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438

answers:

2

Simple question: Can scanf read/accept a "small integer" into an unsigned char in ANSI C?

example code un_char.c:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
    unsigned char character;

    scanf("%hhu", &character);

    return EXIT_SUCCESS;
}

Compiled as:

$ gcc -Wall -ansi -pedantic -o un_char un_char.c
un_char.c: In function ‘main’:
un_char.c:8: warning: ISO C90 does not support the ‘hh’ gnu_scanf length modifier

hh isn't supported by ISO C90. So what scanf conversion can be used in this situation?

+3  A: 

Read it into an unsigned short/int and do some range checking after if you need to.

nos
Unsigned short would be the better datatype as that's 16 bits...just knock off the upper 8 bits.. :)
tommieb75
@tommieb75: There's no guarantee that `unsigned short` is 16 bits. There's no guarantee that `unsigned char` is 8 bits either.
jamesdlin
@jamesdlin: I agree...due to the different processors/compilers/linkers/runtime environments... :O :)
tommieb75
@jamesdlin: I believe there is a guarantee that a char is at least 8 bits, which is enough
Matt Joiner
@Matt Joiner: Yes, but my point was that you can't just knock off the upper 8 bits (as tommieb76 suggested) and that you should do range-checking (as nos suggested).
jamesdlin
+3  A: 

No: C89 (C90) does not support '%hhu' to read a string of digits into an unsigned char. That is a feature in C99.

You would have to read into an unsigned integer ('%u') or unsigned short ('%hu') and then check that the result is with the range of an unsigned char.

Jonathan Leffler