阅读使用scanf()的不好的线路? [英] Reading a line using scanf() not good?
问题描述
scanf(" %[^\n]",line);
我的一个朋友建议使用与fgets()
来读取一行输入会比使用 scanf()的一个更好的主意
如上声明。他是有道理的吗?
A friend of mine suggested that using fgets()
to read a line as input would be a much better idea than using scanf()
as in the statement above. Is he justified?
推荐答案
字符*与fgets(的char * str中,INT NUM,FILE *流);
是安全的使用,因为它避免缓冲区溢出问题,仅扫描 NUM-1
字符数。
char * fgets ( char * str, int num, FILE * stream );
is safe to use because it avoid buffer overflow problem, it scans only num-1
number of char.
从读取字符流,并将它们存储为C字符串str中,直到(NUM-1)个字符已读或达到或者换行或档案结尾,先发生者为准。
Reads characters from stream and stores them as a C string into str until (num-1) characters have been read or either a newline or the end-of-file is reached, whichever happens first.
下面第二个参数 NUM
将被复制到str中的最大字符数(包括终止空字符)。
here second argument num
is Maximum number of characters to be copied into str (including the terminating null-character).
例如,假设你的code字符串数组容量仅仅是 5
如下长字符。
For example suppose in your code a string array capacity is just 5
chars long as below.
char str[5];
fgets (str, 5, fp); //5 =you have provision to avoid buffer overrun
使用上述code,如果从 FP
长于 4
字符,<$ C $输入C>与fgets()将只读取第一个 4
字符,然后追加 \\ 0
(,并丢弃其它多余的输入字符,仅仅存储在海峡[]
的)。
Using above code, if input from fp
is longer then 4
chars, fgets()
will read just first 4
chars then appends \0
(, and discard other extra input chars, just stores five char in str[]
).
而 scanf函数(%[^ \\ n],STR);
将读到 \\ n
不发现,如果输入的字符串长度超过 4
字符 scanf()的
将导致的buffer溢出(如 scanf函数
将尝试访问的内存超越最大指数 4
在海峡[]
)。
Whereas scanf(" %[^\n]",str);
will read until \n
not found and if input string is longer then 4
chars scanf()
will cause of buffer overflow (as scanf
will try to access memory beyond max index 4
in str[]
).
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