Printf Hex Formatting Leads Wrong Output

Question

I have following C code.

char a[] = "\x7f\x80";
printf("0x%02x\n",a[0]);
printf("0x%02x",a[1]);

It should print,

0x7f
0x80

However I'm getting following ?

0x7f
0xffffff80

What changes I have to make to get the output 0x7f 0x80?

Thanks,

Thanks a lot ! I actually tried printf("0x%02x\n",(unsigned)a[0]) . — Dev.K., Apr 28 '16 at 09:47
As Michael said. It is the sign extension during type promotion. See http://stackoverflow.com/questions/9715984/printf-type-promotion-and-sign-extension — Dummy00001, Apr 28 '16 at 09:48

score 3 · Accepted Answer · answered Apr 28 '16 at 09:49

3

Use the correct types and conversion specifiers:

unsigned char a[] = "\x7f\x80";
printf("0x%02hhx\n",a[0]);
printf("0x%02hhx",a[1]);

Conversion specifier x requires an unsigned type, and the length modifier hh is used for unsigned char.

answered Apr 28 '16 at 09:49

2501

1 Answers1