user-0 提问: C语言中scanf为什么没有执行呀？

回复:
user: user-0
content: 哦  是这样呀  谢谢各位老师

user: user-1
content: #include <stdio.h> #include <stdlib.h> #include <string.h> int main() { 	char num[50]; 	scanf("%[^\n]", num); //输入test1回车,此时num读到了test1 	printf("%s\n", num);   	int len = strlen(num); 	scanf("%[^\n]", &num[len]); //这个读取到了空白字符，所以是执行了scanf,运行结果让你以为没执行而已 	printf("%s\n", num);  	system("pause"); 	return 0; }   解决方法 1.在scanf输入个空格 2.使用fflush   #include <stdio.h> #include <stdlib.h> #include <string.h> int main() { 	char num[50]; 	scanf(" %[^\n]", num); //注意有个空格 	printf("%s\n", num);   	int len = strlen(num); 	scanf(" %[^\n]", &num[len]); //注意有个空格 	printf("%s\n", num);  	system("pause"); 	return 0; }   #include <stdio.h> #include <stdlib.h> #include <string.h> int main() { 	char num[50]; 	scanf("%[^\n]", num);  	printf("%s\n", num);   	fflush(stdin); //清空缓冲区 	int len = strlen(num); 	scanf("%[^\n]", &num[len]);  	printf("%s\n", num);   	system("pause"); 	return 0; }

user: user-2
content: 在每个最后不带\n的printf后面加fflush(stdout); 在每个不想受接收缓冲区旧内容影响的scanf前面加rewind(stdin); 另外请检查scanf的返回值。

user: user-3
content: 输入的字符串都在标准输入缓存中，为了避免回车键的干扰，可以在scanf前面加上fflush(stdin); 来清除标准输入的缓存。  #include <stdio.h> #include <string.h> void main() { 	char num[50]; 	scanf("%s",num); 	printf("%s\n",num); 	int len=strlen(num); 	fflush(stdin);  	scanf("%c",&num[len]); 	printf("%s\n",num); }  这样就可以得到你想要的答案

user: user-2
content: 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-0
content: 引用 2 楼 user-4 的回复:第二个scanf指定是执行了的，只不过你的这两个scanf的写法有问题。  你第一个scanf读到 \n 返回了，然后让第二个scanf去读这个\n，不还是直接返回么 恩 你说的对老师，改成scanf（“%s”,num）后就可以执行了，但是我有一点不明白的是在没有修改之前scanf是从哪个地方开始读字符的？从数组开头还是从上次输入的结尾？修改后的scanf输入时我也是用的回车结束输入，第二个scanf为什么没有影响呢？

user: user-4
content: 第二个scanf指定是执行了的，只不过你的这两个scanf的写法有问题。  你第一个scanf读到 \n 返回了，然后让第二个scanf去读这个\n，不还是直接返回么

user: user-0
content: 有三个“hello”，第一个是执行第一个scanf的时候我输入的，后两个是执行了两个printf输出的，唯独第二个scanf没有执行。