Whitespace before %c specification in the format specifier of scanf function in C [duplicate]

问题

When I don\'t include white space between %d and %c specification in the format string of scanf() function in the following program, and give input during run-time as \"4 h\", then the output is \"Integer = 4 and Character= .

How exactly variable \"c\" takes the input in this case and what difference does it make if i include a white space between %d and %c specification ?

Code

#include <stdio.h>

int main()
{
    char c;
    int i;
    printf(\"Enter an Integer and a character:\\n\");
    scanf(\"%d %c\",&i,&c);
    printf(\"Integer = %d and Character = %c\\n\",i,c);
    getch();
}

回答1:

If you read the specification for scanf() carefully, most format specifiers skip leading white space. In Standard C, there are three that do not:

%n — how many characters have been processed up to this point
%[…] — scan sets
%c — read a character.

(POSIX adds a fourth, %C, which is equivalent to %lc.)

Input white-space characters (as specified by isspace) shall be skipped, unless the conversion specification includes a [, c, C, or n conversion specifier.

Adding the space between %d and %c means that optional white space is skipped after the integer is read and before the (not white space) character is read.

Note that literal characters in a format string (other than white space — for example, the X and Y in "X%dY") do not skip white space. Matching such characters does not count as a successful conversion either; they do not affect the return value from scanf() et al.

回答2:

A space before %c specifier in scanf instruct it to skip any number of white-spaces. In other words, read from standard input until and unless a non-white-space character or keyboard interrupt is found.

来源：https://stackoverflow.com/questions/36504135/whitespace-before-c-specification-in-the-format-specifier-of-scanf-function-in

标签

character

scanf

format-specifiers