Trying to understand character arrays.

[drM]

Could anyone please offer some advice:
Trying to play with character arrays in C, the code below generates
this error.
Anyone see something wrong?
thank you in advance.

>>>> #include <stdio.h>

void test(char s[]);
char helloworld[11];

int main () {
printf("%s", test(helloworld )); // <----Invalid use of void
expression
return 0;
}
void test (char t[])
{

t[0]='H';
t[1]='e';
t[2]='l';
t[3]='l';
t[4]='o';
t[5]=' ';
t[6]='W';
t[7]='o';
t[8]='r';
t[9]='l';
t[10]='d';
t[11]='\0';

}
>>>

[Michael Mair]

drM schrieb:

Could anyone please offer some advice:
Trying to play with character arrays in C, the code below generates
this error.
Anyone see something wrong?
thank you in advance.

#include <stdio.h>

void test(char s[]);
char helloworld[11];

int main () {
printf("%s", test(helloworld )); // <----Invalid use of void
expression
1) This nicely illustrates why you should not use // comments in
usenet messages
2) test()'s return type is void -- "nothing".
Nothing cannot be assigned to "anything" (anything non-void).
You can either change test() to
char *test (char *s);
or you have to do it in two steps:
test(helloworld );
printf("%s", helloworld);
3) Note that you should output '\n' as last character in order
to make sure that you see any output at all. return 0;
}
void test (char t[])
{

t[0]='H';
t[1]='e';
t[2]='l';
t[3]='l';
t[4]='o';
t[5]=' ';
t[6]='W';
t[7]='o';
t[8]='r';
t[9]='l';
t[10]='d';
t[11]='\0';

1) If you want to change test() to char *test(char*), then
insert
return &t[0];
here
2) There are also strcpy() and strncpy()...
}

Cheers
Michael
--
E-Mail: Mine is an /at/ gmx /dot/ de address.

[Kenneth Brody]

drM wrote:

Could anyone please offer some advice:
Trying to play with character arrays in C, the code below generates
this error.
Anyone see something wrong? [...] void test(char s[]);
char helloworld[11]; [...] printf("%s", test(helloworld )); // <----Invalid use of void expression

[...]

Hint 1: What does the error say?
Hint 2: What type does test() return, and how does that relate to (1)?
Hint 3: Can the type answered in (2) be passed to a function?
Hint 4: What does printf's "%s" expect?

--
+-------------------------+--------------------+-----------------------------+
| Kenneth J. Brody | www.hvcomputer.com | |
| kenbrody/at\spamcop.net | www.fptech.com | #include <std_disclaimer .h> |
+-------------------------+--------------------+-----------------------------+
Don't e-mail me at: <mailto:Th***** ********@gmail. com>

[drM]

Have just started in C, so your reply is most welcome.

"1) This nicely illustrates why you should not use // comments in
usenet messages "....Point taken.

[Vladimir S. Oka]

drM opined:

Have just started in C, so your reply is most welcome.
Whose reply? To what?

"1) This nicely illustrates why you should not use // comments in
usenet messages "....Point taken.

You should also make a point of quoting what and who you're replying
to. Read the link in my sig, and all the links in it...

--
Parallel lines never meet, unless you bend one or both of them.

<http://clc-wiki.net/wiki/Introduction_to _comp.lang.c>

[drM]

2) test()'s return type is void -- "nothing".
Nothing cannot be assigned to "anything" (anything non-void).
You can either change test() to
char *test (char *s);
or you have to do it in two steps:
test(helloworld );
printf("%s", helloworld);
Michael....If I understand you correctly, the type (void) is producing
an error, not because of the (void) test(helloworld ) function, but
because of the printf function...or am I completely misunderstandin g
you.
tks

[Keith Thompson]

"drM" <md**@comcast.n et> writes:

Have just started in C, so your reply is most welcome.

"1) This nicely illustrates why you should not use // comments in
usenet messages "....Point taken.

Please learn to quote properly. <http://cfaj.freeshell. org/google/>

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.

[drM]

Keith Thompson wrote:

Please learn to quote properly. <http://cfaj.freeshell. org/google/>

Boy...this is a tough crowd!! :-) Thanks for all the input.

[drM]

Michael Mair wrote:

Nothing cannot be assigned to "anything" (anything non-void).
You can either change test() to
char *test (char *s);
or you have to do it in two steps:
test(helloworld );
printf("%s", helloworld);

Michael....If I understand you correctly, the type (void) is producing
an error, not because of the (void) test(helloworld ) function, but
because of the printf function...or am I completely misunderstandin g
you.
tks

[Keith Thompson]

"drM" <md**@comcast.n et> writes:

Michael Mair wrote:

Nothing cannot be assigned to "anything" (anything non-void).
You can either change test() to
char *test (char *s);
or you have to do it in two steps:
test(helloworld );
printf("%s", helloworld);

Michael....If I understand you correctly, the type (void) is producing
an error, not because of the (void) test(helloworld ) function, but
because of the printf function...or am I completely misunderstandin g
you.

It's not really one or the other, it's the combination. Your test
function was declared to return void; you tried to pass the
(nonexistent) result of that function to printf. You called printf
with a "%s" format, which means you need to pass it a char* as its
second argument.

If you had passed, say, an integer value:

printf("%s", 42);

then you would have been invoking undefined behavior. Since the types
for the arguments to printf are determined by the value of the format
string, they can't generally be checked by the compiler. But you
can't pass a nonexistent value (the result of a void function) as an
argument, so the compiler rejects the call.

--
Keith Thompson (The_Other_Keit h) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.