格式化带有多个百分号的字符串 [英] Format string with multiple percent signs

问题描述

我知道 %% 用于转义字符串中的实际％标记，因此 %%% ds 将在以下格式字符串中以％10s 结尾，但是我不知道为什么我需要 %% 5s 是否在此字符串中？

I know %% is used to escape actual % signs in a string, so %%%ds will end up with %10s in the following format string, but I don't know why I need %%5s in this string?

毕竟，只有两个附加参数（BUFFSIZE / 10）。

After all, there are only two additional arguments (BUFFSIZE / 10).

#define BUFFSIZE 100
char buf[100]={0}
sprintf(buf, "%%5s %%%ds %%%ds", BUFFSIZE / 10, BUFFSIZE / 10);

运行上面的代码后，buf将包含字符串，

After running the code above, the buf will contain the string,

%10s %10s 

推荐答案

目的是获取一个格式字符串以在需要诸如 sscanf（）之类的格式字符串的另一个函数中使用它。

The purpose is to get a format string to use it in another function that needs a format string like sscanf().

使用您的代码，您将得到：％5s％10s％10s 写入您的 buf ，参见在线，这意味着它将接受三个带有长度标识符的字符串。

With your code you get: %5s %10s %10s written to your buf, see online, which means it will accept three strings with a length identifier.

%%5s          --> %5s
%%%ds with 10 --> %10s (read it that way: {%%}{%d}{s})

现在可以在 sscanf（）调用中使用缓冲区％5s％10s％10s ，如所示此处。

That buffer %5s %10s %10s could now be used in a sscanf() call like shown here.

但是有一种最佳做法是防止引起缓冲区溢出由 sscanf（）编写，Kernighan和Pike在他们的书编程实践，请参见此处的SO 。

But there is a best practice to prevent a buffer overflow caused by sscanf() which is also described by Kernighan and Pike in their book The Practice of Programming, see here on SO.

您可能无法使用％* s 可能是，请参见在此处找到：

The reason why you maybe can't use %*s may be, see here on SO:

对于 printf ，*允许您通过附加参数指定最小字段宽度，即 printf（％* d ，4，100）; 指定一个国际剑联ld宽度为4。

For printf, the * allows you to specify minimum field width through an extra parameter, i.e. printf("%*d", 4, 100); specifies a field width of 4.

对于 scanf ，*表示要读取但忽略该字段，因此即输入 12 34的 scanf（％* d％d，& i）将忽略12并将34读为整数i。

For scanf, the * indicates that the field is to be read but ignored, so that i.e. scanf("%*d %d", &i) for the input "12 34" will ignore 12 and read 34 into the integer i.

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