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Comments:
<0> maybe for the compilers you use
<1> actually memset sets char values, does it not?
<2> In the general case, memset doesn't cut it - it happens to work for the char array, yes, because char is an integer type
<0> char is integer type?
<1> yeah, it is
<3> yes
<2> hammr: I really think it would be a good idea for you to learn C before you try to teach it.
<4> is work with "" but i can't printf("%s", letter[15]); because the array is not all initialize
<1> [ArKaNgE], that would be out of range anyway
<4> ok
<4> what is the solution?
<0> I got out of college in 90 and have been writing C since then
<2> If you have a 15-element array, there is no letter[15]
<1> what C_Dreamer said
<2> Just letter[0] through letter[14]
<4> yes
<1> hammr, that's a shame -- I knew that and I"m still in college :)
<4> but is not the problem
<2> hammr: I refer you to section 3.5.7 of C89.
<4> letter[14] or letter[15]..
<0> thanks
<0> I deal with military specs, not programming specs
<0> my code is not portable
<2> If you've been writing C for as long as you have, it's about time you learned it.
<0> no need
<4> wait
<2> If you write military code, you really really ought to know the language.
<0> I do
<0> and #1 rules is dont trust the compiler
<2> That is not evident from what you have said here.
<4> http://rafb.net/paste/results/HxPloT53.html
<2> If you don't trust the compiler, you have to inspect every line of machine code output.
<2> So you might as well write the program in machine code.
<0> nah, ***embler
<0> been there done that
<2> You would trust the ***embler, then?
<2> There is no logical distinction between trusting the ***embler and trusting the compiler.
<1> can you trust your disk to give you back the same machine code you put on it? you might have to check it in RAM every time it is loaded :)
<0> yes, in ***embler it is clear that during initiazation variables are initialized as you stated, this is not clear in C
<2> hammr - it is extremely clear from 3.5.7 of the ANSI Standard.
<0> and yes, in military operations, what is in ram is continuously checked agains the prom
<0> again, the militarty has its own standards
<2> Fine, but if its language standard differs from ANSI's standard, then it cannot reasonably be claiming to use C.
<4> for put all 0 on my char array...
<2> It might well be using a very similar language, maybe even a superior language and maybe in a more robust way, but if it ain't ANSI C, it ain't C.
<4> how?
<2> "If an object that has static storage duration is not initialized explicitly, it is initialized implicitly as if every member that has arithmetic type were ***igned 0 and every member that has pointer type were ***igned a null pointer constant." - ANSI C, 3.5.7
<1> [ArKaNgE], <2> char letter[15] = "";
<0> so your answer should have been ' char letter[15];'
<2> "If there are fewer initializers in a list than there are members of an aggregate, the remainder of the aggregate shall be initialized implicitly the same as objects that have static storage duration."
<0> wow
<1> hammr, that's not static storage duration if it's a local variable.
<4> ok is work but i can't printf(%s, letter[15]); because i have letter in letter[2] or letter[8]
<2> No, static char letter[15]; would have done, but he doesn't want static, so a partial initialisation is required.
<1> hammr, at least not unless you explicitly ask it to be.
<0> zzzzzzzzzzzzzz
<2> hammr - if the facts bore you, I suggest finding a lamerz channel.
<0> haha
<0> silly kids
<2> <grin>
<2> I've been using C longer than you, pal
<2> (just)
<0> whatever you say kid
<1> haha
<1> clueless, isn't he?
<4> :#
<2> Don't bother, cn28h - he wouldn't believe you anyway
<1> ;)
<5> C_Dreamer since when?
<0> see www.drs.com
<0> this is where I produce C code
<0> bye bye now
<1> haha, had to get in a spam before he left
<2> jock - C since 1989 - programming in general since 1982
<4> lol
<3> I thought the military and its contractors used Ada.
<2> <meta NAME="ColdFusionMXEdition" CONTENT="ColdFusion DevNet Edition - Not for Production Use."></head>
<2> (From the DRS Web site's HTML)
<1> haha
<2> Nice graphic though
<4> can I write a string which content only 1 letter in letter[3]
<4> **** my english :(
<4> and declaration char letter[15] = "";`
<4> before
<1> [ArKaNgE], it wouldn't make a lot of sense to start your string at &letter[3]
<1> but you *could*
<4> cn28h
<2> Look, it's really easy. If N is a constant, then char letter[N]; gives you an array of N chars. char letter[N] = ""; gives you an array of N chars all set to the null character '\0'. Once you have your array defined, letter[n] = x; /* sets the nth character in the array to the value x; n must be in the range 0 to N-1 */
<4> i have a word...in mot[15] and i compare the word whit letter in char letter;...whit the letter match, i put the letter in match_letter[i]...but if the word is LiGHT, and the letter match is G...the letter go on martch_ltter[2]...
<2> What are you trying to do, compare two strings?
<4> no compare char whit char array
<2> Didn't we cover that before?
<2> I suggested you use strchr
<4> if(m6400_lettre == m6400_mot[b])
<4> {
<4> m6400_mot_joueur[b] = m6400_lettre;
<4> }
<4> ok
<2> that's fine - it's comparing two characters directly
<4> if two letter is same in the word, strchr continu or stop?
<1> you will likely end up with gaps in m6400_mot_joueur
<2> strchr returns a pointer to the /first/ match it finds, or NULL if there is no match
<4> ok
<4> is not good for me...
<4> if the word is letter, have ton T in the word
<1> but you can strchr() from the next location after where the previous letter was found
<4> two...
<1> so you can find the next occurence that way
<4> ok
<4> is to advance for me i think...
<1> are you familiar with "array slicing"?
<2> I give up. I'm sorry, [ArKaNgE] - you know my language much better than I know yours, but the communication barrier is still too high for me to climb.
<4> no
<4> yes i speak bad english hehe
<4> im pratice c and english
<1> it's actually not too bad, considering :)
<1> as C_D said, it's much better than my French as well
<4> hehe
<4> what is array slicing
<1> but are you familiar with pointers? If so, array slicing is easy
<4> yes i understand pointer...
<1> ok, well take this example:
<1> char foo[] = "Hello World!\n"; printf("%s", &foo[6]);
<1> what is printed?
<5> is there a datastructure that has more efficient random access than linked list and doesn't make pointers to other element invalid when you add a new element?
<4> print the adress of foo[6]
<1> [ArKaNgE], no, it's using %s
<4> i dont now
<1> [ArKaNgE], &foo[6] is the address of the 6th element, right?
<4> yes
<1> so it starts reading the string at index 6, and prints "World\n"
<1> try ikt
<4> ok
<1> jock, what od you mean by makes pointers to other elements invalid?
<1> oh
<4> the printf stp at adress ?foo[6]
<1> right, it reads from that address and gets therest of the string
<1> jock, trees
<4> i read from the adress or stop to this adress...
<6> printf %s stops when it finds a \0
<4> i = he...
<1> jock, in a balanced tree, looking up an element is O(lg n) where as in a linked list it's O(n)
<6> you only give it the start address
<4> ok
<1> jock, or you might look at a hash table
<4> what is hash
<5> how does hash table have more efficient random access
<1> jock, because the hash tells you where to look
<5> actually it does
<1> [ArKaNgE], doin't worry about that for now :)
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