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why still use C?

no this is no trollposting and please don't get it wrong but iam very
curious why people still use C instead of other languages especially C++.

i heard people say C++ is slower than C but i can't believe that. in pieces
of the application where speed really matters you can still use "normal"
functions or even static methods which is basically the same.

in C there arent the simplest things present like constants, each struct and
enum have to be prefixed with "struct" and "enum". iam sure there is much
more.

i don't get it why people program in C and faking OOP features(functi on
pointers in structs..) instead of using C++. are they simply masochists or
is there a logical reason?

i feel C has to benefit against C++.

--
cody

[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
--
comp.lang.c.mod erated - moderation address: cl**@plethora.n et
Nov 13 '05
687 23922
CBFalconer wrote:
Keith Thompson wrote:

... snip ...

The macros look fine, but the casts are still unnecessary and
potentially dangerous. Why not just

#define NEW(type) \
(malloc(sizeof( type)))


To make it look more like a C++ new, why not:

#define NEW(var) (var = malloc(sizeof *var))
or
#define NEW(var, n) (var = malloc((n) * sizeof *var))


Maybe because that is not how _C++_ new works? You are thinking of a
different (nice) language.

Kurt Watzka
Nov 13 '05 #411
Dan Pop wrote:
Which prompts for the canonical c.l.c example
of why casting void pointers is evil:

int main(int arc, char* argv[]) {
char* p = (char*)malloc(1 0);
return 0;
}

is a perfect example of undefined behavior in C89.
Drop the cast and a diagnostic is required because an int value
(malloc's return value) is implicitly converted to a pointer value.
You are confused.
cat malloc89.c int main(int arc, char* argv[]) {
char* p = (char*)malloc(1 0);
return 0;
}
gcc -Wall -std=c89 -pedantic -O2 -o malloc89 malloc89.c malloc89.c: In function `main':
malloc89.c:2: warning: implicit declaration of function `malloc'
malloc89.c:2: warning: unused variable `p' ./malloc89
1. A good C compiler will issue a diagnostic message
if malloc has not been declared in the scope where it is used.

2. The C compiler will assume that malloc returns type int
if malloc has not breed declared in the scope where it is used.

3. The link editor will link in malloc which returns a void*.

4. The C compiler is going to apply the conversion
from int to char* in any case.
cat malloc89.c int main(int arc, char* argv[]) {
char* p = malloc(10);
return 0;
}
gcc -Wall -std=c89 -pedantic -O2 -o malloc89 malloc89.c

malloc89.c: In function `main':
malloc89.c:2: warning: implicit declaration of function `malloc'
malloc89.c:2: warning: initialization makes pointer from integer
without a cast
malloc89.c:2: warning: unused variable `p'

The "initializa tion makes pointer from integer without a cast" warning
is cryptic and misleading.

Omitting the explicit cast of the void* value returned by malloc
can only be viewed as an attempt to compensate for a deficiency
in an inferior C compiler.
Stop cobbling your C code. Use a better C compiler.

Nov 13 '05 #412
> I just tried the same test with the first 81 C modules I could find,
all of which are part of the same C89 program, and most of which have
been modified by me personally at least once, and reviewed by me
multiple times. This set was significantly cleaner with respect to C++
compilation than yours:

Yes: 79
No: 2


I've started using C99 features a while ago, so for me it's now
Yes: 0
No: all

But the older code suffers mostly from void *
Nov 13 '05 #413
"E. Robert Tisdale" <E.************ **@jpl.nasa.gov > writes:
Dan Pop wrote:
Which prompts for the canonical c.l.c example
of why casting void pointers is evil:

int main(int arc, char* argv[]) {
char* p = (char*)malloc(1 0);
return 0;
}

is a perfect example of undefined behavior in C89.
Drop the cast and a diagnostic is required because an int value
(malloc's return value) is implicitly converted to a pointer value.
You are confused.
> cat malloc89.c

int main(int arc, char* argv[]) {
char* p = (char*)malloc(1 0);
return 0;
}
> gcc -Wall -std=c89 -pedantic -O2 -o malloc89 malloc89.c

malloc89.c: In function `main':
malloc89.c:2: warning: implicit declaration of function `malloc'
malloc89.c:2: warning: unused variable `p'
> ./malloc89


1. A good C compiler will issue a diagnostic message
if malloc has not been declared in the scope where it is used.


Not all C compilers are good. No diagnostic is required.
2. The C compiler will assume that malloc returns type int
if malloc has not breed declared in the scope where it is used.
Right.
3. The link editor will link in malloc which returns a void*.
Right.
4. The C compiler is going to apply the conversion
from int to char* in any case.


