Why does +e canceled and -e not?
System.out.println(2e+5);
Above line's output is,
200000.0
While below line's,
System.out.println(2e-5);
output is,
2.0e-5
Why does 2e-5 is not solved down with -10^5 giving the output,
0.00002
java numbers double notation scientific-notation
edited Nov 16 '18 at 11:08
Poorna Senani Gamage
96411222
asked Jul 20 '18 at 16:58
Roshana PitigalaRoshana Pitigala
4,86162549
|
show 2 more comments
System.out.println(2e+5);
Above line's output is,
200000.0
While below line's,
System.out.println(2e-5);
output is,
2.0e-5
Why does 2e-5 is not solved down with -10^5 giving the output,
0.00002
java numbers double notation scientific-notation
edited Nov 16 '18 at 11:08
Poorna Senani Gamage
96411222
asked Jul 20 '18 at 16:58
Roshana PitigalaRoshana Pitigala
4,86162549
One reason may be, that in the first case the extra digits may hold significant information, while in the second case they are always zero.
– Henry
Jul 20 '18 at 17:04
Because you're relying on the default formatting rules.2e-5is adoubleliteral - it's64ones and zeroes. There is no format, no notation, no decimal point. How its represented when printed is just that, a representation. It's meaningless.
– Boris the Spider
Jul 20 '18 at 17:05
@Henry the extra digits in2eanything will always be zero...
– Boris the Spider
Jul 20 '18 at 17:06
@BoristheSpider in this case yes, but take 1.23456e5 and 1.23456e-5 for example.
– Henry
Jul 20 '18 at 17:08
In both cases you have shown numbers represented to 6 s.f. @Henry. All the digits are significant - that's what the "s" stands for! I'm really not sure what you mean.
– Boris the Spider
Jul 20 '18 at 17:09
|
show 2 more comments
System.out.println(2e+5);
Above line's output is,
200000.0
While below line's,
System.out.println(2e-5);
output is,
2.0e-5
Why does 2e-5 is not solved down with -10^5 giving the output,
0.00002
java numbers double notation scientific-notation
edited Nov 16 '18 at 11:08
Poorna Senani Gamage
96411222
asked Jul 20 '18 at 16:58
Roshana PitigalaRoshana Pitigala
4,86162549
System.out.println(2e+5);
Above line's output is,
200000.0
While below line's,
System.out.println(2e-5);
output is,
2.0e-5
Why does 2e-5 is not solved down with -10^5 giving the output,
0.00002
java numbers double notation scientific-notation
java numbers double notation scientific-notation
edited Nov 16 '18 at 11:08
Poorna Senani Gamage
96411222
asked Jul 20 '18 at 16:58
Roshana PitigalaRoshana Pitigala
4,86162549
edited Nov 16 '18 at 11:08
Poorna Senani Gamage
96411222
asked Jul 20 '18 at 16:58
Roshana PitigalaRoshana Pitigala
4,86162549
edited Nov 16 '18 at 11:08
Poorna Senani Gamage
96411222
edited Nov 16 '18 at 11:08
Poorna Senani Gamage
96411222
edited Nov 16 '18 at 11:08
Poorna Senani Gamage
96411222
96411222
asked Jul 20 '18 at 16:58
Roshana PitigalaRoshana Pitigala
4,86162549
asked Jul 20 '18 at 16:58
Roshana PitigalaRoshana Pitigala
4,86162549
asked Jul 20 '18 at 16:58
Roshana PitigalaRoshana Pitigala
4,86162549
4,86162549
One reason may be, that in the first case the extra digits may hold significant information, while in the second case they are always zero.
– Henry
Jul 20 '18 at 17:04
Because you're relying on the default formatting rules.2e-5is adoubleliteral - it's64ones and zeroes. There is no format, no notation, no decimal point. How its represented when printed is just that, a representation. It's meaningless.
– Boris the Spider
Jul 20 '18 at 17:05
@Henry the extra digits in2eanything will always be zero...
– Boris the Spider
Jul 20 '18 at 17:06
@BoristheSpider in this case yes, but take 1.23456e5 and 1.23456e-5 for example.
– Henry
Jul 20 '18 at 17:08
In both cases you have shown numbers represented to 6 s.f. @Henry. All the digits are significant - that's what the "s" stands for! I'm really not sure what you mean.
– Boris the Spider
Jul 20 '18 at 17:09
|
show 2 more comments
One reason may be, that in the first case the extra digits may hold significant information, while in the second case they are always zero.
