Printing float such that exponent is marked with "*10^" instead of "e"

Question

I am looking for a possibility within C/C++ to print a float (or double) f, say f = 1.234e-15, such that it is printed as

f = 1.234*10^-15, or, even better, as
f = 1.234*10^{-15}

Can anyone help me? Maybe there is a way to get the exponent "-15" and mantissa "1.234" in base 10. I found the question how can I extract the mantissa of a double, but unfortunately that did not really help, since it only gets the mantissa in base 2.

Seems like log10 rounded(somehow) would get you the exponent. Should be simple enough to get the mantissa once you have that. — , Jun 16 '12 at 08:12
@EthanSteinberg: It's not that simple, depending on whether you want the full precision. — Dietrich Epp, Jun 16 '12 at 08:20
There's no such language as "C/C++", and the answers for the two languages are very different. — R.. GitHub STOP HELPING ICE, Jun 16 '12 at 11:10
What I meant was either C or C++, since I can use C code in my C++ program. — alex, Jun 18 '12 at 08:08

Sergey Kalinichenko · Accepted Answer · 2012-06-16T08:19:38.800

8

You can print to a string using the output string stream, and then replace "e" with "*10^".

ostringstream ss;
ss << scientific << 123456789.87654321;
string s = ss.str();
s.replace(s.find("e"), 1, "*10^");
cout << s << endl;

This snippet produces

1.234568*10^+08

edited Jun 16 '12 at 08:19

answered Jun 16 '12 at 08:14

Sergey Kalinichenko

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Thanks a lot! I did not know the string methods replace() and find(). Thought I will need to use some regular expression library to replace the 'e' and that seemed to complicated. Your solution is really short, which I like. While waiting for replies I actually wrote my own code, see below... – alex Jun 16 '12 at 09:10

score 2 · Answer 2 · answered Jun 16 '12 at 08:14

2

Why not use string parsing? scan the string and replace e with 10^.

answered Jun 16 '12 at 08:14

edison

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score 2 · Answer 3 · answered Jun 16 '12 at 08:22

2

#include <cmath>
#include <iostream>

using namespace std;

template <typename F>
F round_away_from_zero (F x)
{
  return x < 0 ? floor(x) : ceil(x);
}

template <class O, typename F>
O &print_float (O &out, F f) {
    signed ex = round_away_from_zero(log10(f)); // exponent
    F mant = f / pow(10, ex);                   // mantissa
    out << mant << "*10^" << ex;
}

int main () {
    double f = 1.234e-15;
    print_float(cout, f) << endl; // prints 1.234*10^-15
    return 0;
}

answered Jun 16 '12 at 08:22

Kijewski

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Thanks! Also nice mathematical way to get exponent and mantissa! And easy to include my brace brackets into the output. And now I also know floor() and ceil() :)) – alex Jun 16 '12 at 09:25
Just discovered a small error in your code: In print_float() you have to take the log10 of abs(f), otherwise the result will be infinite for negative f. But with this small change it also works for negative numbers. – alex Jun 16 '12 at 10:08

score 0 · Answer 4 · answered Jun 16 '12 at 09:16

While waiting for your solutions I came up with the following idea:

Use sprintf() to print the float or double to an array of char. Parse this to get the exponent and mantissa. Code looks now the following way:

void printDoubleToChar(double d, char* c){
    char valAsChar[256];

    sprintf(valAsChar, "%.12e", d);

    int expStart = 0, expWidth = 0;
    for(int i=0; i<sizeof(valAsChar); i++){
        if(valAsChar[i] == 'e'){
            expStart = i+1;
            break;
        }
    }
    for(int i=expStart; i<sizeof(valAsChar); i++){
        if(valAsChar[i] == '\0'){
            expWidth = i - expStart;
            break;
        }
    }

    char chValExp[32];
    memcpy(chValExp, &valAsChar[expStart], expWidth+1);

    char chValMan[128];
    memcpy(chValMan, valAsChar, expStart-1);
    chValMan[expStart-1] = '\0';

    sprintf(c, "%s*10^{%s}", chValMan, chValExp);
}

int main(){
    double myNewDbl = 3.95743e-5;
    char chDbl[256];
    printDoubleToChar(myNewDbl, chDbl);
    printf("\nchDbl: %s\n", chDbl); // output: chDbl: 3.957430000000*10^{-05}
}

But honestly, I prefer the much simpler solution by dasblinkenlight :)

Thank you all for your help!

Alex

Printing float such that exponent is marked with "*10^" instead of "e"

4 Answers4