Reading a character with scanf_s

橙三吉。 提交于 2019-11-26 21:07:52

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).

According to msdn:

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 [...].

[1] https://msdn.microsoft.com/en-us/library/w40768et.aspx

Destructor

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);
}
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