– Henry
Jul 20 '18 at 17:04
Because you're relying on the default formatting rules.2e-5is adoubleliteral - it's64ones and zeroes. There is no format, no notation, no decimal point. How its represented when printed is just that, a representation. It's meaningless.
– Boris the Spider
Jul 20 '18 at 17:05
@Henry the extra digits in2eanything will always be zero...
– Boris the Spider
Jul 20 '18 at 17:06
@BoristheSpider in this case yes, but take 1.23456e5 and 1.23456e-5 for example.
– Henry
Jul 20 '18 at 17:08
In both cases you have shown numbers represented to 6 s.f. @Henry. All the digits are significant - that's what the "s" stands for! I'm really not sure what you mean.
– Boris the Spider
Jul 20 '18 at 17:09
One reason may be, that in the first case the extra digits may hold significant information, while in the second case they are always zero.
– Henry
Jul 20 '18 at 17:04
One reason may be, that in the first case the extra digits may hold significant information, while in the second case they are always zero.
– Henry
Jul 20 '18 at 17:04
Because you're relying on the default formatting rules.
2e-5 is a double literal - it's 64 ones and zeroes. There is no format, no notation, no decimal point. How its represented when printed is just that, a representation. It's meaningless.– Boris the Spider
Jul 20 '18 at 17:05
Because you're relying on the default formatting rules.
2e-5 is a double literal - it's 64 ones and zeroes. There is no format, no notation, no decimal point. How its represented when printed is just that, a representation. It's meaningless.– Boris the Spider
Jul 20 '18 at 17:05
@Henry the extra digits in
2eanything will always be zero...– Boris the Spider
Jul 20 '18 at 17:06
@Henry the extra digits in
2eanything will always be zero...– Boris the Spider
Jul 20 '18 at 17:06
@BoristheSpider in this case yes, but take 1.23456e5 and 1.23456e-5 for example.
– Henry
Jul 20 '18 at 17:08
@BoristheSpider in this case yes, but take 1.23456e5 and 1.23456e-5 for example.
– Henry
Jul 20 '18 at 17:08
In both cases you have shown numbers represented to 6 s.f. @Henry. All the digits are significant - that's what the "s" stands for! I'm really not sure what you mean.
– Boris the Spider
Jul 20 '18 at 17:09
In both cases you have shown numbers represented to 6 s.f. @Henry. All the digits are significant - that's what the "s" stands for! I'm really not sure what you mean.
– Boris the Spider
Jul 20 '18 at 17:09
|
show 2 more comments
2 Answers
2
active
oldest
votes
Because that's how the designers of the Double class chose to implement it:
[...]
- If m is greater than or equal to 10-3 but less than 107, then it is represented as the integer part of m, in decimal form with no leading zeroes, followed by '.' ('u002E'), followed by one or more decimal digits representing the fractional part of m.
- If m is less than 10-3 or greater than or equal to 107, then it is represented in so-called "computerized scientific notation." [...]
They could have made another choice, of course, but that's simply how it is.
As the documentation says, if you want a specific format, use a NumberFormat to format your double.
answered Jul 20 '18 at 17:15
JB NizetJB Nizet
547k588941021
add a comment |
As already pointed out by @Nizet, when converting double to string, the output will have scientific notation if number of digits are large. And this is not just for decimals.
Consider:
double d = 1500000000;
System.out.println(String.valueOf(d));
This will print 1.5E9
If you want conversion to happen in readable format, use String.format
System.out.println(String.format ("%.0f", d));
will print 1500000000
answered Jul 20 '18 at 17:17
HariUserXHariUserX
1,0881415
add a comment |
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2 Answers
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2 Answers
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active
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Because that's how the designers of the Double class chose to implement it:
[...]
- If m is greater than or equal to 10-3 but less than 107, then it is represented as the integer part of m, in decimal form with no leading zeroes, followed by '.' ('u002E'), followed by one or more decimal digits representing the fractional part of m.
- If m is less than 10-3 or greater than or equal to 107, then it is represented in so-called "computerized scientific notation." [...]
They could have made another choice, of course, but that's simply how it is.
As the documentation says, if you want a specific format, use a NumberFormat to format your double.
answered Jul 20 '18 at 17:15
JB NizetJB Nizet
547k588941021
add a comment |
Because that's how the designers of the Double class chose to implement it:
[...]
- If m is greater than or equal to 10-3 but less than 107, then it is represented as the integer part of m, in decimal form with no leading zeroes, followed by '.' ('u002E'), followed by one or more decimal digits representing the fractional part of m.
- If m is less than 10-3 or greater than or equal to 107, then it is represented in so-called "computerized scientific notation." [...]
They could have made another choice, of course, but that's simply how it is.
