将Unicode存储在char中

Question

I have a program I made to test I/O from a terminal:

#include <stdio.h>
int main()
{
    char *input[100];
    scanf("%s", input);
    printf("%s", input);
    return 0;

}

It works as it should with ASCII characters, but it also works with Unicode characters and emoji.

Why is this?

Answer 1

Your code works because the input and output stream have the same encoding, and you do not do anything with c .

Basically, you type something, which is converted into a sequence of bytes, which are then stored in c , then you send back that sequence of bytes to stdout which convert them back to readable characters.

As long as the encoding and decoding process are compatible, you will get the "expected" result.

Now, what happens if you try to use standard "string" C functions? Let's assume you typed "♠Hello" in your terminal, you will get the expected output but:

strlen(c) -> 8
c[0] -> Some strange character
c[3] -> H

You see? You may be able to store whatever you want in a char array, it does not mean you should. If you want to deal with extended character sets, use wchar_t instead.

Answer 2

You're probably running on Linux, with your terminal set to UTF-8 so scanf produces UTF-8, and printf can output it. UTF-8 is designed such that char[] can store it. I explicitly use char[] and not char because non-ASCII characters need more than one byte.

Answer 3

Your program is undefined as it has undefined behavior.

scanf("%s", input);

expects a pointer to string, but

char *input[100];

input is pointer to pointer to char , char * .

Your program may work because the buffer you pass to scanf is of sufficient size to store unicode character and a characters you pass don't have a NULL byte in between them, but it may not work as well because the implementation of C on your (and any other) machine is allowed to do anything in cases of UB.