How to set format of std::scientific - c++

Question

I want to print 12345.0 as 0.1235e+05. But the below code gives 1.23450e+04. Is there a way to control the format of std::scientific?
#include <iostream>
#include <string>
#include <sstream>
int main()
{
double a = 12345.0;
std::stringstream s;
s.precision(3);
s << std::scientific;
s << a;
std::cout << s.str() << std::endl;
return 0;
}
The output is
1.235e+04
I also tried the code here Print exponential notation with one leading zero with C++ but there is a problem about rounding. If I set precision to 4 it does not round and gives 0.1234E5 however it should have been 0.1235E5.
Any solution to this?

Answer 1

Then that isn't scientific form. Scientific form must produce a number between 1-10 and multiplied by a power of 10
Source: http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch1/scinot.html

Answer 2

A problem with re-inventing double to string conversion is the many corner cases.
Consider printing values near 99995.0 to 4 places
void foo(double value, int precision) {
if (value == 0.) {
printf("%g\n", 0.0);
}
int exponent = (int) floor(log10(fabs(value)));
double base = value / pow(10, exponent);
base /= 10.0;
int p = precision;
base = round(base * pow(10, p)) / pow(10, p);
exponent++;
printf("%.*fE%d\n", p, base, exponent);
bar(value, precision);
}
foo(99994.99, 4);
foo(99995.01, 4);
Prints
0.9999E5
1.0000E5
Rather than the hoped for
0.9999E5
0.1000E6
Rather than try to out-compute the library conversion of floating point to a string, simply post process the string to the desired format.
Below is C, yet OP is looking for C++, so take this as a guide.
void bar(double value, int precision) {
precision--;
char buf[400];
snprintf(buf, sizeof buf, "%.*e", precision, value);
if (isfinite(value)) {
char *e = strchr(buf, 'e');
char *first_digit = e - precision - 2;
int expo = atoi(e + 1) + 1;
printf("%.*s0.%c%.*sE%d\n", !isdigit((unsigned char )buf[0]), buf,
*first_digit, precision, first_digit + 2, expo);
} else {
printf("%s\n", buf);
}
}
bar(99994.99, 4);
bar(99995.01, 4);
Prints
0.9999E5
0.1000E6
"If I set precision to 4 it does not round and gives 0.1234E5 however it should have been 0.1235E5."
This is a result of the default rounding mode is "round to nearest, ties to even" and OP is hoping for "round to nearest, ties to away".
Code such as base = round(base * pow(10, p)) / pow(10, p); may achieve OP's goal in select cases as the multiplication and division here can incur rounding due to imprecision, sometimes in the desired direction. This is not reliable across all double to achieve "round to nearest, ties to away".

Answer 3

I corrected the error in the solution given in the link Print exponential notation with one leading zero with C++ by adding two lines of code. I also repeated the process for correct formatting:
int p = stream.precision();
base = round(base * pow(10, p)) / pow(10, p);
It now makes correct rounding and gives 0.1235E5. Here is the whole code:
#include <iostream>
#include <string>
#include <sstream>
#include <iomanip>
#include <cmath>
class Double {
public:
Double(double x) : value(x) {}
const double value;
};
std::ostream& operator<< (std::ostream& stream, const Double& x) {
// So that the log does not scream
if (x.value == 0.) {
stream << 0.0;
return stream;
}
int exponent = floor(log10(std::abs(x.value)));
double base = x.value / pow(10, exponent);
// Transform here
base /= 10.0;
int p = stream.precision(); // This line is added to correct the code
base = round(base * pow(10, p)) / pow(10, p); // This line is added to correct the code
exponent += 1;
double x2 = base * pow(10, exponent);
exponent = floor(log10(std::abs(x2)));
base = x2 / pow(10, exponent);
// Transform here
base /= 10.0;
base = round(base * pow(10, p)) / pow(10, p); // This line is added to correct the code
exponent += 1;
stream << base << "+e" << exponent;
return stream;
}
int main()
{
//double a = 12345.0;
double a = 99995.01;
std::stringstream s;
s.precision(4);
s << Double(a);
std::cout << s.str() << std::endl;
return 0;
}