Size of characters in *argv[] when passed argument from command line

Question

I have simple program.

#include <stdio.h>
#include <string.h

int main(int argc, char *argv[])
{   
    for (int i = 0; i < strlen(argv[1]); ++i)
        printf("%x ", argv[1][i]);
    printf("\n");
}

I run it like

$ ./program 111
31 31 31

But when I run it like

$ ./program ●●●
ffffffe2 ffffff97 ffffff8f ffffffe2 ffffff97 ffffff8f ffffffe2 ffffff97 ffffff8f

Here each ● is should be encoded by 3 bytes (UTF-8): e2 97 8f, but looks like it is encoded by 3 unsigned. I don't understand where the ffffff comes from if sizeof(char) is always 1 byte.

pmg · Accepted Answer · 2019-04-27T12:26:51.713

4

printf() is a function accepting a variable number of arguments.

Any integer argument of a type shorter than int is automatically converted to type int.

Apparently, in your implementation, the "character" little-round-thing is composed of 3 chars, all with a negative value.

Try these

printf("%x ", (unsigned char)argv[1][i]);
printf("%hhx ", argv[1][i]); // thanks to Jonathan Leffler

edited Apr 27 '19 at 12:26

answered Apr 27 '19 at 12:02

pmg

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Rather, on his implementation, `char` has a sign, and all multi-codeunit codepoints only contain codeunits >127, thus negative on 8bit 2s-complement. – Deduplicator Apr 27 '19 at 12:08
@Deduplicator: you mean `strlen("little-round-thing") == 3` means `1` multi-codeunit codepoint or `3` characters? I changed my answer :) – pmg Apr 27 '19 at 12:13
yes, but that's not really the important part. – Deduplicator Apr 27 '19 at 12:15
1

Or `%hhx`as the conversion specification. – Jonathan Leffler Apr 27 '19 at 12:21
Or `#include ` and `printf ("%" PRIx8 " ", (uint8_t)argv[1][i]);` using exact-width types. – David C. Rankin Apr 27 '19 at 12:53

bruno · Answer 2 · 2019-04-27T12:29:55.763

I don't understand where the ffffff comes from if sizeof(char) is always 1 byte.

by definition sizeof(char) is 1, but '●' is not a char in the C sense and produces 3 char

your char are visibly signed (a char is a signed char by default in your case), each the input ● produce each 3 negative codes, because your char is converted to an int (32b in your case) and the format %x consider the argument unsigned you have these output

you will have the same output doing printf("%x", -30); -> ffffffe2

note to do for (int i = 0; i < strlen(argv[1]); ++i) is expensive for nothing, the length doesn't change, better to save it or to just do for (int i = 0; argv[1][i] != 0; ++i)

it was also better to check argc is at least 1 before to look into argv[1]

Thansk for valuable advice. – Kirill Bugaev Apr 27 '19 at 12:21 — Kirill Bugaev, Apr 27 '19 at 12:21

score 2 · Answer 3 · answered Apr 27 '19 at 12:20

UTF-8 codeunits for multi-codeunit codepoints (everything but ASCII) are all from 128 to 255, meaning outside the ASCII range.

printf() is a vararg function, and all the arguments passed to the vararg part (all but the format-string) are subject to the standard promotions.

As your implementation's bare char is 8bit signed 2s-complement, meaning the UTF-8 codeunit-value is negative, and between -1 and -128, after promotion you have an int with that value.

Then you lie to printf() by asserting it's an unsigned (%x is for unsigned int), and 2s-complement results in your Undefined Behavior printing a very big unsigned int.

You could get the right result by using %hhx, though strictly speaking you should cast the argument to unsigned char.

Size of characters in *argv[] when passed argument from command line

3 Answers3