user-0 提问: scanf的连续输入

回复:
user: user-1
content: 讨论千句道听途说的标准、想法、…… 不如调通一句自己编写的代码！

user: user-1
content: 不要使用 while (条件) 更不要使用 while (组合条件) 要使用 while (1) {  if (条件1) break;  //...  if (条件2) continue;  //...  if (条件3) return;  //... } 因为前两种写法在语言表达意思的层面上有二义性，只有第三种才忠实反映了程序流的实际情况。 典型如： 下面两段的语义都是当文件未结束时读字符 while (!feof(f)) {  a=fgetc(f);  //...  b=fgetc(f);//可能此时已经feof了！  //... } 而这样写就没有问题： while (1) {  a=fgetc(f);  if (feof(f)) break;  //...  b=fgetc(f);  if (feof(f)) break;  //... } 类似的例子还可以举很多。  Format Specification Fields: scanf and wscanf Functions A format specification has the following form:  % [width] [{h | l | I64 | L}]type  The format argument specifies the interpretation of the input and can contain one or more of the following:   White-space characters: blank (' '); tab ('\t'); or newline ('\n'). A white-space character causes scanf to read, but not store, all consecutive white-space characters in the input up to the next non–white-space character. One white-space character in the format matches any number (including 0) and combination of white-space characters in the input.   Non–white-space characters, except for the percent sign (%). A non–white-space character causes scanf to read, but not store, a matching non–white-space character. If the next character in stdin does not match, scanf terminates.   Format specifications, introduced by the percent sign (%). A format specification causes scanf to read and convert characters in the input into values of a specified type. The value is assigned to an argument in the argument list.  The format is read from left to right. Characters outside format specifications are expected to match the sequence of characters in stdin; the matching characters in stdin are scanned but not stored. If a character in stdin conflicts with the format specification, scanf terminates, and the character is left in stdin as if it had not been read.  When the first format specification is encountered, the value of the first input field is converted according to this specification and stored in the location that is specified by the first argument. The second format specification causes the second input field to be converted and stored in the second argument, and so on through the end of the format string.  An input field is defined as all characters up to the first white-space character (space, tab, or newline), or up to the first character that cannot be converted according to the format specification, or until the field width (if specified) is reached. If there are too many arguments for the given specifications, the extra arguments are evaluated but ignored. The results are unpredictable if there are not enough arguments for the format specification.  Each field of the format specification is a single character or a number signifying a particular format option. The type character, which appears after the last optional format field, determines whether the input field is interpreted as a character, a string, or a number.   The simplest format specification contains only the percent sign and a type character (for example, %s). If a percent sign (%) is followed by a character that has no meaning as a format-control character, that character and the following characters (up to the next percent sign) are treated as an ordinary sequence of characters, that is, a sequence of characters that must match the input. For example, to specify that a percent-sign character is to be input, use %%.  An asterisk (*) following the percent sign suppresses assignment of the next input field, which is interpreted as a field of the specified type. The field is scanned but not stored.

user: user-2
content: 引用 4 楼 qq_25987103 的回复:那么输入 3   4    +     敲一个回车 怎么写break的if条件呢？scanf最后还会读取‘\n’?我发现敲了一个回车之后还要求继续输入？为什么？ 输入回车并不是输入“EOF”啊，满足循环的条件那可不继续执行吗。

user: user-2
content: 用%s接收数据时，是不接收空格的，你输入4  5 + ，这三个字符之间有空格，第一次循环接收4 这个字符，然后接收数据结束，满足条件，输出一个a，紧接着第二次循环，由于输入缓冲区中（这里涉及到键盘缓冲区的问题）还有5这个字符等待输入，所以直接把5读进去，依然不读空格，在输出一个a，“+”这个字符也是这样的。

user: user-1
content: 作为一个C程序员，对  scanf,sscanf,fscanf  printf,sprintf,fprintf 这类函数的用法，还是要做到“拳不离手，曲不离口”的。  #include <stdio.h> #include <string.h> int main() {     char tmp[30];     int a,b;     char s[2],op,c;     while (1) {         printf("Input # # op(example:4 5 +):");fflush(stdout);         rewind(stdin);         fgets(tmp,30,stdin);         if (tmp[0]=='\n') break;         if (4==sscanf(tmp,"%d%d%1s%c",&a,&b,s,&c)) {             if (c=='\n') {                 op=s[0];                 if (op=='+'                  || op=='-'                  || op=='*'                  || op=='/') {                     if (op!='/' || (op=='/' && b!=0)) {                         printf("You input is %d %d %c\n",a,b,op);                     }                 }             }         }     }     return 0; } //Input # # op(example:4 5 +):4 5 + //You input is 4 5 + //Input # # op(example:4 5 +):1 2 - //You input is 1 2 - //Input # # op(example:4 5 +):3 4 * //You input is 3 4 * //Input # # op(example:4 5 +):5 6 / //You input is 5 6 / //Input # # op(example:4 5 +):7 0 / //Input # # op(example:4 5 +):4 5 +! //Input # # op(example:4 5 +): //

user: user-0
content: 那么输入 3   4    +     敲一个回车 怎么写break的if条件呢？scanf最后还会读取‘\n’?我发现敲了一个回车之后还要求继续输入？为什么？

user: user-3
content: 另外我再补充一下，你可以这样实现你的功能while(（tmp[i++]=getchar()） != EOF&&i<10 )

user: user-3
content: scanf返回的是非0值，所以你的程序不管输入什么一直在while循环中，你输入几个数就输出几个a