What is the storage size of each address?

Question

I had expected each memory address contain 1-byte data. But on below code, I getting 4-byte hex value. Why this happened?

_{I'm aware of negative int representation.}

char num = -25;
char *ptr = &num;
printf("%02X", *ptr);

Output : FFFF FFE7

Expected : E7

If `I'm aware of negative int representation.` why are you asking the question? `*ptr` is being converted to a 4-byte `int`. If you don't want this, do this: `printf("%02X", (uint8_t)*ptr);`. — Fiddling Bits, Aug 13 '23 at 23:51
%X prints unsigned int as a hexadecimal number. x uses lower-case letters and X uses upper-case. — Jarod42, Aug 13 '23 at 23:51
@EtiennedeMartel: with C-variadic ellipsis, there are some convertions happening. see [cppreference](https://en.cppreference.com/w/cpp/language/variadic_arguments#Default_conversions) for details — Jarod42, Aug 13 '23 at 23:54
@EtiennedeMartel no, there is no UB. char is being converted to int as it as variadic function — 0___________, Aug 13 '23 at 23:55
You shouldn't make assumptions about the signedness of `char`. That is implementation-defined. `char` is a distinct type, separate from `signed char` and `unsigned char`, unlike other integral types that are implicitly signed. — paddy, Aug 13 '23 at 23:59
@paddy it does not matter in this case https://godbolt.org/z/vn5YYzajE as if it is unsigned the program will behave exactly the same — 0___________, Aug 14 '23 at 00:10

score 5 · Accepted Answer · answered Aug 13 '23 at 23:52

5

printf("%02X", *ptr);

*ptr is being converted to int and printed as int

You need to use the correct printf format:

printf("%02hhX", *ptr);

answered Aug 13 '23 at 23:52

0___________

1 Answers1