scanf and its left white-space character problem

C c

创建时间:2019-10-04 14:13

字数:618 阅读:

Several tricks of scanf

Several tricks of scanf

The scanf function is used to scan input from the stdin stream according to the format specification. In some cases, the collocation of other input functions(e.g., gets) will cause some problems. Assuming we will read below contents from the stdin to the program:

3
cap
to
cat

A common method is taking advantage of the scanf function to read the integer firstly; thereafter, read the string for each line. The codes are shown below:

int main()
{
    int n;
    scanf("%d", &n);
    printf("%d\n", n);
    while(n--)
    {
        char str[100];
        gets(str);
        printf("%s\n", str);
    }
    /* results:
    3

    cap
    to
    */
}

Obviously, there is an unexpected \n after the first scanf. The reason is the scanf("%d") will stop reading when it encountered the WSC(white-space characters, i.e. white-space, tab, newline). The first input(i.e. 3) in the input file can be presented as 3\n. Therefore, the scanf has only read the 3, but the \n has been left in the stream buffer. However, the gets() function read the string from the buffer until encountering the \n. Therefore, the first gets read the \n which left in the first scanf.

Consume the left ‘\n’

In conclusion, to get the correct input, we need to clean the left \n in the buffer. There are two methods to do this:

int main()
{
    int n;
    scanf("%d", &n);
    getchar();  // use getchar() to consume the left '\n'
    /* 
    char ch;
    scanf("%c", ch); // or use scanf to consume the left '\n'
                     // scanf("%c") or scanf("%d") can recognize the WSC
    */
    printf("%d\n", n);
    while(n--)
    {
        char str[100];
        gets(str);
        printf("%s\n", str);
    }
}

fflush(stdin) cannot clean the buffer in Linux system because it is not a standard function and supported by all compiler.

Ignore the left ‘\n’

Alternatively, we can force the scanf to ignore the left ‘\n’:

int main()
{
    int n;
    scanf("%d", &n);
    printf("%d\n", n);
    while(n--)
    {
        char str[100];
        scanf("%s", str);
        printf("%s\n", str);
    }
}

The core idea is that the scanf("%s") will match a sequence of non-white-space characters. In other words, it can skip the left \n.

int main()
{
    int n;
    scanf("%d\n", &n); // equal to scanf("%d ", &n);
    printf("%d\n", n);
    while(n--)
    {
        char str[100];
        gets(str);
        printf("%s\n", str);
    }
}

An alternative is adding the WSC to the format string of scanf.

Note: the meaning of the WSC in the format string of scanf is to explicitly read and ignore as many whitespace characters as it can[1], instead of expecting a \n new line sign. Therefore, the scanf("%d\n") means after reading an integer, it will continue to read characters, discarding all whitespace until it sees a non-whitespace character on the input. That non-whitespace character will be left as the next character to be read by an input function.

Therefore, the first scanf("%d\n") will read the integer 3 and discard the \n, then encounter the next non-white-space character(i.e. c) and stop. But the c will be left to the buffer to the next input function(e.g. gets()).

Note: gets() is unsafe because it is potential to cause overflow for the string. Therefore, we can try to use fgets(str,MAXLEN,stdin);

References and relative reading:

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