scanf（）如何读取未格式化的输入？

Question

Explain the output in this case when string is not formatted but still scanf() is able to read integers from an input string.

#include <stdio.h>
int main(void)
{
int n;
while (scanf("%d", & n))
    printf("%d\n", n);
return 0;
}

Example Input:

1 2 3 4 5 54 34 abcd

Output:

1
2
3
4
5
54
34

Input is so messy but output is still clean. How scanf() is working?

Answer 1

For the "%d" specifier it will ignore whitespace characters, and then when it reaches the invalid characters non-digit it will return 0 , for each numeric value it returns 1 which is the number of specifiers successfully matched.

Read here for more information.

Answer 2

scanf("%d", &n)对于整数将返回1（成功读取的值数），并且在遇到非数字时将失败，并且会破坏while

Answer 3

scanf ignore the blank characters such as:

\t(ASCII:0x09) 
\n(ASCII:0x0a)
blank(ASCII:0x20)

ref: scanf

Whitespace character: the function will read and ignore any whitespace characters encountered before the next non-whitespace character (whitespace characters include spaces, newline and tab characters -- see isspace). A single whitespace in the format string validates any quantity of whitespace characters extracted from the stream (including none).

Answer 4

In c programming language data type is strictly defined . You have to enter proper data in scanf() function . In you enter any wrong type then it will return 0 that means false . Try to do like this . The mechanism will be clear to you .

#include <stdio.h>
int n,m;
int main()
{
    do{
        m = scanf("%d",&n);
        printf("Legality: %d  Value: %d\n",m,n);
    }while(m);
    return 0;
}