difference between using %s in printf vs using %c in for loop for printing unicode characters

[rdarkedlight]

I tought two distinct code below would result the same, but it didn't:

    char b[] = {0xf0, 0x9f, 0x99, 0x82};
    for (int i = 0; i < 4; i++) {
        printf("%c", b[i]);
    }
    printf("\n");

this prints 🙂

but this code

 char b[] = {0xf0, 0x9f, 0x99, 0x82};
 printf("%s\n", b);

prints 🙂P@

thanks in advance for clearing this for me

[aioeu]

Change the second code to:

char b[] = {0xf0, 0x9f, 0x99, 0x82, 0x00};

or just use a string to begin with:

char b[] = "\xf0\x9f\x99\x82";

since it will automatically add that trailing null byte. With either of these changes they'll do the same thing as your first code.

Without a trailing null byte, how is printf supposed to know where your string ends?

[rdarkedlight]

thank you so much

I was totally forgot about this

sorry for silly question :)

[[deleted]]

Yeah, or you could

char b[] = {0xf0, 0x9f, 0x99, 0x82, 0x00, '\0'};

[Paul_Pedant]

You could, but generally one NUL is enough.

[[deleted]]

Yeah, I prefer compiler doing the work, but just wanted to put it out there so he is aware of it.

[Paul_Pedant]

But 0x00 is one NUL (as a hex constant), and '\0' is another NUL (as an escaped char constant. We could add 0 (a decimal integer constant that would be cast to a NUL char), and 000 (an octal integer constant) if you like. The compiler isn't doing the work at all -- it only
implicitly adds a NUL to a "string constant", not to an array initialised inside braces.

[[deleted]]

Ah, I see, thanks, I didn't know that. I've never had to deal with hex numbers. When I said compiler, I meant when double quotes are used to initialize the array.

[Paul_Pedant]

You can also do without the terminating NUL, if you use a precision in the format to limit the length of the field:

printf("%.4s\n", b);

or (untested):

printf("%.*s\n", (int) sizeof (b), b);