I was just messing around with C and ran into this small problem. As you can see from my output I getting '╠' this character.
#include <stdio.h>
int main(void)
{
char c;
printf("Do you want to be X's or O's?\n");
scanf_s("%c", &c);
printf("You chose %c\n", c);
}
See program output
You are misusing scanf_s()
. Microsoft compilers may warn you to use their secure extensions (aka c11 annex k). But, be careful if you do so. scanf_s()
is not a direct replacement for scanf()
.
In this case you have to pass the size of the output buffer as an extra argument.
char c;
scanf_s("%c", &c, 1);
Having to put a 1 as the size of a single character may seem a bit pedantic. That's because %c
can read any number of character. %c
is just an alias for %1c
(a single character).
By knowing the buffer size scanf_s()
can prevent buffer overflow (a security risk).
Unlike scanf and wscanf, scanf_s and wscanf_s require the buffer size to be specified for all input parameters of type c, C, s, S, or string control sets that are enclosed in []. The buffer size in characters is passed as an additional parameter immediately following the pointer to the buffer or variable.
…
In the case of characters, a single character may be read as follows:
char c;
scanf_s("%c", &c, 1);
With scanf_s
you must supply a length [1] :
char c;
scanf_s("%c", &c, 1);
In the case of scanf_s
think of %c
to be a special shortcut for %1c
, which makes this more clear.
MSDNAA states [1]:
Unlike scanf and wscanf, scanf_s and wscanf_s require the buffer size to be specified for all input parameters of type
c
,C
,s
,S
[...].
The documentation of scanf_s says that:
In the case of characters, a single character may be read as follows:
char c;
scanf_s("%c", &c, 1);
So following should work ( See live demo here )
#include <stdio.h>
int main(void)
{
char i;
printf("Do you want to be X's or O's?\n");
scanf_s("%c",&i,1);
printf("You chose %c\n", i);
}
来源:https://stackoverflow.com/questions/41199504/reading-a-character-with-scanf-s