Why was the pattern string not followed in this code?
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box;
}
When following code is executed:
DecimalFormat eNotation1 = new DecimalFormat("#0.###E0");
System.out.println(eNotation1.format(123.456789));
My expected output is;
1.235E2
instead,
1.2346E2
was printed.
Why was the output 1.2346E2?
I know I could have used a method like setMaximumFractionDigits() if all I wanted was to set the maximum number of digits after the decimal point, but I just really want to understand why the output was 1.2346E2
java decimalformat scientific-notation
edited 43 mins ago
Strazan
53112
asked 1 hour ago
Lenovo GenoviaLenovo Genovia
485
add a comment |
When following code is executed:
DecimalFormat eNotation1 = new DecimalFormat("#0.###E0");
System.out.println(eNotation1.format(123.456789));
My expected output is;
1.235E2
instead,
1.2346E2
was printed.
Why was the output 1.2346E2?
I know I could have used a method like setMaximumFractionDigits() if all I wanted was to set the maximum number of digits after the decimal point, but I just really want to understand why the output was 1.2346E2
java decimalformat scientific-notation
edited 43 mins ago
Strazan
53112
asked 1 hour ago
Lenovo GenoviaLenovo Genovia
485
add a comment |
When following code is executed:
DecimalFormat eNotation1 = new DecimalFormat("#0.###E0");
System.out.println(eNotation1.format(123.456789));
My expected output is;
1.235E2
instead,
1.2346E2
was printed.
Why was the output 1.2346E2?
I know I could have used a method like setMaximumFractionDigits() if all I wanted was to set the maximum number of digits after the decimal point, but I just really want to understand why the output was 1.2346E2
java decimalformat scientific-notation
edited 43 mins ago
Strazan
53112
asked 1 hour ago
Lenovo GenoviaLenovo Genovia
485
When following code is executed:
DecimalFormat eNotation1 = new DecimalFormat("#0.###E0");
System.out.println(eNotation1.format(123.456789));
My expected output is;
1.235E2
instead,
1.2346E2
was printed.
Why was the output 1.2346E2?
I know I could have used a method like setMaximumFractionDigits() if all I wanted was to set the maximum number of digits after the decimal point, but I just really want to understand why the output was 1.2346E2
java decimalformat scientific-notation
java decimalformat scientific-notation
edited 43 mins ago
Strazan
53112
asked 1 hour ago
Lenovo GenoviaLenovo Genovia
485
edited 43 mins ago
Strazan
53112
asked 1 hour ago
Lenovo GenoviaLenovo Genovia
485
edited 43 mins ago
Strazan
53112
edited 43 mins ago
Strazan
53112
edited 43 mins ago
Strazan
53112
53112
asked 1 hour ago
Lenovo GenoviaLenovo Genovia
485
asked 1 hour ago
Lenovo GenoviaLenovo Genovia
485
asked 1 hour ago
Lenovo GenoviaLenovo Genovia
485
485
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
If you want that exact format, then use the pattern 0.000E0:
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
System.out.println(eNotation1.format(123.456789));
1.235E2
As to why you are seeing your current behavior, the # placeholders are optional digits, which means that DecimalFormat is not obligated to actually use them exactly as you used them in the pattern. The only requirement appears to be that the total number of digits appearing in the scientific notation output matches. In this case, the total number of digits is five, so we get the output 1.2346E2.
answered 54 mins ago
Tim BiegeleisenTim Biegeleisen
242k13103163
add a comment |
Seems like this pattern will work as you expect
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
edited 50 mins ago
Pshemo
96.6k15135194
answered 1 hour ago
NayaNaya
693515
Yes DecimalFormat("0.000E0") will print 1.235E2, but I'm still wondering why the output wasn't 1.235E2 with the code segment given in the question.
– Lenovo Genovia
53 mins ago
1
Because the difference between # and 0 in decimal Format is in existing of this digit. If you use # here there are several options of presenting this number and that's why it will work wrongly with some numbers.
– Naya
51 mins ago
add a comment |
In order to fit your first pattern better i would just remove the first #. Like this you fit international unit system while keeping optional numbers after point.
DecimalFormat eNotation1 = new DecimalFormat("0.###E0");
System.out.println(eNotation1.format(123.456789));
answered 26 mins ago
PopHipPopHip
1037
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
If you want that exact format, then use the pattern 0.000E0:
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
System.out.println(eNotation1.format(123.456789));
1.235E2
As to why you are seeing your current behavior, the # placeholders are optional digits, which means that DecimalFormat is not obligated to actually use them exactly as you used them in the pattern. The only requirement appears to be that the total number of digits appearing in the scientific notation output matches. In this case, the total number of digits is five, so we get the output 1.2346E2.
answered 54 mins ago
Tim BiegeleisenTim Biegeleisen
242k13103163
add a comment |
If you want that exact format, then use the pattern 0.000E0:
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
System.out.println(eNotation1.format(123.456789));
1.235E2
As to why you are seeing your current behavior, the # placeholders are optional digits, which means that DecimalFormat is not obligated to actually use them exactly as you used them in the pattern. The only requirement appears to be that the total number of digits appearing in the scientific notation output matches. In this case, the total number of digits is five, so we get the output 1.2346E2.
answered 54 mins ago
Tim BiegeleisenTim Biegeleisen
242k13103163
add a comment |
If you want that exact format, then use the pattern 0.000E0:
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
System.out.println(eNotation1.format(123.456789));
1.235E2
As to why you are seeing your current behavior, the # placeholders are optional digits, which means that DecimalFormat is not obligated to actually use them exactly as you used them in the pattern. The only requirement appears to be that the total number of digits appearing in the scientific notation output matches. In this case, the total number of digits is five, so we get the output 1.2346E2.
