Why are there garbled characters in this C program output?

Question

#include <stdio.h>
#include <locale.h>
#include <wchar.h>

int main() {
//    setlocale("LC_ALL","");
    unsigned char utf[]={0xe4,0xb8,0x80,0x0a};
    printf("%s",utf);
    return 0;
}

The first four bytes of output are correct. The second line in the console is not expected

`%s` expects a nullterminated string. `utf` is not. – Raildex Mar 27 '22 at 10:15 — Raildex, Mar 27 '22 at 10:15

score 3 · Accepted Answer · answered Mar 27 '22 at 10:18

3

Your array is missing the null terminator strings require, so printf keeps on printing bytes beyond the end of the array until it happens upon a null byte. (Or when your program crashes due to an out of bounds access.)

Add the null byte:

unsigned char utf[]={0xe4,0xb8,0x80,0x0a,0x00};

answered Mar 27 '22 at 10:18

AKX

152,115
15
115
172

3

Alternatively: `unsigned char utf[] = "\xe4\xb8\x80\x0a";` will also require 5 bytes. – pmg Mar 27 '22 at 10:19
thank you bro,It seems that I still need to learn more c language details – user8911572 Mar 27 '22 at 10:20

score 1 · Answer 2 · answered Mar 27 '22 at 10:26

The format %s expects a pointer to a string as the corresponding argument. That is the sequence of outputted characters shall be ended with the terminating zero character '\0'.

This array

unsigned char utf[]={0xe4,0xb8,0x80,0x0a};

does not contain a string. So you need to specify explicitly how many characters you are going to output. For example

printf("%.*s", 4, utf);

Why are there garbled characters in this C program output?

2 Answers2