So if we put that together we get
$0 10000010 110110101100...0$
The failure is, that I have $0.83984375$ equals $110101100...0$ and the solution says $0.83984375$ equals $110101110...0$.
Where does the $1$ come from?
Thank you in advance.
You didn't convert 0.83984375 correctly; it is 0.11010111.
Related
I've just started learning. I am trying to understand this thoroughly and deeply. I somewhat understand it as long as it's subtraction between a Smaller number from a Larger number or a Larger number from a Smaller number, but when the Minuend in the question is also Negative it confuses me. Which value do I get the complement of?
When asked to do a problem such as, 0101 - 1100, when do I treat the 1 as a negative bit instead of just an unsigned bit? When would I read it as 5-12 instead of 5-(-3)?
How do you solve 0101 - 1100 using One's Complement? Is it possible? Do I treat the question as asking me to subtract 12 from 5 instead of -3 from 5.
How do you subtract a number from a negative number? For example, -5-7? How do you do this using One's Complement only? Do you get the complement of 7 or 5 or both and add them? How would it change in Two's Complement?
Please someone can clarify this for me, I want to understand this and move on from this.
So after scouring other sources and getting some help, I have cleared all this up. I will answer it helps anyone that comes across this.
1. Identifying whether to read 0101 - 1100 as 5-12 or 5-(-3) can only be done if there is context given as to whether it's signed or unsigned.
2. With Ones complement, 0101 - 1100 = 5-(-3) = 5+3.
We get Ones complement of 1100 by negating/inverting = 0011 = 3. Then we do addition:
5+3 = 0101 + 0011 = 1000
The answer is not 8 and is a negative result.
This means it has overflowed, because we added two values of the same sign and got a result with the opposite sign. The overflow occurs because 4 bits Ones complement has a range of -7 to +7. The 1 in 1000 occupies the positional value -(2^(4-1))-1 which is -7. So it has gone from +7 to -7 with the extra 1 value (8-7 = +1), meaning the -7 is essentially +7 + 1 and that additional 1 has overflowed into -7. This because the 8th value in the range starting from 0 is -7 in 4 bits Ones complement representation. Consider this:
Values from -7 to +7 can form a wheel. -7,-6,-5,-4,-3,-2,-1,-0,0,1,2,3,4,5,6,7
If you keep adding 1 to each value above it will take you to the next value and
keep going in a circle in binary form.
Example:
+7 = 0111
0111 + 1 = 1000 = -7
1000 + 1 = 1001 = -6
1001 + 1 = 1010 = -5 and on on and on.
So the calculation is correct, just that the 4 bit ones complement representation cannot represent a +8 and overflows to a -7.
**3.**In this case it underflows, where the negative number of the -5-7 is too low to be represented with Ones Complement and Twos Complement in 4 bits representation.
-5 in Ones' Complement = 1010, -7 in Ones' Complement = 1000
-5-7 = -a-b = -(a+b)
So mathematically you can look at it as 1010+1000 or -(0101+0111)
1010 0101
+1000 +0111
=Carry->1 0010 = -(1100)= -(-3)= 3
+ 1
= 0011= 3
So we end up with the same answer regardless, I'd recommend the first method though so that it doesn't lead to confusion.
The result is +3 after adding two negative values, and the value itself is 3, which shows that it has underflowed because -12 cannot be represented in 4 bit Ones Complement as the values are only from -7 to +7 as mentioned above.
So referring to the list of the range of values above. If you count backwards 7 units you end up at +3. If the range of values contained -12 value then counting back from -5 will have you end up at -12.
With Twos complement, the difference is that you add 1 to the ones complement values to get Two's Complement. Then add them the same way.
-5 in Ones' Complement = 1010, -7 in Ones' Complement = 1000
Two's Complement = 1011, Two's Complement = 1001
1011
+1001
= Overflow bit->1 0100 = +4
In Two's Complement calculation the Overflow bit is ignored if you get it when adding two values of different signs, or when adding the same sign you get an answer with the same sign. In the above case however, it is 0100 which is +4, after adding two negative values. So it indicates overflow.
The reason the value is +4 instead of +3 like in Ones' Complement is because there is an extra negative value in Two's Complement range compared to One's Complement as Two's Complement doesn't have a negative 0.
The range of values for Two's Complement 4 bit representation is:
-2^(4-1) to +(2^(4-1))-1 which is equal to -8 to +7
-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7
If you count backwards 7 units from -5, you end up at +4. Since -12 cannot still
be represented between -8 to +7, +4 is where you end up with the overflow.
I hope this clears up things and helps someone. Please let me know if there are any mistakes and I'll correct them.
