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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

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++.


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comp.lang.c.mod erated - moderation address: cl**@plethora.n et
Nov 13 '05
687 23922
Chris Hills <ch***@phaedsys .org> wrote in message news:<cl******* *********@pleth ora.net>...

Being power hungry is not the reason why an 8051 does not (normally) run
at 2Ghz.... Many 8051's work in areas where power is not a factor. They
run code from 2Kbytes to 16Mega bytes.

Yes, the usual reasons for slower clock speeds are (in my business):
- Slower rated chips are cheaper
- You don't usually need that speed
- Heat dissipation becomes a problem at high speeds
There are also many 32 and 64 bit embedded systems running that have
megabytes of code.... virtually any car radio or engine management
system. I did have the figures for the line of code in a car radio. It
was in the 100's of thousand lines! It surprised me. Engine management
systems have the odd megabyte of code.

and when things are programmed in assembler, an 8K processor can still
hold 6-8 KLOC!.

Keith Derrick (ACCU member)
PS Don't bother sending mail to my posting address, it's a spam trap.
comp.lang.c.mod erated - moderation address: cl**@plethora.n et
Nov 13 '05 #241

"Fergus Henderson" <fj*@cs.mu.oz.a u> wrote in message
news:cl******** ********@pletho ra.net...
Jerry Feldman <ga********@blu .org> writes:
Keith Thompson <ks*@cts.com> wrote:
C++ has never been a strict superset of any version of C. C++ has
several keywords that are not reserved in C; that alone makes prevents
it from being a superset.
I agree that "C++ has never been a strict superset of any version of C",
but I disagree with your logic. A superset can define new keywords (and
comment operators).

A strict superset can't, if those keywords have names which are not
already reserved for use by the implementation, because adding new
keywords will invalidate existing C programs that happen to use those
keywords as identifiers.

I don't know why people mold terms to fit things how they would like to see
C++ is not a superset, period. ``strict superset'' throw it out of the
When something is a ``superset'' of something else, that something else is
considered to be a ``subset''.
Now inorder for C to be a subset of C++, you would be able to compile _any_
C program with a C++ compiler. This can't be done, because C is not a subset
of C++, and that is because C++ is not a superset of C. It is a
``derivative'' of C. Nothing more and nothing less.
Adding new comment operators can also change the meaning of existing code.
Consider the following program:

int main() {
? "fail\n" : "pass\n");
return 0;

On a conforming C89 implementation, this will print "pass",
but on a C++ or C99 implementation, it will print "fail".
However there are many constructs in standard C that
are illegal in C++. One example that comes to mine is:
In C:
void foo(void);
In C++ this will cause a syntax error because C++ requires fully
prototyped functions.
That's not correct either. C++ allows "void foo(void);".

Fergus Henderson <fj*@cs.mu.oz.a u> | "I have always known that the

pursuit The University of Melbourne | of excellence is a lethal habit"
WWW: <http://www.cs.mu.oz.au/~fjh> | -- the last words of T. S. Garp.
comp.lang.c.mod erated - moderation address: cl**@plethora.n et

Nov 13 '05 #242
On Mon, 13 Oct 2003 22:18:56 +0200 in comp.std.c, Alexander
Terekhov <te******@web.d e> wrote:

th*@cs.ucr.e du wrote:
C is pretty much, but not quite, a sublanguage of C++. C programmers
who don't use the non-C++ features of C are programming in C++ whether
they claim to or not. They are restricting themselves to an older,
more established, more easily learned, and more easily implemented
subset of C++. But they are writing in C++ --- non-idiomatic C++, but
C++ nevertheless. AFAIK, a C++ compile is free to generate the same
code for those programs as would a C compiler, so there is no
intrinsic difference in performance.

In C without exceptions (stuff like MS or DEC/HP SEH extensions), all
functions are kinda "throw()".

I take exception to the object of that last block! ;^>
In idiomatic C, each function is like a try block, each return is
like a throw, and each call is like a catch.

