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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 23872
In article <0Y************ ********@comcas t.com>,
"Douglas A. Gwyn" <DA****@null.ne t> wrote:
Keith Thompson wrote:
Personally, I would actually prefer it if character constants
had type char rather than int, but my preference isn't strong enough
to suggest changing the language.


There are a large number of things about C that deserve
to be changed, if we were to go about cleaning up the
language. Unfortunately, nearly all of these would
cause existing carefully written programs to behave
differently. Legacy is a severe constraint.

I'm thinking seriously about using my copious free time
(when I finally get some!) to go back to ~7th Edition
Unix and explore an alternate evolution using a clean
C-inspired language that takes into account lessons
learned over decades of C programming. I don't think
it would look much like Limbo, but it might borrow some
ideas such as tuples: (a;b) := (b;a); // swap a & b


With a slightly different syntax

(a;b) = (b;a);

this could be added to C. The semantics would be interesting. I assume
that you would want this to have defined behaviour. Tuple r-values would
be obvious, and something like (a;a;a) is no problem as an r-value. I
guess modifying a tuple l-value that contains the same object twice is
undefined behavior, for example (a;a) = (1;2); would be undefined
behavior. Apart from that, if a tuple l-value is used in an assignment
statement, then all the components of the tuple form one object which is
the only object that is modified.

(a;b) += (b;a);

would be no problem because it is by definition the same as

(a;b) = (a;b) + (b;a);
Nov 13 '05 #221
"E. Robert Tisdale" <E.************ **@jpl.nasa.gov > wrote:
Christian Bau wrote:
If you post in comp.std.c and don't understand the C language
then you deserve everything that is coming.


The only thing that cody deserved was to be ignored.


What, and allow other newbies to believe that his erroneous statements
are correct? _That_ would be far less than courteous, perhaps not to
cody (though a good programmer is aware of his own failings and grateful
for correct information), but certainly to all other beginners trying to
find good C information here.

Richard
Nov 13 '05 #222
> 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?
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.


How long have you been programming in C++?


Was there something incorrect?

--
cody

[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
Nov 13 '05 #223
"Douglas A. Gwyn" <DA****@null.ne t> schrieb im Newsbeitrag
news:FZ******** ************@co mcast.com...
cody wrote:
But I tried it using the microsoft C Compiler and
printf("%i",siz eof('a'));
... By the way, you don't need the extra parentheses,
which obscure the distinction between the two forms
of sizeof.

You mean sizeof for types and sizeof for variables?
In C++ sizeof for variables needs parantheses, sizeof for types don't.

--
cody

[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
Nov 13 '05 #224
> Cody doesn't need anybody's help to look foolish.
If he would stop asserting misinformation and instead
ask questions, he would get a different response.

The problem is that i believe that my assertions are correct. But iam very
very
thankful for the many many valuable responses i got here.

--
cody

[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
Nov 13 '05 #225
cody wrote:
In C++ sizeof for variables needs parantheses, sizeof for types don't.


First, you're posting in C newsgroups, not C++.
Second, I doubt very much that that is true for C++;
it is certainly wrong for C.
Third, they're not variables, they're expressions.

Nov 13 '05 #226
cody wrote:
The problem is that i believe that my assertions are correct.


Yes, that is a problem. You should work on a better
understanding of what it is that you definitely know
versus what it is that you're just guessing about.

Nov 13 '05 #227
OK i tried it and:

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 them.
This seems to be true for both C and C++ compilers (I tried both).

--
cody

[Freeware, Games and Humor]
www.deutronium.de.vu || www.deutronium.tk
Nov 13 '05 #228
"Douglas A. Gwyn" wrote:

cody wrote:
In C++ sizeof for variables needs parantheses, sizeof for types don't.


First, you're posting in C newsgroups, not C++.
Second, I doubt very much that that is true for C++;
Third, they're not variables, they're expressions.


Section 5.3p1 of the C++ standard says pretty much the same thing that
6.5.3 says in the C standard:

_unary-expression_:
...
sizeof _unary-expression_
sizeof ( _type-id_ )

Paragraph 3 says the same thing in words: "... The operand is either an
expression, ..., or a parenthesized _type-id_. ..."

So cody has it almost exactly backwards. It's only "almost", not because
there's some truth in his claim, but because he used the wrong words.
This seems to fit his track record. I wonder when he'll decide to start
checking the validity of his claims before posting them?
Nov 13 '05 #229
cody wrote:
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?


No, and that wasn't his only mistake.

int test(int size)
{
int *new = malloc(size);
return 1 //*
-1 + 1 /* 0 in C90, 1 in C++ and C99 */
;
}
> 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.


How long have you been programming in C++?


Was there something incorrect?


Yes. void foo(void); /is/ a full prototype, in both C and C++.

It is as well for your real-world reputation that you post under a
pseudonym.

--
Richard Heathfield : bi****@eton.pow ernet.co.uk
"Usenet is a strange place." - Dennis M Ritchie, 29 July 1999.
C FAQ: http://www.eskimo.com/~scs/C-faq/top.html
K&R answers, C books, etc: http://users.powernet.co.uk/eton
Nov 13 '05 #230

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