No. If there's a prototype in scope, the value returned by malloc()
will be treated as void*, and will be implicitly (if there's no cast)
or explicitly (if there is) converted from void* to char*. If there's
no prototype in scope, the compiler will simply assume that the value
returned by malloc() is of type int, and, in the presence of a cast,
will apply some (possibly non-trivial) conversion from int to char*.
On many systems, this happens to work (for example, if int and char*
happen to be of the same size, and the calling convention returns
integers and pointers the same way).

--
Keith Thompson (The_Other_Keit h) ks*@cts.com <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://www.sdsc.edu/~kst>
Schroedinger does Shakespeare: "To be *and* not to be"
Nov 13 '05 #414
"E. Robert Tisdale" wrote:
....
> cat malloc89.c int main(int arc, char* argv[]) {
char* p = (char*)malloc(1 0);
return 0;
}

.... 4. The C compiler is going to apply the conversion
from int to char* in any case.


Without a valid prototype in scope, it converts fom int to char*. With a
valid prototype in scope, it converts void* to char*, which can be a
very different conversion (on the other hand, it can also be the exact
same conversion).
> cat malloc89.c

int main(int arc, char* argv[]) {
char* p = malloc(10);
return 0;
}
> gcc -Wall -std=c89 -pedantic -O2 -o malloc89 malloc89.c

malloc89.c: In function `main':
malloc89.c:2: warning: implicit declaration of function `malloc'
malloc89.c:2: warning: initialization makes pointer from integer
without a cast
malloc89.c:2: warning: unused variable `p'

The "initializa tion makes pointer from integer without a cast" warning
is cryptic and misleading.


Unless you're suggesting that the compiler special-case the standard
library functions, I don't see any alternative way to handle that code.
It's warned you about the implicit declaration, then it attempts to use
it. In the process, an int is implicitly converted to a pointer type,
and that's pretty much the only other feature of the code on that line
that it could use as a basis for giving a warning message.

--
James Kuyper
MODIS Level 1 Lead
Science Data Support Team
(301) 352-2150
Nov 13 '05 #415
"E. Robert Tisdale" wrote:

Dan Pop wrote:
Which prompts for the canonical c.l.c example
of why casting void pointers is evil:

int main(int arc, char* argv[]) {
char* p = (char*)malloc(1 0);
return 0;
}

is a perfect example of undefined behavior in C89.
Drop the cast and a diagnostic is required because an int value
(malloc's return value) is implicitly converted to a pointer value.

You are confused.


[ Trollsdale nonsense deleted ]
Just the very idea of Trollsdale saying Dan Pop is confused makes my
brane hurt.

Brian Rodenborn
Nov 13 '05 #416
Keith Thompson wrote:
E. Robert Tisdale writes:
4. The C compiler is going to apply the conversion
from int to char* in any case.

No. If there's a prototype in scope, the value returned by malloc()
will be treated as void*, and will be implicitly (if there's no cast)
or explicitly (if there is) converted from void* to char*. If there's
no prototype in scope, the compiler will simply assume that the value
returned by malloc() is of type int, and, in the presence of a cast,
will apply some (possibly non-trivial) conversion from int to char*.
On many systems, this happens to work (for example, if int and char*
happen to be of the same size, and the calling convention returns
integers and pointers the same way).


Can you show us an example of a "system"
on which this happens *not* to work?

