what's the different between char[] and char* as parameters

[Virtual_X]

if we use char[] in a function parameter like that

void st(char x[])
{cout << x;}

and char* as a parameter in the same function instead of char x[]

what would be different

in my experiments i conclude that
with char[] i can change the assigned value unlike char*

but in another question

why the value assigned to char[] when i change it in the function
it will also change the passed variable

ie:

void st(char x[])
{
x[0]= 'e';
cout << x;
}

int main()
{
char q[]= "CPP"

st(q);
cout << q;
}

in st(q) it will return "qPP" that logical

but
the second line will return the same result

HOW

i pass a variable not a reference and the function also don't accept
references
why the value changed in the result of the second line

[Default User]

Virtual_X wrote:

if we use char[] in a function parameter like that

void st(char x[])
{cout << x;}

and char* as a parameter in the same function instead of char x[]

Nothing, they are two ways of saying the same thing.

what would be different

in my experiments i conclude that
with char[] i can change the assigned value unlike char*

I don't know where you got that idea, but it's untrue.

but in another question

why the value assigned to char[] when i change it in the function
it will also change the passed variable

Because it's a pointer.

Your textbook should explain all of this.

Brian

[Marcus Kwok]

Virtual_X <C.*******@gmai l.comwrote:

if we use char[] in a function parameter like that

void st(char x[])
{cout << x;}

and char* as a parameter in the same function instead of char x[]

what would be different

Nothing. When used as a function parameter, arrays will decay into
pointers to the first element. You cannot pass an array by value to a
function.

in my experiments i conclude that
with char[] i can change the assigned value unlike char*

What do you mean? In both cases you will be passing a pointer to the
first element of the array. You cannot change the value (of the
pointer) passed to the function, but you can change what it points to.

but in another question

why the value assigned to char[] when i change it in the function
it will also change the passed variable

ie:

void st(char x[])
{
x[0]= 'e';
cout << x;
}

int main()
{
char q[]= "CPP"

st(q);
cout << q;
}

in st(q) it will return "qPP" that logical

No, it should change q to be "ePP".

but
the second line will return the same result

HOW

i pass a variable not a reference and the function also don't accept
references
why the value changed in the result of the second line

Because you passed a pointer. See above.

For these basic questions, maybe you would be better posting in
alt.comp.lang.l earn.c-c++

--
Marcus Kwok
Replace 'invalid' with 'net' to reply

[Virtual_X]

but

the second line will return the same result

HOW

i pass a variable not a reference and the function also don't accept
references
why the value changed in the result of the second line

Because you passed a pointer. See above.

how i pass a pointer to char[]

[Victor Bazarov]

Virtual_X wrote:

>>but
the second line will return the same result

>>HOW

>>i pass a variable not a reference and the function also don't accept
references
why the value changed in the result of the second line

Because you passed a pointer. See above.

how i pass a pointer to char[]

You need to understand that 'char[]' is the same as 'char*' in C++.
If you don't mean 'char[]', please be more explicit (like use the word
"array" or something).

A pointer to an array of char has the type 'char (*)[N]' where 'N' has
to be a compile-time constant expression. A pointer to a pointer to
char has the type 'char**'.

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Marcus Kwok]

Virtual_X <C.*******@gmai l.comwrote:

but
the second line will return the same result

HOW

i pass a variable not a reference and the function also don't accept
references
why the value changed in the result of the second line

Because you passed a pointer. See above.

how i pass a pointer to char[]

Re-read my first post and Brian (Default User)'s post. The function
signatures

void foo(char* bar);

and

void foo(char[] bar);

mean exactly the same thing.
If you are asking how to pass a pointer to an array, then the size must
be known at compile time, or you must make it a template that can deduce
the size at compile time:
#include <iostream>

template <int N>
void foo(char (*arr)[N])
{
std::cout << *arr << '\n';
}

int main()
{
char cstr[] = "foo";
foo(&cstr);
}

--
Marcus Kwok
Replace 'invalid' with 'net' to reply

[Victor Bazarov]

Marcus Kwok wrote:

[..]
Re-read my first post and Brian (Default User)'s post. The function
signatures

void foo(char* bar);

and

void foo(char[] bar);

void foo(char bar[]);

the syntax 'char[] bar' does not exist in C++.

>
mean exactly the same thing.

[..]

V
--
Please remove capital 'A's when replying by e-mail
I do not respond to top-posted replies, please don't ask

[Marcus Kwok]

Victor Bazarov <v.********@com acast.netwrote:

Marcus Kwok wrote:

> void foo(char[] bar);

void foo(char bar[]);

the syntax 'char[] bar' does not exist in C++.

Thanks, sorry about that. I've just had to start learning some C# for
work, and it's the subtle differences in the languages (like this one)
that I find are confusing me more than the major differences.

--
Marcus Kwok
Replace 'invalid' with 'net' to reply

[Virtual_X]

On Jul 13, 1:25 pm, ricec...@gehenn om.invalid (Marcus Kwok) wrote:

Victor Bazarov <v.Abaza...@com acast.netwrote:

Marcus Kwok wrote:

void foo(char[] bar);

void foo(char bar[]);

the syntax 'char[] bar' does not exist in C++.

Thanks, sorry about that. I've just had to start learning some C# for
work, and it's the subtle differences in the languages (like this one)
that I find are confusing me more than the major differences.

--
Marcus Kwok
Replace 'invalid' with 'net' to reply

thank's for all
i got it

[BobR]

Virtual_X <C.*******@gmai l.comwrote in message...

but
the second line will return the same result
HOW i pass a variable not a reference and the function also
don't accept references
why the value changed in the result of the second line

Because you passed a pointer. See above.

how i pass a pointer to char[]

Learn to experiment, test things. Try this:

#include <iostream>

void st( char x, std::ostream &out ){
out<<"char x="<< x;
}

void st( char x[], std::ostream &out ){
out<<"char x[]="<< x;
}

void st( char *x, std::ostream &out ){
out<<"char *x="<< x;
}

void st( char *&x, std::ostream &out ){
out<<"char *&x="<< x;
}
// ....etc.

int main(){
char q[]= "CPP";
st( q, std::cout );
std::cout <<std::endl;
std::cout << q <<std::endl;

st( q[1], std::cout );
std::cout <<std::endl;
std::cout << q <<std::endl;

return 0;
}

What did your compiler say when you tried to compile that?

Then comment-out /* void st( char x[], std::ostream &out ){} */.
Compile and run.
Then comment-out /* void st( char *x, std::ostream &out ){} */,
Un-comment the first (char x[]), compile and run.

Your findings?

--
Bob R
POVrookie