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| Rev 264 | Rev 266 | ||
|---|---|---|---|
| Line 47... | Line 47... | ||
| 47 | int counter; |
47 | int counter; |
| 48 | int exponent,exponenttmp; |
48 | int exponent,exponenttmp; |
| 49 | unsigned char buf[DEFAULT_DOUBLE_BUFFER_SIZE]; |
49 | unsigned char buf[DEFAULT_DOUBLE_BUFFER_SIZE]; |
| 50 | unsigned long in1,in2; |
50 | unsigned long in1,in2; |
| 51 | 51 | ||
| 52 | /* |
52 | |
| 53 | if (fmath_is_nan(num)) { |
53 | if (fmath_is_nan(num)) { |
| 54 | print_str("NaN"); |
54 | print_str("NaN"); |
| 55 | return; |
55 | return; |
| 56 | } |
56 | } |
| 57 | */ |
- | |
| 58 | 57 | ||
| 59 | if (fmath_is_negative(num)) { |
58 | if (num<0.0) { |
| 60 | putchar('-'); |
59 | putchar('-'); |
| - | 60 | num=num*-1.0; |
|
| - | 61 | } |
|
| - | 62 | ||
| - | 63 | ||
| - | 64 | if (fmath_is_infinity(num)) { |
|
| - | 65 | print_str("Inf"); |
|
| - | 66 | return; |
|
| 61 | } |
67 | } |
| 62 | - | ||
| 63 | num=fmath_abs(num); |
- | |
| 64 | 68 | ||
| 65 | if ((modifier=='E')||(modifier=='e')) { |
69 | if ((modifier=='E')||(modifier=='e')) { |
| 66 | intval2=fmath_fint(fmath_get_decimal_exponent(num),&intval); |
70 | intval2=fmath_fint(fmath_get_decimal_exponent(num),&intval); |
| 67 | exponent=intval; |
71 | exponent=intval; |
| 68 | if ((intval2<0.0)&&(exponent<0)) exponent--; |
72 | if ((intval2<0.0)&&(exponent<0)) exponent--; |
| Line 76... | Line 80... | ||
| 76 | } |
80 | } |
| 77 | print_number(exponent,10); |
81 | print_number(exponent,10); |
| 78 | return; |
82 | return; |
| 79 | } |
83 | } |
| 80 | 84 | ||
| 81 | - | ||
| 82 | /* |
- | |
| 83 | if (fmath_is_infinity(num)) { |
- | |
| 84 | print_str("Inf"); |
- | |
| 85 | } |
- | |
| 86 | */ |
- | |
| 87 | //TODO: rounding constant - when we got fraction >= 0.5, we must increment last printed number |
85 | //TODO: rounding constant - when we got fraction >= 0.5, we must increment last printed number |
| 88 | 86 | ||
| 89 | /* Here is problem with cumulative error while printing big double values -> we will divide |
87 | /* Here is problem with cumulative error while printing big double values -> we will divide |
| 90 | the number with a power of 10, print new number with better method for small numbers and then print decimal point at correct position */ |
88 | the number with a power of 10, print new number with better method for small numbers and then print decimal point at correct position */ |
| 91 | 89 | ||