As the documentation says, if you want a specific format, use a NumberFormat to format your double.
answered Jul 20 '18 at 17:15
JB NizetJB Nizet
547k588941021
add a comment |
Because that's how the designers of the Double class chose to implement it:
[...]
- If m is greater than or equal to 10-3 but less than 107, then it is represented as the integer part of m, in decimal form with no leading zeroes, followed by '.' ('u002E'), followed by one or more decimal digits representing the fractional part of m.
- If m is less than 10-3 or greater than or equal to 107, then it is represented in so-called "computerized scientific notation." [...]
They could have made another choice, of course, but that's simply how it is.
As the documentation says, if you want a specific format, use a NumberFormat to format your double.
answered Jul 20 '18 at 17:15
JB NizetJB Nizet
547k588941021
Because that's how the designers of the Double class chose to implement it:
[...]
- If m is greater than or equal to 10-3 but less than 107, then it is represented as the integer part of m, in decimal form with no leading zeroes, followed by '.' ('u002E'), followed by one or more decimal digits representing the fractional part of m.
- If m is less than 10-3 or greater than or equal to 107, then it is represented in so-called "computerized scientific notation." [...]
They could have made another choice, of course, but that's simply how it is.
As the documentation says, if you want a specific format, use a NumberFormat to format your double.
answered Jul 20 '18 at 17:15
JB NizetJB Nizet
547k588941021
answered Jul 20 '18 at 17:15
JB NizetJB Nizet
547k588941021
answered Jul 20 '18 at 17:15
JB NizetJB Nizet
547k588941021
answered Jul 20 '18 at 17:15
JB NizetJB Nizet
547k588941021
547k588941021
add a comment |
add a comment |
As already pointed out by @Nizet, when converting double to string, the output will have scientific notation if number of digits are large. And this is not just for decimals.
Consider:
double d = 1500000000;
System.out.println(String.valueOf(d));
This will print 1.5E9
If you want conversion to happen in readable format, use String.format
System.out.println(String.format ("%.0f", d));
will print 1500000000
answered Jul 20 '18 at 17:17
HariUserXHariUserX
1,0881415
add a comment |
As already pointed out by @Nizet, when converting double to string, the output will have scientific notation if number of digits are large. And this is not just for decimals.
Consider:
double d = 1500000000;
System.out.println(String.valueOf(d));
This will print 1.5E9
If you want conversion to happen in readable format, use String.format
System.out.println(String.format ("%.0f", d));
will print 1500000000
answered Jul 20 '18 at 17:17
HariUserXHariUserX
1,0881415
add a comment |
As already pointed out by @Nizet, when converting double to string, the output will have scientific notation if number of digits are large. And this is not just for decimals.
Consider:
double d = 1500000000;
System.out.println(String.valueOf(d));
This will print 1.5E9
If you want conversion to happen in readable format, use String.format
System.out.println(String.format ("%.0f", d));
will print 1500000000
answered Jul 20 '18 at 17:17
HariUserXHariUserX
1,0881415
As already pointed out by @Nizet, when converting double to string, the output will have scientific notation if number of digits are large. And this is not just for decimals.
Consider:
double d = 1500000000;
System.out.println(String.valueOf(d));
This will print 1.5E9
If you want conversion to happen in readable format, use String.format
System.out.println(String.format ("%.0f", d));
will print 1500000000
answered Jul 20 '18 at 17:17
HariUserXHariUserX
1,0881415
answered Jul 20 '18 at 17:17
HariUserXHariUserX
1,0881415
answered Jul 20 '18 at 17:17
HariUserXHariUserX
1,0881415
answered Jul 20 '18 at 17:17
HariUserXHariUserX
1,0881415
1,0881415
add a comment |
add a comment |
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One reason may be, that in the first case the extra digits may hold significant information, while in the second case they are always zero.
– Henry
Jul 20 '18 at 17:04
Because you're relying on the default formatting rules.
2e-5is adoubleliteral - it's64ones and zeroes. There is no format, no notation, no decimal point. How its represented when printed is just that, a representation. It's meaningless.– Boris the Spider
Jul 20 '18 at 17:05
@Henry the extra digits in
2eanything will always be zero...– Boris the Spider
Jul 20 '18 at 17:06
@BoristheSpider in this case yes, but take 1.23456e5 and 1.23456e-5 for example.
– Henry
Jul 20 '18 at 17:08
In both cases you have shown numbers represented to 6 s.f. @Henry. All the digits are significant - that's what the "s" stands for! I'm really not sure what you mean.
– Boris the Spider
Jul 20 '18 at 17:09