用％c或％s扫描 [英] Scanning with %c or %s
问题描述
我必须为大学做一个计划，在这个计划中，我应该将一定数量的人，喜欢的人和不喜欢某些东西的人分开. 所以我这样做了:
I had to do a program for college in which I should separate, between a certain amount of people, those who liked and the ones who disliked something. so I did this:
char like[100];
printf("Like? Y or N \n");
scanf ("%c", like);
该程序已编译，但未按应有的方式运行.当询问喜欢吗?"时，用户无法写"y或n"
The program compiled, but didn't work the way it should. The user was not able to write "y or n" when asked "Like?"
所以我尝试了这个:
char like[100];
printf("Like? Y or N \n");
scanf ("%s", like);
它奏效了.但是我不知道它为什么起作用.有人可以解释一下scanf中%c和%s之间的区别吗?
And it worked. But I don't know why it worked. Can somebody please explain me the difference between %c and %s in a scanf?
推荐答案
首先，请先进行一些基础研究，然后再访问此处-通常可以通过快速Google搜索或查看方便的C参考手册来回答此类问题.
First, please do some basic research before coming here - questions like this can usually be answered with a quick Google search or checking your handy C reference manual.
char inputChar; // stores a single character
char inputString[100] = {0}; // stores a string up to 99 characters long
scanf( " %c", &inputChar ); // read the next non-whitespace character into inputChar
// ^ Note & operator in expression
scanf( "%s", inputString ); // read the next *sequence* of non-whitespace characters into inputString
// ^ Note no & operator in expression
当您要从输入流中读取单个字符并将其存储到char对象时，将使用%c. %c转换说明符不会跳过任何前导空格，因此，如果您想读取下一个 non -空白字符，则在格式字符串中%c说明符之前需要一个空格，例如如上所示.
You would use %c when you want to read a single character from the input stream and store it to a char object. The %c conversion specifier will not skip over any leading whitespace, so if you want to read the next non-whitespace character, you need a blank before the %c specifier in your format string, as shown above.
如果要从输入流中读取非空格字符的序列并将它们存储到char的 array 中，则可以使用%s.您的目标数组必须足够大，以存储输入字符串 plus 一个终止于0值的字符. %s转换说明符会跳过所有前导空格，并在非空格字符之后的第一个空格字符处停止读取.
You would use %s when you want to read a sequence of non-whitespace characters from the input stream and store them to an array of char. Your target array must be large enough to store the input string plus a terminating 0-valued character. The %s conversion specifier skips over any leading whitespace and stops reading at the first whitespace character following the non-whitespace characters.
%c和%s都期望它们对应的参数具有类型char *(指向char的指针)；但是，在第一种情况下，假定指针指向单个对象，而在第二种情况下，假定指针指向数组的第一个元素.对于inputChar，我们必须使用一元&运算符来获取指针值.对于inputString，我们不这样做，因为在大多数情况下，类型为"T的数组"的表达式将被转换(衰变")为将指针指向
Both %c and %s expect their corresponding argument to have type char * (pointer to char); however, in the first case, it's assumed that the pointer points to a single object, whereas in the second case, it's assumed that the pointer points to the first element of an array. For inputChar, we must use the unary & operator to obtain the pointer value. For inputString, we don't, because under most circumstances an expression of type "array of T" will be converted ("decay") to an expression of type "pointer to T", and the value of the expression will be the address of the first element of the array.
您的代码可以正常工作，但是读取单个字符并将其存储到数组有点混乱.
Your code works fine as it is, but it's a bit confusing to read a single character and store it to an array.
在没有明确字段宽度的情况下使用%s是有风险的；如果有人键入了100个以上的非空白字符，则scanf将在inputString之后愉快地将这些多余的字符存储到内存中，从而可能破坏重要内容.通常写这样的东西比较安全
Using %s without an explicit field width is risky; if someone types in more than 100 non-whitespace characters, scanf will happily store those extra characters to memory following inputString, potentially clobbering something important. It's generally safer to write something like
scanf( "%99s", inputString ); // reads no more than 99 characters into inputString
或改用fgets()读取输入字符串:
or to use fgets() to read input strings instead:
fgets( inputString, sizeof inputString, stdin );
请检查 C语言标准的在线草案，以完整描述*scanf函数的所有转换说明符.
Please check §7.21.6.2 of the online draft of the C language standard for a complete description of all of the conversion specifiers for the *scanf functions.
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