[英]whitespace in the format string (scanf)
Consider the following code: 考虑以下代码:
#include<stdio.h>
int main() {
int i=3, j=4;
scanf("%d c %d",&i,&j);
printf("%d %d",i,j);
return 0;
}
It works if I give 2c3
or 2 c 3
or 2c 3
as input if I have to change the value of variables. 如果我必须更改变量的值,则当我输入2c3
或2 c 3
或2c 3
作为输入时,它将起作用。 What should I do if I want the user to enter the same pattern as I want means if %dc%d
then only 2c3
is acceptable and not 2 c 3
and vice versa if it is %dc %d
? 如果我希望用户输入与我想要的相同的模式,该怎么办意味着如果%dc%d
只能接受2c3
而不是2 c 3
,反之亦然,如果它是%dc %d
则相反?
Whitespace in the format string matches 0 or more whitespace characters in the input. 空白格式字符串匹配在输入0或多个空格字符。
So "%dc %d"
expects number, then any amount of whitespace characters, then character c
, then any amount of whitespace characters and another number at the end. 因此, "%dc %d"
期望数字,然后是任意数量的空白字符,然后是字符c
,然后是任意数量的空白字符,最后是另一个数字。
"%dc%d"
expects number, c
, number. "%dc%d"
期望数字c
。
Also note, that if you use *
in the format string, it suppresses assignment: 另请注意,如果在格式字符串中使用*
,它将禁止分配:
%*c
= read 1 character, but don't assign it to any variable %*c
=读取1个字符,但不要将其分配给任何变量
So you can use "%d%*cc%*c %d"
if you want to force user to enter: number, at least 1 character followed by any amount of whitespace characters, c
, at least 1 character followed by any amount of whitespace characters again and number. 因此,如果要强制用户输入,可以使用"%d%*cc%*c %d"
:数字,至少1个字符,后跟任意数量的空白字符, c
,至少1个字符,后跟任意数量的空白再次输入空格字符和数字。
If you want to accept 1c2
but not 1 c 2
, use the pattern without the space: 如果要接受1c2
而不是1 c 2
,请使用不带空格的模式:
scanf("%dc%d", &x, &y);
If you want to accept 1c2
and 1 c 2
(and also 1 \\t \\tc \\t 2
etc), use the pattern with the space: 如果要接受1c2
和1 c 2
(以及1 \\t \\tc \\t 2
等),请使用带有空格的模式:
scanf("%d c %d", &x, &y);
If you want to accept 1 c 2
but not 1c2
, add a fake string containing whitespace: 如果要接受1 c 2
而不是1c2
,请添加包含空格的假字符串:
scanf("%d%*[ \t]c%*[ \t]%d", &x, &y);
Here the format string %[ \\t]
would mean "read a string that contains any number of space and tab characters"; 这里的格式字符串%[ \\t]
表示“读取了包含任意数量的空格和制表符的字符串”; but using the additional *
, it becomes " expect a string that contains any number of space and tab characters; then discard it " 但是使用附加的*
,它将变成“ 期望包含任意数量的空格和制表符的字符串;然后将其丢弃 ”
I think I would read the scanf result into different variables (ie not reuse i
and j
) as "%d%s%d"
. 我想我会将scanf结果读入不同的变量(即不重用i
和j
)作为"%d%s%d"
。 Then check the string you got from the %s and if it matches your requirements, use the other variables to overwrite i and j. 然后检查从%s获得的字符串,如果它符合您的要求,请使用其他变量覆盖i和j。
Force a string parsing first : 首先强制字符串解析:
char a[100], b[100];
scanf("%99s c %99s", a, b);
Then use sscanf() to convert the strings to int. 然后使用sscanf()将字符串转换为int。
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