answered 54 mins ago
Tim BiegeleisenTim Biegeleisen
242k13103163
If you want that exact format, then use the pattern 0.000E0:
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
System.out.println(eNotation1.format(123.456789));
1.235E2
As to why you are seeing your current behavior, the # placeholders are optional digits, which means that DecimalFormat is not obligated to actually use them exactly as you used them in the pattern. The only requirement appears to be that the total number of digits appearing in the scientific notation output matches. In this case, the total number of digits is five, so we get the output 1.2346E2.
answered 54 mins ago
Tim BiegeleisenTim Biegeleisen
242k13103163
answered 54 mins ago
Tim BiegeleisenTim Biegeleisen
242k13103163
answered 54 mins ago
Tim BiegeleisenTim Biegeleisen
242k13103163
answered 54 mins ago
Tim BiegeleisenTim Biegeleisen
242k13103163
242k13103163
add a comment |
add a comment |
Seems like this pattern will work as you expect
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
edited 50 mins ago
Pshemo
96.6k15135194
answered 1 hour ago
NayaNaya
693515
Yes DecimalFormat("0.000E0") will print 1.235E2, but I'm still wondering why the output wasn't 1.235E2 with the code segment given in the question.
– Lenovo Genovia
53 mins ago
1
Because the difference between # and 0 in decimal Format is in existing of this digit. If you use # here there are several options of presenting this number and that's why it will work wrongly with some numbers.
– Naya
51 mins ago
add a comment |
Seems like this pattern will work as you expect
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
edited 50 mins ago
Pshemo
96.6k15135194
answered 1 hour ago
NayaNaya
693515
Yes DecimalFormat("0.000E0") will print 1.235E2, but I'm still wondering why the output wasn't 1.235E2 with the code segment given in the question.
– Lenovo Genovia
53 mins ago
1
Because the difference between # and 0 in decimal Format is in existing of this digit. If you use # here there are several options of presenting this number and that's why it will work wrongly with some numbers.
– Naya
51 mins ago
add a comment |
Seems like this pattern will work as you expect
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
edited 50 mins ago
Pshemo
96.6k15135194
answered 1 hour ago
NayaNaya
693515
Seems like this pattern will work as you expect
DecimalFormat eNotation1 = new DecimalFormat("0.000E0");
edited 50 mins ago
Pshemo
96.6k15135194
answered 1 hour ago
NayaNaya
693515
edited 50 mins ago
Pshemo
96.6k15135194
edited 50 mins ago
Pshemo
96.6k15135194
edited 50 mins ago
Pshemo
96.6k15135194
96.6k15135194
answered 1 hour ago
NayaNaya
693515
answered 1 hour ago
NayaNaya
693515
answered 1 hour ago
NayaNaya
693515
693515
Yes DecimalFormat("0.000E0") will print 1.235E2, but I'm still wondering why the output wasn't 1.235E2 with the code segment given in the question.
– Lenovo Genovia
53 mins ago
1
Because the difference between # and 0 in decimal Format is in existing of this digit. If you use # here there are several options of presenting this number and that's why it will work wrongly with some numbers.
– Naya
51 mins ago
add a comment |
Yes DecimalFormat("0.000E0") will print 1.235E2, but I'm still wondering why the output wasn't 1.235E2 with the code segment given in the question.
– Lenovo Genovia
53 mins ago
1
Because the difference between # and 0 in decimal Format is in existing of this digit. If you use # here there are several options of presenting this number and that's why it will work wrongly with some numbers.
– Naya
51 mins ago
Yes DecimalFormat("0.000E0") will print 1.235E2, but I'm still wondering why the output wasn't 1.235E2 with the code segment given in the question.
– Lenovo Genovia
53 mins ago
Yes DecimalFormat("0.000E0") will print 1.235E2, but I'm still wondering why the output wasn't 1.235E2 with the code segment given in the question.
– Lenovo Genovia
53 mins ago
1
1
Because the difference between # and 0 in decimal Format is in existing of this digit. If you use # here there are several options of presenting this number and that's why it will work wrongly with some numbers.
– Naya
51 mins ago
Because the difference between # and 0 in decimal Format is in existing of this digit. If you use # here there are several options of presenting this number and that's why it will work wrongly with some numbers.
– Naya
51 mins ago
add a comment |
In order to fit your first pattern better i would just remove the first #. Like this you fit international unit system while keeping optional numbers after point.
DecimalFormat eNotation1 = new DecimalFormat("0.###E0");
System.out.println(eNotation1.format(123.456789));
answered 26 mins ago
PopHipPopHip
1037
add a comment |
In order to fit your first pattern better i would just remove the first #. Like this you fit international unit system while keeping optional numbers after point.
DecimalFormat eNotation1 = new DecimalFormat("0.###E0");
System.out.println(eNotation1.format(123.456789));
answered 26 mins ago
PopHipPopHip
1037
add a comment |
In order to fit your first pattern better i would just remove the first #. Like this you fit international unit system while keeping optional numbers after point.
DecimalFormat eNotation1 = new DecimalFormat("0.###E0");
System.out.println(eNotation1.format(123.456789));
answered 26 mins ago
PopHipPopHip
1037
In order to fit your first pattern better i would just remove the first #. Like this you fit international unit system while keeping optional numbers after point.
DecimalFormat eNotation1 = new DecimalFormat("0.###E0");
System.out.println(eNotation1.format(123.456789));
answered 26 mins ago
PopHipPopHip
1037
answered 26 mins ago
PopHipPopHip
1037
answered 26 mins ago
PopHipPopHip
1037
answered 26 mins ago
PopHipPopHip
1037
1037
add a comment |
add a comment |
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