user-0 提问: while ( scanf ("%d %d %d", &x, &y, &z ) != EOF ) 的相关问题

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user: user-0
content: 谢谢各位的解答，已经解决问题。。谢谢

user: user-1
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-2
content: 引用 3 楼 user-3 的回复:[quote=引用 1 楼 zjq9931 的回复:] 因为要输入3个数，所以要3个Ctrl+Z才可以。  #include <stdio.h>  int main() { 	int      x, y, z; 	int	    vnum = 0; 	 	while ( scanf("%d %d %d", &x, &y, &z)!=EOF )	//因为这个函数要输入3个数字，所以，要3个文件结束符，才能结束 	{ 		printf("x=%d, y=%d, z=%d", x, y, z); 	} 	//如果和结构体无关，建议用这种形式： 	while ( 1 )  	{ 		if(scanf("%d", &x)==EOF) 		{ 			break; 		} 		if(scanf("%d", &y)==EOF) 		{ 			break; 		} 		if(scanf("%d", &z)==EOF) 		{ 			break; 		} 		printf("x=%d, y=%d, z=%d", x, y, z); 	} 	puts("end"); 	return 0; }  别胡说  不是什么个数问题  [/quote] 啊，我试验过的啊，感觉确实是个数的问题。 我个人在这方面确实也有些疑惑，请指教。

user: user-3
content: 引用 1 楼 zjq9931 的回复:因为要输入3个数，所以要3个Ctrl+Z才可以。  #include <stdio.h>  int main() { 	int      x, y, z; 	int	    vnum = 0; 	 	while ( scanf("%d %d %d", &x, &y, &z)!=EOF )	//因为这个函数要输入3个数字，所以，要3个文件结束符，才能结束 	{ 		printf("x=%d, y=%d, z=%d", x, y, z); 	} 	//如果和结构体无关，建议用这种形式： 	while ( 1 )  	{ 		if(scanf("%d", &x)==EOF) 		{ 			break; 		} 		if(scanf("%d", &y)==EOF) 		{ 			break; 		} 		if(scanf("%d", &z)==EOF) 		{ 			break; 		} 		printf("x=%d, y=%d, z=%d", x, y, z); 	} 	puts("end"); 	return 0; }  别胡说  不是什么个数问题

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。

user: user-3
content: 引用 4 楼 zjq9931 的回复:[quote=引用 3 楼 user-3 的回复:] [quote=引用 1 楼 zjq9931 的回复:] 因为要输入3个数，所以要3个Ctrl+Z才可以。  #include <stdio.h>  int main() { 	int      x, y, z; 	int	    vnum = 0; 	 	while ( scanf("%d %d %d", &x, &y, &z)!=EOF )	//因为这个函数要输入3个数字，所以，要3个文件结束符，才能结束 	{ 		printf("x=%d, y=%d, z=%d", x, y, z); 	} 	//如果和结构体无关，建议用这种形式： 	while ( 1 )  	{ 		if(scanf("%d", &x)==EOF) 		{ 			break; 		} 		if(scanf("%d", &y)==EOF) 		{ 			break; 		} 		if(scanf("%d", &z)==EOF) 		{ 			break; 		} 		printf("x=%d, y=%d, z=%d", x, y, z); 	} 	puts("end"); 	return 0; }  别胡说  不是什么个数问题  [/quote] 啊，我试验过的啊，感觉确实是个数的问题。 我个人在这方面确实也有些疑惑，请指教。[/quote]不同环境要求不一样  这里是ctrl+z 别的环境可能是ctrl+d 然后 ctrl+z 好像有一个特殊的要求 不能是 1 2 3 ctrl+z 回车 应该是 1 2 3 回车 ctrl+z 回车

user: user-2
content: 因为要输入3个数，所以要3个Ctrl+Z才可以。  #include <stdio.h>  int main() { 	int      x, y, z; 	int	    vnum = 0; 	 	while ( scanf("%d %d %d", &x, &y, &z)!=EOF )	//因为这个函数要输入3个数字，所以，要3个文件结束符，才能结束 	{ 		printf("x=%d, y=%d, z=%d", x, y, z); 	} 	//如果和结构体无关，建议用这种形式： 	while ( 1 )  	{ 		if(scanf("%d", &x)==EOF) 		{ 			break; 		} 		if(scanf("%d", &y)==EOF) 		{ 			break; 		} 		if(scanf("%d", &z)==EOF) 		{ 			break; 		} 		printf("x=%d, y=%d, z=%d", x, y, z); 	} 	puts("end"); 	return 0; }