Nov 13 '05 #417
"E. Robert Tisdale" wrote:

Keith Thompson wrote:
E. Robert Tisdale writes:
[Re: the following code]

int main(int arc, char* argv[]) {
char* p = malloc(10);
return 0;
}

....
4. The C compiler is going to apply the conversion
from int to char* in any case.

No. If there's a prototype in scope, the value returned by malloc()
will be treated as void*, and will be implicitly (if there's no cast)
or explicitly (if there is) converted from void* to char*. If there's
no prototype in scope, the compiler will simply assume that the value
returned by malloc() is of type int, and, in the presence of a cast,
will apply some (possibly non-trivial) conversion from int to char*.
On many systems, this happens to work (for example, if int and char*
happen to be of the same size, and the calling convention returns
integers and pointers the same way).


Can you show us an example of a "system"
on which this happens *not* to work?


Well, I can't even get it to compile with the compiler I use most
frequently, the SGI compiler for IRIX 6.5. It complains

"malloc89.c ", line 2: error(1140): a value of type "int" cannot be used
to
initialize an entity of type "char *"
char* p = malloc(10);

I'm not sure what it said in the C89 standard, but this is correct in
terms of the C99 standard. 6.7.8p11 says that for initialization: "the
same type constraints and conversions as for simple assignment apply". A
left operand of type char* and a right operand that is an int which is
not a null pointer constant, doesn't match any of the options listed in
6.5.16.1p1, the constraints section for simple assignment.

However, even if the compiler did allow implicit conversion of 'int' to
char*, I wouldn't expect this code to work. In the 64 bit mode that I
normally use it in, 'void*' and size_t are both 8 bytes, while int is 4
bytes. I don't know any of the details of how SGI C handles function
calls, but malloc() is going to be expecting to recieve four more bytes
than main() thinks it needs to send; and malloc() is going to return
four more bytes than main() expects to see. I don't see how that could
possibly work.
Nov 13 '05 #418
On Fri, 24 Oct 2003 12:45:10 -0700, "E. Robert Tisdale"
<E.************ **@jpl.nasa.gov > wrote:
Keith Thompson wrote:
E. Robert Tisdale writes:
4. The C compiler is going to apply the conversion
from int to char* in any case.

No. If there's a prototype in scope, the value returned by malloc()
will be treated as void*, and will be implicitly (if there's no cast)
or explicitly (if there is) converted from void* to char*. If there's
no prototype in scope, the compiler will simply assume that the value
returned by malloc() is of type int, and, in the presence of a cast,
will apply some (possibly non-trivial) conversion from int to char*.
On many systems, this happens to work (for example, if int and char*
happen to be of the same size, and the calling convention returns
integers and pointers the same way).


Can you show us an example of a "system"
on which this happens *not* to work?


Large model MS-DOS. sizeof(int)==2 && sizeof(char *)==4. Functions
return int in AX, void * in ES:AX

If you fail to include the proper prototype, but explicitly cast the
return value, the compiler converts the int (in AX) to a pointer by
assuming segment 0. For example, if malloc returns 1234:5678, the
pointer would be assigned the value 0000:5678.

The cast tells the computer, "Shut up, I know what I'm doing." Even
when you don't. I learned this the hard way, back around 1988.

Regards,

-=Dave
--
Change is inevitable, progress is not.
Nov 13 '05 #419
In article <3F************ **@jpl.nasa.gov >,
"E. Robert Tisdale" <E.************ **@jpl.nasa.gov > wrote:
Keith Thompson wrote:
E. Robert Tisdale writes:
4. The C compiler is going to apply the conversion
from int to char* in any case.

No. If there's a prototype in scope, the value returned by malloc()
will be treated as void*, and will be implicitly (if there's no cast)
or explicitly (if there is) converted from void* to char*. If there's
no prototype in scope, the compiler will simply assume that the value
returned by malloc() is of type int, and, in the presence of a cast,
will apply some (possibly non-trivial) conversion from int to char*.
On many systems, this happens to work (for example, if int and char*
happen to be of the same size, and the calling convention returns
integers and pointers the same way).


Can you show us an example of a "system"
on which this happens *not* to work?


Several versions of the Metrowerks C compiler.
Nov 13 '05 #420

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