c - 使用变量设置 float 宽度和精度

Question

我正在尝试打印具有用户可变宽度和精度的 float 。我遇到的问题是只读取精度字段，宽度被完全忽略。

int width = 4; int precision = 12;
printf("%*.*lf ", width , precision, value);

605.32364879299 5980.79287087619 802.32966093936
5839.78856776635 5355.77379680776 6602.20148319956

本应填充空格以确保小数位对齐，但就好像甚至没有读取宽度变量一样。

Answer 1

width is being completely ignored.

肯定是对字段宽度的误解。这是要打印的最小字符数，包括符号、数字、小数点等。

An optional minimum field width. If the converted value has fewer characters than the field width, it is padded with spaces ... C11dr §7.21.6.1 4

int width = 4; int precision = 12;
double value =  605.32364879299;
printf("%*.*lf\n", width , precision, value);
width = 21;
printf("%*.*lf\n", width , precision, value);
/*
123456789012345678901 */

输出

605.323648792990       // Minimum width =  4, 16 characters in numeric text
     605.323648792990  // Minimum width = 21, 5 spaces + 16 characters in numeric text