I'm using c++ and I'm trying to convert a string to double, the string is given to me as an input in the form: "xx.yy", and it should stay that way, meaning no more than two digits after the dot.
I've tried atof and strod but outcome was no good, plus i can't use anything to spectacular' because it's for a course and no one will belive me
str_db = strtod(ptr, NULL);
item->price = str_db;
str_db is type double.
Thank you.
Related
I read that in Matlab it is possible to include a function call inside a regex transformation like this $1double$2[${doubleTextNumber($4)}], assuming 1, 2, 3 to be some regex groups, and 4 to be a purely numeric group. The exact thing I want to do is to catch all arrays consisting of the type creal_T, replace the type with double and double the length of the array.
codeText = "typedef struct {
double tolRob;
creal_T Mt2o[704];
creal_T Ho2o[704];
creal_T Ht2t[704];
creal_T Zo2t[704];
creal_T Ztd[64];
} testType;"
So, I want the struct above to become:
typedef struct {
double tolRob;
double Mt2o[1408];
double Ho2o[1408];
double Ht2t[1408];
double Zo2t[1408];
double Ztd[128];
} SpdEstType;
In Matlab I have made a function to convert a number to text and double it:
function [doubleValue] = doubleTextNumber(inputNumber)
doubleValue = string(str2double(inputNumber)*2.0);
end
I also have a regex that I expect would find the number in each declaration and feed it to the function:
resultString = regexprep(
codeText,
'(?m)^(\W*)creal_T(\s*\w*)(\[([^\]]*\d+)\])',
"$1double$2[${doubleTextNumber($4)}]");
However, as I run this peace of code, Matlab gives me the following error msg:
Error using regexprep
Evaluation of 'doubleTextNumber($4)' failed:
Undefined function 'doubleTextNumber' for input arguments of type 'char'.
As far as I understand, I have made the method do conversion from char, and expect it also to accept this value from my regex. I have tested that it works when I input '704' or "704" directly, and also that the regex works appart from this insertion.
Why does not Matlab find the function from my regex? (they are in the same m file)
It looks like I had 3 issues with my original approach:
In order for regexprep() to recognize my function, it had to be moved to its own m-file. Simply calling a method from inside the same file did not work.
I was using https://regex101.com/ to edit the search expression, but even though it seemed to be selecting the number inside the brackets, group 4 did not get populated by regexprep() in Matlab. A new version did work, and populated group 3 with the numbers I wanted: (?m)^(\W*)creal_T(\s*\w*).([^\]]*\d*)\]
I also added more conversion options to my multiplication method in case the input was a combination of numbers and char arrays.
The final version of my regex call becomes:
resultString = regexprep(
codeText,
'(?m)^(\W*)creal_T(\s*\w*).([^\]]*\d*)\]',
"$1double$2[${multiplyTextNumbers($3,2)}]");
where multiplyTextNumbers() is defined in its own m file as
function [productText] = multiplyTextNumbers(inputFactorText1,inputFactorText2)
%MULTIPLY This method takes numbers as input, and acepts either string,
%char or double or any combination of the three. Returns a string with the
%resulting product.
if (isstring(inputFactorText1) || ischar(inputFactorText1))
inputFactor1 = str2double(inputFactorText1);
else
inputFactor1 = inputFactorText1;
end
if (isstring(inputFactorText2) || ischar(inputFactorText2))
inputFactor2 = str2double(inputFactorText2);
else
inputFactor2 = inputFactorText2;
end
productText = sprintf('%d',inputFactor1*inputFactor2);
end
Hope this can be helpefull to others facing similar issues.
Suppose I have a string that contains a necessary numeric character but it is not terminated by '/0', it has garbage characters instead. Actually, the string has garbage characters after the number. So how to deal with the garbage character while storing that numerical character in another string or variable?
So how to deal with the garbage character while storing that numerical character in another string or variable?
Only copy a substring. Example:
std::string example "garbage1garbage";
char numerical = example[7];
We got the numerical character excluding the garbage entirely.
If the text be converted is in a std::string, then you can extract a number from the front as follows:
#include <sstream>
...
std::string input = "128734garbage";
std::istringstream iss{input};
int num;
if (iss >> num)
...use_num...
else
std::cerr << "wasn't able to parse an int from input\n";
Just change int to double, uint64_t, ... - whatever suits your data.
If you have only a pointer to the text and know it's not null-terminated, just getting the text into a std::string is problematic. You could instead use a function that converts text to a number, but stops at the first invalid character. std::stol et al, and the other unsigned and floating point variants linked from the same reference page, are good candidates for that.
