inpfile>>ch;
if(ch<16) outfile<<"0×0"<<std::hex<<setprecision(2)<<(int)ch<<" ";
what does std::hex<<setprecision(2) mean?
iostreams can be manipulated to achieve the desired formatting - this is done by what at first sight looks like outputting predefined values to them as shown in our subject line of code.
std::hex displays following integer values in base16.
setprecision sets the precision for display of following floating values.
Fur further info on manipulators, start here
This line is the same as:
char ch;
inpfile>>ch;
if(ch<16)
{
outfile << "0×0" // Prints "0x0" (Ox is the standard prefix for hex numbers)
/*outfile*/ << std::hex // Tells the stream to print the next number in hex format
/*outfile*/ << setprecision(2) // Does nothing. Presumably they were trying to indicate print min of 2 characters
/*outfile*/ << (int)ch // Covert you char to an integer (so the stream will print it in hex
/*outfile*/ << " "; // Add a space for good measure.
}
Rather than setprecision(2) what was probably intended was setw(2) << setfill('0')
what does std::hex<<setprecision(2) mean?
std::hex and std::setprecision are both so-called manipulators. Applied to a stream (done by outputting them) they manipulate the stream, usually to change the stream's formatting. In particular, std::hex manipulates the stream so that values are written in hexadecimal, and std::setprecision(x) manipulates it to output numbers with x digits.
(A rather popular manipulator which you might already know about is std::endl.)
As you can see, there are manipulators that take arguments and those that take none. Also, most (in principle all) manipulators are sticky, which means their manipulation of the stream lasts until it is explicitly changed. Here is an extensive discussion about this topic.
std::hex sets the output base to hexadecimal.
setprecision has no effect on this line since it affects floating point only.
Related
Is there any elegant solution using the std C++ or Boost libraries to output a double to std::cout in a way that the following conditions are met:
scientific notation is disabled
the precision for the decimal part is 6
however, trailing 0's (for the decimal part) are not printed out
For example:
double d = 200000779998;
std::cout << `[something]` << d;
should print out exactly 200000779998. [something] should possibly be a noexcept combination of some existing manipulators.
This is not a solution to the problem:
std::cout << std::setprecision(6) << std::fixed << d;
because it prints out 200000779998.000000 with trailing 0's
Instead of using the fixed manipulator, you can try to use (abuse?) defaultfloat. As far as I understand, it chooses either fixed or scientific based on the ability to put the number within the specified precision. As a result you can set the precision to the number of digits of the integral part + the requested fractional precision (6 in your case).
double d = 200000779998;
std::cout << std::setprecision(integralDigits(d) + 6) << d << std::endl;
You can try it here.
Hard to prove a negative, but I would assume no.
The requirements are inconsistent with any normal use. Space efficiency dictates a binary format. 6 digits (decimal) of precision suggests a format intended for human readers, who can't churn through lots of data. And humans have no issue dealing with a consistent 6 digit format.
So, you're basically targeting a format that has no obvious audience, and that is why I would be surprised if there is support for that.
I have been searching and experimenting for many, many, hours now, and so far I have not been able to adapt any of the solutions I have come across to do what I want.
My goal is to take an integer (538214658) and convert it into an 8 character hex string (020148102). Then I want to drop the first two characters (0148102) and convert it back into an integer(1343746) which I am using as a key in a map array.
The solutions I've seen so far just convert an integer into hex string, but don't take into account the desired digit length.
I am able to print out just the first 6 characters using the following code:
Console_Print("%06X", form ? form->refID : 0)
So I thought that maybe I could use that technique to store it into a string, and then use iostream or sstream to convert it back to an integer, but none of my searches turned up anything I could use. And all of my experiments have failed.
Some help would be greatly appreciated.
EDIT: Below is my solution based on Klaus' suggestion:
uint32_t GetCoreRefID(TESForm* form)
{
uint32_t iCoreRefID = 0;
if (form)
{
uint32_t iRefID = (uint32_t)(form->refID);
iCoreRefID = iRefID & 0x00ffffff;
}
return iCoreRefID;
}
There is no need to convert to a string representation.
Look the following example:
int main()
{
uint32_t val = 538214658 & 0x00ffffff;
std::cout << std::hex << val << std::endl;
std::cout << std::dec << val << std::endl;
}
You have to learn that a value is still only a value and is not dependent on the representation like decimal or hex. The value stored in a memory area or a register is still the same.
As you can see in the given example I wrote your decimal value representation and remove the first two hexadecimal digits simply by do a bitwise and operation with the hexadecimal representation of a mask.
Furthermore you have to understand that the printing with cout in two different "modes" did not change the value at all and also not the internal representation. With std::dec and std::hex you tell the ostream object how to create a string from an int representation.
I have some old C code I'm trying to replicate the behavior of in C++. It uses the printf modifiers: "%06.02f".
I naively thought that iomanip was just as capable, and did:
cout << setfill('0') << setw(6) << setprecision(2)
When I try to output the test number 123.456, printf yields:
123.46
But cout yields:
1.2+e02
Is there anything I can do in iomanip to replicate this, or must I go back to using printf?
[Live Example]
Try std::fixed:
std::cout << std::fixed;
Sets the floatfield format flag for the str stream to fixed.
When floatfield is set to fixed, floating-point values are written using fixed-point notation: the value is represented with exactly as many digits in the decimal part as specified by the precision field (precision) and with no exponent part.
The three C format specifiers map to corresponding format setting in C++ IOStreams:
%f -> std::ios_base::fixed (fixed point notation) typically set using out << std::fixed.
%e -> std::ios_base::scientific (scientific notation) typically set using out << std::scientific.
%g -> the default setting, typically set using out.setf(std::fmtflags(), std::ios_base::floatfield) or with C++11 and later out << std::defaultfloat. The default formatting is trying to yield the "best" of the other formats assuming a fixed amount of digits to be used.
The precision, the width, and the fill character match the way you already stated.
The following code will print value of a and b:
double a = 3.0, b=1231231231233.0123456;
cout.setf(std::ios::fixed);
cout.unsetf(std::ios::scientific);
cout << a << endl << b << endl
The output is:
3.000000
1231231231233.012451
You can see that a is outputed with fixed 6 count of decimals.
But I want the output like this:
3
1231231231233.012451
How can i set flags only once, and output the above result.
The stream inserts 0s following the double because the stream's default precision for the output of floating-point values is 6. Unfortunately there is no straightforward way of checking if the double represents a whole number (so you could then only print the integral part). What you could do however is cast the value to an integer.
std::cout << static_cast<int>(a);
The default formatting for floating point numbers won't support the formats as requested. There are basically three settings you could use:
std::fixed which will use precision() digits after the decimal point.
std::scientific which will use scientific notation with precision() digits.
std::defaultfloat which will choose the shorter of the two forms.
(there is also std::hexfloat but that just formats the number in an form which is conveniently machine readable).
What you could do is to create you own std::num_put<char> facet which formats the value into a local buffer using std::fixed formatting an strips off trailing zero digits before sending the values one.
I am writing a code which involves converting a decimal number to binary and store the binary number. I'm not able to store the leading zeros in some of the binary number e.g 001101011 and instead it prints and stores -> 1101011. Any help would be appreciated. thanks
With my mind reading powers I'm deducing that this will help you.
printf("%08x", number);
To the best of my knowledge there is no standard data type for binary numbers in c++. So i guess you are using integers to store the binary number. So to print the leading zeroes just use this .
std::cout << std::setw(5) << std::setfill('0') << binary_number << std::endl;
See http://www.daniweb.com/software-development/cpp/threads/114864/setw-and-setfill.