C语言 scanf() 跳过变量
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scanf() skip variable
提问by ross
In C, using scanf()with the parameters, scanf("%d %*d", &a, &b)acts differently. It enters value for just one variable not two!
在 C 中，scanf()与参数一起使用，scanf("%d %*d", &a, &b)作用不同。它只为一个变量而不是两个变量输入值！
Please explain this!
请解释一下！
scanf("%d %*d", &a, &b);
回答by jpalecek
The *basically means the specifier is ignored (integer is read, but not assigned).
所述*基本上意味着说明符被忽略（整数被读取，但是未分配）。
Quotation from man scanf:
引自man scanf：
* Suppresses assignment. The conversion that follows occurs as usual, but no pointer is used; the result of the conversion is simply discarded.
* Suppresses assignment. The conversion that follows occurs as usual, but no pointer is used; the result of the conversion is simply discarded.
回答by Tomek Szpakowicz
Asterisk (*) means that the value for format will be read but won't be written into variable. scanfdoesn't expect variable pointer in its parameter list for this value. You should write:
星号 (*) 表示将读取 format 的值，但不会将其写入变量。scanf不希望在其参数列表中为该值提供变量指针。你应该写：
scanf("%d %*d",&a);
回答by reader_1000
http://en.wikipedia.org/wiki/Scanf#Format_string_specifications
http://en.wikipedia.org/wiki/Scanf#Format_string_specifications
An optional asterisk (*) right after the percent symbol denotes that the datum read by this format specifier is not to be stored in a variable.
百分比符号后面的可选星号 (*) 表示该格式说明符读取的数据不存储在变量中。
