You can use this non-regex method:

num = 0.112345678
pow = floor(log10(abs(num)));
sprintf('%.8fe%+.3d', num/10^pow, pow)

ans =
1.12345678e-001

For multiple inputs use this:

num= [.123 .456 .789];
pow = floor(log10(abs(num)));
sprintf('%.8fe%+.3d ', [num./10.^pow; pow])

What you see is a consequence of how *floating point arithmetic* works. See: * and read * |1.05*100| evaluates to a whole number (flint). The other two don't.

you can use:

sprintf('%.10f', yourNumber)

Or a more sophisticated option is to use Java-based formatting (see more info , credit to Yair Altman for showing this method), for example:

char(java.text.DecimalFormat('#.00000000').format(yourNumber));

There is no way to use directly fprintf format specifier for the format you require. A way around is to use the output of disp as a string to be printed. But disp doesn't return a string, it writes directly to the standard output. So, how to do this?

Here's where evalc (eval with capture of output) comes to the rescue:

%// Create helper function
sdisp = @(x) strtrim(evalc(sprintf('disp(%g)', x)));

%// Test helper function
format ShortEng;
a = 123e-12;
fprintf(1, 'Test: %s', sdisp(a));

This is a workaround, of course, and can backfire in multiple ways because of the untested inputs of the helper functions. But it illustrates a point, and is one of the rare occasions where the reviled eval function family is actually irreplaceable.

format long or sprintf('%16.f',2332456943534324)

In your code, you are assigning a string to a double-float array. It looks like MATLAB automatically converts the string to a double to store it there. Thus, your formatting gets lost:

>> sprintf("%.1e", 1/1000)
ans = 
    "1.0e-03"
>> a=0;
>> a(1) = sprintf("%.1e", 1/1000)
a =
   1.0000e-03
>> class(a)
ans =
    'double'

Instead, use a string array:

a = strings(19,1);
%...
   a(i-1) = sprintf("%.1e", w);

---

I'm not used to the new strings, and this behavior surprises me. Assigning a number to a string converts the number to a string, and assigning the string to a number converts it back to a number. This does not happen with the "old-fashioned" char arrays:

>> a=0;
>> a(1) = sprintf('%.1e', 1/1000);
Unable to perform assignment because the left and right sides have a different number of elements.

When using char arrays, store them in a cell array:

a = cell(19,1);
%...
   a{i-1} = sprintf('%.1e', w);

For variables that you disp(), command format long or format long g For the variable editor, see these instructions:

The *|fprintf|* function _optionally_ requires a *|‘fileID’|* variable as its first argument, with *|1|* indicating *|‘stdout’|*, that being the Command Window. Otherwise it will be to the file you want to write to. (I used it in the first *|fprintf|* call but not in the second.) To print the values of *|‘x’|* _and_ *|f(x)|*, this works: x = [5.201, -8323.6, 0.0003]; f = @(x) sin(x.^3); fprintf(1, 'f(%f) = %.8f\n', [x; f(x)]) fprintf('f(%f) = %.10e\n', [x; f(x)]) f(5.201000) = 0.63076123 f(-8323.600000) = -0.52794696 f(0.000300) = 0.00000000 f(5.201000) = 6.3076122538e-01 f(-8323.600000) = -5.2794696417e-01 f(0.000300) = 2.7000000000e-11 The *|(\n)|* is a newline character.

It looks like this might have something to do with flintmax, which would mean there would be no way to change this. This value is 2^53 for doubles, which works out to 9007199254740992 (so just under 17 digits). If I try writing more digits, my copy of Matlab starts rounding. I also observe the conversion to exponential notation at 20 digits. Hidden in the large doc for fprintf, you can find this: If you specify a conversion that does not fit the data, such as a text conversion for a numeric value, MATLAB overrides the specified conversion, and uses %e. Example: '%s' converts pi to 3.141593e+00. So apparently, Matlab does not consider such large values to be integers anymore, which makes them not fit the d flag, which causes the conversion to exponential. What you can do, is either preproces to a string, or dynamically choosing a format specificiation based on your value. You can either compare to flintmax, or use log10() to find the number of digits. (These two are not equivalent.) %replace this: %fprintf(not,'%21d', step(k,l)); %before the loop: FormatSpec={'%21d','%21e'}; NumberOfDigits=@(x) 1+floor(log10(x)); %in the loop val=step(k,l) fprintf(not,FormatSpec{1+(val>flintmax)}, val); %or: %fprintf(not,FormatSpec{1+(NumberOfDigits(val)>19)}, val);