From your "another string or variable" - the above addresses storing into a numeric variable. You can then create a new std::string from the number using std::to_string, or a std::ostringstream, if that's what you want to do. This will standardise the output format though, so input like say "1E4" might end up looking like say 1000.0. Alternatively, with the stol-type functions you can use the pointer-to-the-end-of-the-number to work out the length of the numeric part, and use std::string::substr() to extract the leading number as a new std::string object.
You should also be aware that the distinction between number and garbage is not always what you might expect. For example "0XBEFHJQ" might be split by some of the above functions as 0xBEF hex and HJQ garbage.
Or which type do I need to use?
I have string and I try to convert it into double
NFR_File.ReadString(sVal); // sVal = " 0,00003"
dbl = _wtof(sVal);
and get:
3.0000000000000001e-05
And I need 0,00003, because then I should write it into the file as "0,00003" but not as 3e-05.
If the number greater then 0,0009 everything works.
displaying:
sOutput.Format(_T("%9g"),dbl);
NFP1_File.WriteString(sOutput);
I need it without trailing zeros and also reserve 9 digits (with spaces)
When you write using printf you can specify the number of significant digits you want by using the .[decimals]lf.
For example, in your case you want to print with 5 decimals, so you should use
printf("%.5f", yourNumber);
If you can use C++11
try use http://en.cppreference.com/w/cpp/string/basic_string/to_string
std::string to_string( double value );
CString::Format
Call this member function to write formatted data to a CString in the same way that sprintf formats data into a C-style character array.
It is same as c sprintf format. You may check other answer's format usage.
I have a class that has a long double vector:
MyClass{
vector<long double> myvec;
public:
MyClass(){ //Constructor }
// Some memeber functions that operate on the vector
};
I have overloaded the input operator an I'm taking input from a user that are then pushed into the vector. The problem that I'm having is if the user inputs a number that is out of range of double, the code should append append the long double suffix to the input with out the user having too. This is what I have tried so far:
long double input;
...
input = (long double)(input + "L");
myvec.push_back(input);
I thought of using scanf, but I'm not sure how safe that is to use when overloading the input operator.
Use std::stold to convert input text to long double. There is no need for a suffix; stold will do it right. The suffix is needed in source code to tell the compiler what type the text represents. When you're reading from an external source the compiler isn't involved, so you have to sort out the type yourself.
Suffixes are only for literal values, e.g. auto x = 12345.0L. You use them to prevent implicit conversions since the default type of a floating point literal is double.
You can't use them on a named variable.
The question is how you get your input?
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
How to convert a single char into an int
Well, I'm doing a basic program, wich handles some input like:
2+2
So, I need to add 2 + 2.
I did something like:
string mys "2+2";
fir = mys[0];
sec = mys[2];
But now I want to add "fir" to "sec", so I need to convert them to Int.
I tried "int(fir)" but didn't worked.
There are mulitple ways of converting a string to an int.
Solution 1: Using Legacy C functionality
int main()
{
//char hello[5];
//hello = "12345"; --->This wont compile
char hello[] = "12345";
Printf("My number is: %d", atoi(hello));
return 0;
}
Solution 2: Using lexical_cast(Most Appropriate & simplest)
int x = boost::lexical_cast<int>("12345");
Solution 3: Using C++ Streams
std::string hello("123");
std::stringstream str(hello);
int x;
str >> x;
if (!str)
{
// The conversion failed.
}
Alright so first a little backround on why what you attempted didn't work. In your example, fir is declared as a string. When you attempted to do int(fir), which is the same as (int)fir, you attempted a c-style cast from a string to an integer. Essentially you will get garbage because a c-style cast in c++ will run through all of the available casts and take the first one that works. At best your going to get the memory value that represents the character 2, which is dependent upon the character encoding your using (UTF-8, ascii etc...). For instance, if fir contained "2", then you might possibly get 0x32 as your integer value (assuming ascii). You should really never use c-style casts, and the only place where it's really safe to use them are conversions between numeric types.
If your given a string like the one in your example, first you should separate the string into the relevant sequences of characters (tokens) using a function like strtok. In this simple example that would be "2", "+" and "2". Once you've done that you can simple call a function such as atoi on the strings you want converted to integers.
Example:
string str = "2";
int i = atoi(str.c_str()); //value of 2
However, this will get slightly more complicated if you want to be able to handle non-integer numbers as well. In that case, your best bet is to separate on the operand (+ - / * etc), and then do a find on the numeric strings for a decimal point. If you find one you can treat it as a double and use the function atof instead of atoi, and if you don't, just stick with atoi.
Have you tried atoi or boost lexical cast?