printf：如何對齊小數點上的數字

Question

考慮有一組數字：

my @array =   (
   1.788139343e-007, 0.0547055073198, -0.703213036125,
   -0.583665880391, -1.41198285298, +0.171879081676,
   -0.58966025098, -86.0627173425, -0.84449797709,
 3.49876623321e-005, 3.02660429162, -0.256948695361);

我想將小數點對齊在總寬度為m的第n列（可能n = 6和m = 25）

如果使用%f我得到了很好的對齊數字，但是需要科學記數法的數字正在被破壞。 usging %g將點后面的precision參數解釋為絕對精度，導致小數點后面的小數點不同。 由於大多數數字在（-10,10）范圍內，我不想采用科學記數法%e

是否有我忽略的標志或格式屬性（或組合）？

預期的結果將是：

foreach my $f (@array){
  printf("[%+25.12g]$/", $f);
}


[   +1.788139343e-007     ]
[   +0.0547055073198      ]
[   -0.703213036125       ]
[   -0.583665880391       ]
[   -1.41198285298        ]
[   +0.0171879081676      ]
[   -0.58966025098        ]
[  -86.0627173425         ]
[   -0.84449797709        ]
[   +3.49876623321e-005   ]
[   +3.02660429162        ]
[   -0.256948695361       ]

甚至更好

[   +1.7881393430000e-007 ]
[   +0.0547055073198      ]
[   -0.7032130361250      ]
[   -0.5836658803910      ]
[   -1.4119828529800      ]
[   +0.0171879081676      ]
[   -0.5896602509800      ]
[  -86.0627173425000      ]
[   -0.8444979770900      ]
[   +3.4987662332100e-005 ]
[   +3.0266042916200      ]
[   -0.2569486953610      ]

（問題是關於Perl但是s?printf的格式字符串是相當獨立的，所以我也添加了C標簽）

Answer 1

[*]printf函數允許您：

• 得到多少字符，
• 添加一些填充。

因此，如果您知道點之前有多少字符（ d = sprintf(buf, "%.0f", ar[i]); ），您可以使用（ printf("[%*s %g", 4-d, "", ar[i]); ）。

然后使用相同的邏輯來對齊右括號：

#include <stdio.h>

int main()
{
    double ar[] = {
        1.788139343e-007,   0.0547055073198, -0.703213036125,
       -0.583665880391,    -1.41198285298,    0.171879081676,
       -0.58966025098,    -86.0627173425,    -0.84449797709,
        3.49876623321e-005, 3.02660429162,   -0.256948695361};

    for (int i = 0; i < 12; ++i)
    {
        /* buffer to count how much character are before the dot*/
        char buf[64];

        /* how much before the dot? */
        int d = sprintf(buf, "%+.0lf", ar[i]);

        /* write float with aligned dot and store second padding */
        int e = printf("[%*s %+.15lg", 4-d, "", ar[i]);
        printf("%*s]\n", 25-e, "");

    }
    return 0;
}

得到：

[   +1.788139343e-07     ]
[   +0.0547055073198     ]
[   -0.703213036125      ]
[   -0.583665880391      ]
[   -1.41198285298       ]
[   +0.171879081676      ]
[   -0.58966025098       ]
[  -86.0627173425        ]
[   -0.84449797709       ]
[   +3.49876623321e-05   ]
[   +3.02660429162       ]
[   -0.256948695361      ]