Thanks. Take care, Brian Inglis Calgary, Alberta, Canada
Br**********@CS i.com (Brian dot Inglis at SystematicSw dot ab dot ca)
fake address use address above to reply
Nov 13 '05 #243
On Wed, 15 Oct 2003 00:21:21 +0200 in comp.std.c, "cody"
<do************ *********@gmx.d e> wrote:
The type of 'a' is int. The C99 standard, section paragraph 2,

An integer character constant is a sequence of one or more
multibyte characters enclosed in single-quotes, as in 'x'.

Was is the same in other versions of C besides C99?

As far back as K&R1 -- common C idiom was 'ab' (16 bit) or 'abcd'
(32 bit) to handle packed chars -- undefined behaviour as of C89
-- but still allowed and works?

Thanks. Take care, Brian Inglis Calgary, Alberta, Canada
Br**********@CS i.com (Brian dot Inglis at SystematicSw dot ab dot ca)
fake address use address above to reply
Nov 13 '05 #244
Arthur J. O'Dwyer wrote:

[C89/C99/C++ tests]
Now, what silly mistakes have I made *this* time? :-)

Failure to use the standard macros? ;-)

#include "stdio.h"
int main(void)
#ifdef __cplusplus
i = 2;
#elif defined __STDC_VERSION_ _ && __STDC_VERSION_ _ == 199901L
i = 1;
i = 0;
puts(i == 2 ? "C++" : i == 1 ? "C99" : "C89");
return 0;


Nov 13 '05 #245
Jirka Klaue wrote:
#include "stdio.h"
int main(void)
#ifdef __cplusplus
i = 2;
#elif defined __STDC_VERSION_ _ && __STDC_VERSION_ _ == 199901L
i = 1;
i = 0;
puts(i == 2 ? "C++" : i == 1 ? "C99" : "C89");
return 0;

A C89 implementation is allowed to define a __cplusplus macro, AFAICT.
(You also classify C0X as C89).

Nov 13 '05 #246
On Wed, 15 Oct 2003 23:42:23 +0000 (UTC), Richard Heathfield
<do******@addre ss.co.uk.invali d> wrote:
cody wrote: [...]
This seems to be true for both C and C++ compilers (I tried both).

It really is time to stop guessing at this stuff. The C language is defined
by an international standard, not by your apparently random guesses.

"It works on my compiler" is never a good argument in c.s.c.


Change is inevitable, progress is not.
Nov 13 '05 #247
"Mark Gordon" <sp******@fla sh-gordon.me.uk> schrieb im Newsbeitrag
news:2003101523 2209.128f9a60.s p******@flash-gordon.me.uk...
On Wed, 15 Oct 2003 22:17:35 +0200
"cody" <do************ *********@gmx.d e> wrote:

Please don't trim the attributions since it prevents people from seeing
who said what.
int test(int size)
int *new = malloc(size); /* two non-C++isms in this line */
return 1 //*
-1 + 1; /* 0 in C99, 1 in C++ */

is the missing semicolon at the end of a block allowed in C?

A semicolon is required to terminate every C statement, however in C99
there is no missing semicolon in the above.

HINT: C99 added a form of // commenting.

int test(int size)
int *new = malloc(size); /* two non-C++isms in this line */

return 1 //* <-- the c version misses a semicolon here,
additionally //* will imho produce no valid comment in c.

-1 + 1; /* 0 in C99, 1 in C++ */


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Nov 13 '05 #248
> > int i = sizeof (int);
// int j = sizeof int; // invalid in both C and C++
int k = sizeof i;
int l = sizeof (i);

The form of sizeof that gets a type as parameter has to use parantheses,
while the sizeof that gets non-expressions as parameter do not need
Wrong. In your example, i is an unary-expression, not a non-expression.

I don't know why i wrote non-expression. My example clearly showed how to
use sizeof:

The form of sizeof that gets a type (NON-EXPRESSION) as parameter has to use
while the sizeof that gets expressions (NON-TYPES) as parameter do not need

Is it now correctly stated?


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Nov 13 '05 #249
"Richard Heathfield" <do******@addre ss.co.uk.invali d> schrieb im Newsbeitrag
news:bm******** **@titan.btinte rnet.com...
Douglas A. Gwyn wrote:
cody wrote:
The problem is that i believe that my assertions are correct.

Yes, that is a problem.

I hope the sig-monsters are awake.

What is a sig-monster?


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Nov 13 '05 #250

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