C 输入错误后如何刷新缓冲区
我必须在输入中得到一个int,为了验证它,我写了:C 输入错误后如何刷新缓冲区,c,validation,io,int,scanf,C,Validation,Io,Int,Scanf,我必须在输入中得到一个int,为了验证它,我写了: do { scanf("%d", &myInt); if (myInt >= 2147483648 || myInt <= -2147483648) printf("I need an integer between -2147483647 and 2147483647: "); } while (myInt >= 2147483648 || myInt <= -21474836
do {
scanf("%d", &myInt);
if (myInt >= 2147483648 || myInt <= -2147483648)
printf("I need an integer between -2147483647 and 2147483647: ");
} while (myInt >= 2147483648 || myInt <= -2147483648);
do{
scanf(“%d”、&myInt);
如果(myInt>=2147483648 | | myInt=2147483648 | | myInt使用的返回值来实现此目的:
int myInt;
while (scanf("%d", &myInt) != 1) {
// scanf failed to extract int from the standard input
}
// TODO: integer successfully retrieved ...
这就是为什么我通常建议不要使用scanf
进行交互式输入;为了使其真正防弹,您最好使用fgets()
并使用strtod
或strtol
转换为数字类型
char inbuf[MAX_INPUT_LENGTH];
...
if ( fgets( inbuf, sizeof inbuf, stdin ))
{
char *newline = strchr( inbuf, '\n' );
if ( !newline )
{
/**
* no newline means that the input is too large for the
* input buffer; discard what we've read so far, and
* read and discard anything that's left until we see
* the newline character
*/
while ( !newline )
if ( fgets( inbuf, sizeof inbuf, stdin ))
newline = strchr( inbuf, '\n' );
}
else
{
/**
* Zero out the newline character and convert to the target
* data type using strtol. The chk parameter will point
* to the first character that isn't part of a valid integer
* string; if it's whitespace or 0, then the input is good.
*/
newline = 0;
char *chk;
int tmp = strtol( inbuf, &chk, 10 );
if ( isspace( *chk ) || *chk == 0 )
{
myInt = tmp;
}
else
{
printf( "%s is not a valid integer!\n", inbuf );
}
}
}
else
{
// error reading from standard input
}
C语言中的交互式输入可以是简单的异或输入,也可以是健壮的输入,两者都不能兼有
有人真的需要修改IE上的格式。如果这些数字是INT\u MAX和INT\u MIN,那么If语句不是永远都是真的吗?@CharlieBurns我想是的。除非这些是在64位INT
架构上硬编码的32位限制(即OP使用64位INT
但需要32位限制),这似乎有点毫无意义。@WhozCraig,我在想。他有=所以INT_MIN和INT_MAX也是如此。尽管如此,我怀疑他是这么想的。不是相反吗?while(scanf(“%d”,&myInt)!=1){//TODO:scanf无法检索整数…}
@fvdacin:这一点很好。唯一的问题是,如果用户输入类似于12w3
的内容,12
将被转换并分配给myInt
,而scanf
将返回1,将w3
留在输入流中,以干扰下一次读取。