Numbers under 1 are currently being represented with a leading zero before the decimal point (example: 0.50). Because I'm working with baseball statistics (which almost never have the zero before the decimal) I would like to remove that. I want to keep the number before the decimal if its greater than 1 though. How would I do that?
For instance if I'm working with this measure. Is there something I can add to that?
AVG = SUM(Batter[H])/sum(Batter[AB])
Thanks. I appreciate the help.
Here is some sample data
Name AB H
Gleyber Torres 546 152
Brett Gardner 491 123
Aaron Judge 378 103
Adam Ottavino 0 0
Aroldis Chapman 0 0
The NAN error is occurring because you are dividing by 0. You should add an IF condition to avoid that:
AVG = IF(sum(Batter[AB])=0,BLANK(),SUM(Batter[H])/sum(Batter[AB]))
To tackle the formatting issue you can use the FORMAT function as mentioned by Andrey:
AVG = IF(sum(Batter[AB])=0,BLANK(),FORMAT(SUM(Batter[H])/sum(Batter[AB]),"###.0#"))
Hope this helps.
Unfortunately, it isn't directly possible. However, in the last step (the visualization of the data), you can convert the decimal number to text and format it as you want. For example, your measure could be like this:
AVG = FORMAT(SUM(Batter[H])/SUM(Batter[AB]), "#,###.00")
This will give you 2 decimal places (0 means that there will be a digit displayed at this position), but the digits before the decimal are optional (# means it will show a digit, but will omit the leading zeros) or here are some examples:
I am sorry if it's already answered in here, but I couldn't find the exact solution I am looking for.
I am trying to output the result of a division of two variable which is continuously changing. The result may vary from 0.00 to 100.12314235234523 (not exactly this specific, I just wanted to give you an idea). I want to print the result only 2 digits after the decimal point, if there is nothing after decimal, it should print 2 zeroes after decimal.
For example:
10 / 5 = 2.00
23 / 6 = 3.83
I don't need to round up the result for example, if the output is: 73.4869999, I don't need it to be 73.49, 73.48 is fine with me.
What I have written so far is:
packet_loss[2]->setText(QString("%1%2%3").arg(((data.packet_loss_tx) * 100) / data.packets_tx).
arg(locale.decimalPoint()).
arg((((data.packet_loss_tx) * 100) % data.packets_tx), 2, 10, QChar('0')));
But this prints all the values after the decimal point. I can divide this part arg((((data.packet_loss_tx) * 100) % data.packets_tx) with 10, 100 or 1000 to reduce the number of decimals after the decimal point but this is a variable which changes every seconds. So if the output is 3 digits after decimal and I divide it by 10, I will get proper output, but the next value may be 5 digits after decimal and division by 10 will give me 4 digits after decimal. I want the final output to show only 2 digits after decimal.
You could try to use QString::number() function with specific formatting options (two digits precision). For example:
auto s = QString::number(100.12914235234523, 'f', 2); // Gives "100.13"
Besides, if you use floating point numbers, it's better to multiply them with floating point numbers too. I.e. you need to perform your calculations using 100.0 instead of integer value 100.
I have a regexp to check for a decimal with 2 numbers, but I want to check both the integer and the decimal part for their length.
/^\s*-?[1-9]\d*(\.\d{1,2})?\s*$/;
The above code is decimal with length 2 (ex: 12.23) but I want 10 integer value and 2 decimal value (10,2) like,
1234567890.12
Use /^(?![.])\d{0,10}(\.\d{1,2})?$/
It allows 1.23, 1.2 0.2
Invalid values ., 1.
Depending on what you exactly want, you can use:
/^\s*-?(\d{1,10}(\.\d{1,2})?)\s*$/
for input like: 12.23, 3.4, 1234567890.34, 4, 456, etc., or:
/^\s*-?(\d{10}(\.\d{1,2})?)\s*$/
for: 9087654321, 1234567890.1, 1234567890.23 (10 digits, and optional point and one or two digits), or:
/^\s*-?(\d{10}\.\d{2})\s*$/
for exactly 10 digits fallowed by point and 2 digits, like: 9087654321.12, etc. Its all depends on what kind of numbers you want to filter.
I am trying to find a regular expression that works for a decimal with a max of 2 digits before the decimal point and 2 digits after the decimal point. The decimal point and decimal places are optional. So these values would be accepted :
90
5.4
45.21
0.5
0
And the would be rejected :
100
105.56
05.6
55.543
78.
Can any regex gurus help?
This should work:
^[1-9]\d?(?:\.\d{1,2})?$
If you want to accept even 0.5, try this:
^(?:[1-9]\d?|0)(?:\.\d{1,2})?$
I hope this will work for you
\d{0,2}(\.\d{1,2})?
/\A[1-9]?\d(?:\.\d{1,2})?\z/
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