我有一个问题，因为当我这样做时我的代码不工作：
#include <stdio.h>
int main() {
int age;
printf("Enter your age: ");
scanf("%d", &age);
while(age == (what do i put here)) {
printf("\nYou have missed something\n");
printf("Enter your age: ");
scanf("%d", &age);
}
}
做类似的事情
while(age == (put something here)) {
没有意义，因为如果
scanfage不是检查
agescanf#include <stdio.h>
int main( void )
{
int age;
//repeat until input is valid
for (;;) //infinite loop, equivalent to while(1)
{
printf( "Enter your age: " );
if ( scanf( "%d", &age ) != 1 )
{
int c;
printf( "Invalid input! Try again.\n" );
//discard remainder of line
do
{
c = getchar();
} while ( c != EOF && c != '\n' );
continue;
}
//input was ok, so break out of the infinite loop
break;
}
printf( "The input was successfully converted to %d.\n", age );
}
这个程序有以下行为：
Enter your age: abc
Invalid input! Try again.
Enter your age: 25
The input was successfully converted to 25.
但是，这个解决方案并不完美，因为它接受输入
25abc25Enter your age: 25abc
The input was successfully converted to 25.
在这种情况下，最好拒绝输入并提示用户输入新的输入。
另一个问题是
scanfEnter your age:
abc
Invalid input! Try again.
Enter your age:
25
The input was successfully converted to 25.
这是因为当使用
scanf%d在处理基于行的用户输入时，通常最好不要使用
scanffgetsstrtol这里是一个示例程序，它修复了上述所有问题，并添加了一些额外的输入验证：
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>
#include <limits.h>
#include <errno.h>
//forward declaration
int get_int_from_user( const char *prompt );
int main( void )
{
int age;
age = get_int_from_user( "Enter your age: " );
printf( "The input was successfully converted to %d.\n", age );
}
//This function will attempt to read one integer from the user. If
//the input is invalid, it will automatically reprompt the user,
//until the input is valid.
int get_int_from_user( const char *prompt )
{
//loop forever until user enters a valid number
for (;;)
{
char buffer[1024], *p;
long l;
//prompt user for input
fputs( prompt, stdout );
//get one line of input from input stream
if ( fgets( buffer, sizeof buffer, stdin ) == NULL )
{
fprintf( stderr, "Unrecoverable input error!\n" );
exit( EXIT_FAILURE );
}
//make sure that entire line was read in (i.e. that
//the buffer was not too small)
if ( strchr( buffer, '\n' ) == NULL && !feof( stdin ) )
{
int c;
printf( "Line input was too long!\n" );
//discard remainder of line
do
{
c = getchar();
if ( c == EOF )
{
fprintf( stderr, "Unrecoverable error reading from input!\n" );
exit( EXIT_FAILURE );
}
} while ( c != '\n' );
continue;
}
//attempt to convert string to number
errno = 0;
l = strtol( buffer, &p, 10 );
if ( p == buffer )
{
printf( "Error converting string to number!\n" );
continue;
}
//make sure that number is representable as an "int"
if ( errno == ERANGE || l < INT_MIN || l > INT_MAX )
{
printf( "Number out of range error!\n" );
continue;
}
//make sure that remainder of line contains only whitespace,
//so that input such as "6abc" gets rejected
for ( ; *p != '\0'; p++ )
{
if ( !isspace( (unsigned char)*p ) )
{
printf( "Unexpected input encountered!\n" );
//cannot use `continue` here, because that would go to
//the next iteration of the innermost loop, but we
//want to go to the next iteration of the outer loop
goto continue_outer_loop;
}
}
return l;
continue_outer_loop:
continue;
}
}
这个程序有以下行为：
Enter your age:
Error converting string to number!
Enter your age: test
Error converting string to number!
Enter your age: 25abc
Unexpected input encountered!
Enter your age: 25
The input was successfully converted to 25.
如何知道用户是否没有输入任何内容？
简短的回答是：通过检查
scanf成功
scanfscanf在代码中这将是：
int items_scanned = scanf("%d", &age);
if (items_scanned == EOF)
{
// Input error. Could not read any input
// Add error handling
}
else if (items_scanned == 1)
{
// All good. Scanned one item, i.e. age now contains an integer
printf("Age is %d\n", age);
}
else
{
// The input did not represent a valid integer
// Add error handling
}
这就提出了一个问题：“错误处理”部分怎么办？
当
scanferrnoexit当
scanf1abgetcharscanf还有：如何继续阅读直到
age使用某种循环。它可以通过许多不同的方式来完成。下面是一个非常简单的方法。
while(1)
{
int items_scanned = scanf("%d", &age);
if (items_scanned == EOF)
{
// Input error. Could not read any input
// Add error handling
exit(1);
}
else if (items_scanned == 1)
{
// All good. Scanned one item, i.e. age now contains an integer
// Stop the loop
break;
}
else
{
// The input did not represent a valid integer
// Add error handling
// Remove the next input character
int tmp = getchar();
if (tmp == EOF)
{
// Input error. Could not read any input
// Add error handling
exit(1);
}
}
}
// When arriving here, age is a valid integer from the user
printf("Age is %d\n", age);
就是说...
获取用户输入并不像听起来那么容易。在编写任何代码之前，需要进行一些设计工作来指定程序应认为哪种输入有效。
上面的小程序将例如接受“abc12”作为整数值 12。程序设计者必须决定是否可以接受。
同样，输入“3.14”作为整数值3。程序设计者必须决定是否可以接受。
scanf%dscanfscanfscanf