C语言 在 printf 中设置可变文本列宽
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Set variable text column width in printf
提问by Alaa M.
In order to determine the size of the column in C language we use %<number>d.
For instance, I can type %3dand it will give me a column of width=3.
My problem is that my number after the %is a variable that I receive, so I need something like %xd(where xis the integer variable I received sometime before in my program).
But it's not working.
为了确定 C 语言中列的大小，我们使用%<number>d. 例如，我可以打字%3d，它会给我一列宽度=3。我的问题是%我收到的数字是一个变量，所以我需要类似的东西%xd（x我之前在程序中收到的整数变量在哪里）。但它不起作用。
Is there any other way to do this?
有没有其他方法可以做到这一点？
回答by Oliver Charlesworth
You can do this as follows:
您可以按如下方式执行此操作：
printf("%*d", width, value);
From Lee's comment:
You can also use a * for the precision size:
来自 Lee 的评论：
您还可以使用 * 作为精确尺寸：
printf("%*.*f", width, precision, value);
回答by jetset
Just for completeness, wanted to mention that with POSIX-compliant versions of printf()you can also put the actual field width (or precision) value somewhere else in the parameter list and refer to it using the 1-based parameter number followed by a dollar sign:
只是为了完整性，想提一下，使用 POSIX 兼容版本，printf()您还可以将实际字段宽度（或精度）值放在参数列表中的其他位置，并使用基于 1 的参数编号后跟美元符号来引用它：
A field width or precision, or both, may be indicated by an asterisk ‘?' or an asterisk followed by one or more decimal digits and a ‘$' instead of a digit string. In this case, an int argument supplies the field width or precision. A negative field width is treated as a left adjustment flag followed by a positive field width; a negative precision is treated as though it were missing. If a single format directive mixes positional (nn$) and non-positional arguments, the results are undefined.
字段宽度或精度，或两者，可以用星号“?”表示。或星号后跟一个或多个十进制数字和一个“$”而不是数字字符串。在这种情况下， int 参数提供字段宽度或精度。负场宽被视为左调整标志，然后是正场宽；负精度被视为缺失。如果单个格式指令混合了位置 (nn$) 和非位置参数，则结果未定义。
E.g., printf ( "%1$*d", width, value );
例如， printf ( "%1$*d", width, value );
