Re: "'char' type is unsigned" seems to be broken
Re: "'char' type is unsigned" seems to be broken
- Subject: Re: "'char' type is unsigned" seems to be broken
- From: "email@hidden" <email@hidden>
- Date: Tue, 10 May 2005 14:07:01 -0400
David Leimbach wrote:
On 5/10/05, Clark Cox <email@hidden> wrote:
On 5/9/05, Brian Barnes <email@hidden> wrote:
Turning on "'char' type is unsigned" and running this code:
char txt[256];
unsigned char txt2[256];
strcpy(txt,"abc");
strcpy(txt2,"abc");
gets you:
warning: pointer targets in passing argument 1 of 'strcpy' differ in
signedness
For the first line.
"-funsigned-char" is in the compile arguments. Once again, I'm using
-fast, this might be breaking it.
Though the message is a bit off, I'd argue that this is actually
correct. Even if 'char' is unsigned, it is still a type that is
distinct from 'unsigned char'. 'unsigned char*' is not implicitly
convert-able to 'char*'. In fact, the following code:
Right and in the C standard "char", "unsigned char" and "signed char"
are distinct types.
Char is specified to behave in an implementation defined way.
"int" and "signed int", for example, are synonyms in the C standard
and do not require such warnings.
The compiler flag just sets what the implementation does for "char"
but doesn't change the semantics of the declaration "char" vs "signed
char" or "unsigned char".
Alright, that sounds reasonable. I suggest a better naming for this, as
I assumed (wrongly) that it was replacing all 'char' types with
'unsigned char', instead of just dealing with how the compiler handles
the math.
[>] Brian
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