I'm trying to get printf to diplay number of a constant length (in characters) in OS X Terminal. So if I specify three character length, then the number will be
I'd like to achieve this:
100.4125 -> 100
1.5131 -> 1.51
59.9159 -> 59.9
print substr(0,4) would nearly achieve this, but I would be left with a decimal pointer after the 100.
Using the specifier g or G will format to significant digits. The following example will format to 3 significant figures.
printf("ans: %.3g", 100.4125)
Related
Friends,
I am doing SAS to COBOL conversion.I am stuck with below declaration and conversion.So I am getting SOC7 in COBOL run.Please provide some solution.
IP in SAS - PD3.5
OP in SAS - z6.5
My COBOL declaration below.
IP s9.9(5);
OP .9(5);
Please suggest some solution..
Thanks a lot!!
Packed Decimal is stored one digit per nibble, which is two digits per byte, with the last nibble storing the sign. The sign nibbles C, A, F, and E are treated as positive; the sign nibbles B and D are treated as negative. Sign nibbles C and D are referred to as "preferred sign". A sign nibble of F is considered "unsigned," meaning it is neither positive nor negative, though pragmatically you can think of it as positive for arithmetic purposes. +123 is stored in two bytes as x'123C', -456 is stored as x'456D'.
The SAS PD informat specifies PDw.d where w is the width of the field in bytes and d is the number of decimal places to the right within the field. So PD3.5 is a 3 byte field (which would store 5 digits and a sign) with all 5 digits to the right of the decimal point.
To obtain the COBOL declaration for a SAS PDw.d declaration...
a = (w * 2) - 1
b = a - d
if b = 0
PIC SVd Packed-Decimal
else
PIC S9(b)Vd Packed-Decimal
The SAS Z format specifies Zw.d where w is the width of the field in bytes and d is the number of decimal places to the right within the field. The field will be padded with zeroes on the left to make it w bytes wide. So Z6.5 specifies a 6 byte output field with 5 bytes to the right of the decimal point. One byte is taken by the decimal point itself, and unfortunately there is no room for the sign, which may be a bug or may be intentional (perhaps all the data is known to be positive).
IP PIC Sv99999 Packed-Decimal.
OP PIC .99999.
When you MOVE IP TO OP the conversion from Packed Decimal to Zoned Decimal will be done for you by COBOL.
When I calculate log(8) / log(2) I get 3 as one would expect:
?log(8)/log(2)
3
However, if I take the int of this calculation like this the result is 2 and thus wrong:
?int(log(8)/log(2))
2
How and why does this happen?
Likely because the actual number returned is of type double. Because floats and doubles cannot accurately represent most base 10 rational numbers the number returned is something like 2.99999999999. Then when you apply int() the .999999999 is truncated.
How floating-point number works: it dedicates a bit for the sign, a few bits to store an exponent, and the rest for the actual fraction. This leads to numbers being represented in a form similar to 1.45 * 10^4; except that instead of the base being 10, it's two.
I am currently trying to figure out how to multiply two numbers in fixed point representation.
Say my number representation is as follows:
[SIGN][2^0].[2^-1][2^-2]..[2^-14]
In my case, the number 10.01000000000000 = -0.25.
How would I for example do 0.25x0.25 or -0.25x0.25 etc?
Hope you can help!
You should use 2's complement representation instead of a seperate sign bit. It's much easier to do maths on that, no special handling is required. The range is also improved because there's no wasted bit pattern for negative 0. To multiply, just do as normal fixed-point multiplication. The normal Q2.14 format will store value x/214 for the bit pattern of x, therefore if we have A and B then
So you just need to multiply A and B directly then divide the product by 214 to get the result back into the form x/214 like this
AxB = ((int32_t)A*B) >> 14;
A rounding step is needed to get the nearest value. You can find the way to do it in Q number format#Math operations. The simplest way to round to nearest is just add back the bit that was last shifted out (i.e. the first fractional bit) like this
AxB = (int32_t)A*B;
AxB = (AxB >> 14) + ((AxB >> 13) & 1);
You might also want to read these
Fixed-point arithmetic.
Emulated Fixed Point Division/Multiplication
Fixed point math in c#?
With 2 bits you can represent the integer range of [-2, 1]. So using Q2.14 format, -0.25 would be stored as 11.11000000000000. Using 1 sign bit you can only represent -1, 0, 1, and it makes calculations more complex because you need to split the sign bit then combine it back at the end.
Multiply into a larger sized variable, and then right shift by the number of bits of fixed point precision.
Here's a simple example in C:
int a = 0.25 * (1 << 16);
int b = -0.25 * (1 << 16);
int c = (a * b) >> 16;
printf("%.2f * %.2f = %.2f\n", a / 65536.0, b / 65536.0 , c / 65536.0);
You basically multiply everything by a constant to bring the fractional parts up into the integer range, then multiply the two factors, then (optionally) divide by one of the constants to return the product to the standard range for use in future calculations. It's like multiplying prices expressed in fractional dollars by 100 and then working in cents (i.e. $1.95 * 100 cents/dollar = 195 cents).
Be careful not to overflow the range of the variable you are multiplying into. Your constant might need to be smaller to avoid overflow, like using 1 << 8 instead of 1 << 16 in the example above.
Why vb prints out 1??? when d is a double aproximation to 1? shoudnt be 0.99999 or something similar? what if I really need the float value? and how could I print it?
Dim d As Double
For i = 1 To 10
d = d + 0.1
Next
MsgBox(d)
Console.WriteLine(d)
thanks
When using MsgBox or Console.WriteLine, double.ToString() is called in order to convert the double to a string.
By default this uses the G format specifier.
The general ("G") format specifier converts a number to the most compact of either fixed-point or scientific notation, depending on the type of the number and whether a precision specifier is present. The precision specifier defines the maximum number of significant digits that can appear in the result string. If the precision specifier is omitted or zero, the type of the number determines the default precision, as indicated in the following table.
And:
However, if the number is a Decimal and the precision specifier is omitted, fixed-point notation is always used and trailing zeros are preserved.
When converting the infinite 0.9999999.... to a string, since it goes forever, rounding occurs, this results in 1.
A simple test is to run this:
MsgBox((0.9999999999999999999999999).ToString())
I would like to display the a number value to the max number of decimal places. If you don't format the float to:
#"%.1f"
then it will display the number as e.g. 1.000000. What I would like is that the number would have the max number of decimal places it needs e.g.
1 would not need any
1.5 would need 1 decimal place
1.24 would need 2 decimal places
Is there some sort of code that formats the number to the max number of decimal places?
Replace "f" with "g".
From printf(3):
gG The double argument is converted in style f or e (or F or E for G conver-
sions). The precision specifies the number of significant digits. If the
precision is missing, 6 digits are given; if the precision is zero, it is
treated as 1. Style e is used if the exponent from its conversion is less
than -4 or greater than or equal to the precision. Trailing zeros are removed
from the fractional part of the result; a decimal point appears only if it is
followed by at least one digit.