I need some basic C++ help

[Psykarrd]

I am trying to declare a string variable as an array of char's. the
code looks like this.

char name[80];

then when i try to use the variable it dosn't work, however i am not
sure you can use it the way i am trying to. Could some one please tell
me what i am doing wrong, or another way of doing the same thing. i am
trying to use the variable like this.

name = "modem";
thanks for any help you can give me.

[Dave]

"Psykarrd" <ps******@hotma il.com> wrote in message
news:19******** *************** ***@posting.goo gle.com...

I am trying to declare a string variable as an array of char's. the
code looks like this.

char name[80];

then when i try to use the variable it dosn't work, however i am not
sure you can use it the way i am trying to. Could some one please tell
me what i am doing wrong, or another way of doing the same thing. i am
trying to use the variable like this.

name = "modem";
thanks for any help you can give me.

strcpy(name, "modem");

An expression such as "modem" is viewed by the compiler as a const char *
(which points to an address in memory where the string literal "modem" is
stored).

So, the expression you gave was trying to assign an address to array. As
such an operation does not make sense to do, your compiler yelped!

[Andrey Tarasevich]

Psykarrd wrote:

I am trying to declare a string variable as an array of char's. the
code looks like this.

char name[80];

then when i try to use the variable it dosn't work, however i am not
sure you can use it the way i am trying to. Could some one please tell
me what i am doing wrong, or another way of doing the same thing. i am
trying to use the variable like this.

name = "modem";
...

Both 'name' and "modems" are _arrays_. Arrays in C++ are not assignable.
In order to copy one array to another in general case you have to copy
them "manually" - element by element. The good news is that in most
cases there's a library function that can do it for you.

In this particular case you can use 'strcpy':

strcpy(name, modem);

--
Best regards,
Andrey Tarasevich

[Dave]

"Dave" <be***********@ yahoo.com> wrote in message
news:vr******** ****@news.super news.com...

"Psykarrd" <ps******@hotma il.com> wrote in message
news:19******** *************** ***@posting.goo gle.com...

I am trying to declare a string variable as an array of char's. the
code looks like this.

char name[80];

then when i try to use the variable it dosn't work, however i am not
sure you can use it the way i am trying to. Could some one please tell
me what i am doing wrong, or another way of doing the same thing. i am
trying to use the variable like this.

name = "modem";
thanks for any help you can give me.

strcpy(name, "modem");

An expression such as "modem" is viewed by the compiler as a const char *
(which points to an address in memory where the string literal "modem" is
stored).

So, the expression you gave was trying to assign an address to array. As
such an operation does not make sense to do, your compiler yelped!

Oops, sorry, I should have said "array of const char" instead of "const char
*". So, it's really an array, and *then* that array undergoes a conversion
to a pointer when the compiler encounters "modem". Finally then, as I said
before, a pointer cannot be assigned to an array...

[Mike Wahler]

"Psykarrd" <ps******@hotma il.com> wrote in message
news:19******** *************** ***@posting.goo gle.com...

I am trying to declare a string variable as an array of char's. the
code looks like this.

char name[80];

then when i try to use the variable it dosn't work, however i am not
sure you can use it the way i am trying to. Could some one please tell
me what i am doing wrong, or another way of doing the same thing. i am
trying to use the variable like this.

name = "modem";

Arrays are not assignable.

strcpy(name, "modem"); /* 'strcpy()' is declared by <cstring>
or <string.h>

Better yet, use the std::string type, which *is* assignable,
and also has the advantage of not being restricted to a fixed
size like an array:

std::string name; /* 'std::string' declared by <string> */
name = "modem";

Or if "modem" is to be the inital value:

std::string name("modem");

-Mike

[John Carson]

"Dave" <be***********@ yahoo.com> wrote in message
news:vr******** ****@news.super news.com

Oops, sorry, I should have said "array of const char" instead of
"const char
*". So, it's really an array, and *then* that array undergoes a
conversion to a pointer when the compiler encounters "modem".
Finally then, as I said before, a pointer cannot be assigned to an
array...

It is not the chars that are const --- if they were you couldn't change the
char values. Rather, it is the location in memory of the chars that cannot
be changed. In pointer terms, it is like a const pointer to char, i.e.,
char * const
--
John Carson
1. To reply to email address, remove donald
2. Don't reply to email address (post here instead)

[Dave]

"John Carson" <do***********@ datafast.net.au > wrote in message
news:3f******@u senet.per.parad ox.net.au...

"Dave" <be***********@ yahoo.com> wrote in message
news:vr******** ****@news.super news.com

Oops, sorry, I should have said "array of const char" instead of
"const char
*". So, it's really an array, and *then* that array undergoes a
conversion to a pointer when the compiler encounters "modem".
Finally then, as I said before, a pointer cannot be assigned to an
array...
It is not the chars that are const --- if they were you couldn't change

the char values. Rather, it is the location in memory of the chars that cannot
be changed. In pointer terms, it is like a const pointer to char, i.e.,
char * const
--
John Carson
1. To reply to email address, remove donald
2. Don't reply to email address (post here instead)

2.13.4-1: An ordinary string literal has type "array of n const char"

And indeed, you may not change the characters of a string literal.

[jbruno4000]

Looking at the way you're wanting to use the variable, perhaps you'll be better
off using a string variable instead of char. i.e.
#include <string>. You'll them be able to do what you seem to have in mind.

string name;

name = "modem";

and you can still access the individual components of name:

for(int count = 0; count < strlen(name); count++)
cout << name[count] << "," ;
cout << endl;

outputs: m, o, d, e, m,

[rajanikanth]

ps******@hotmai l.com (Psykarrd) wrote in message news:<19******* *************** ****@posting.go ogle.com>...

I am trying to declare a string variable as an array of char's. the
code looks like this.

char name[80];

then when i try to use the variable it dosn't work, however i am not
sure you can use it the way i am trying to. Could some one please tell
me what i am doing wrong, or another way of doing the same thing. i am
trying to use the variable like this.

name = "modem";
thanks for any help you can give me.

hi!

It is better to use the string class

#include<string >

string s = "modem";
cout << s << "\n";

Best,

Raj

[Andrey Tarasevich]

John Carson wrote:

"Dave" <be***********@ yahoo.com> wrote in message
news:vr******** ****@news.super news.com

Oops, sorry, I should have said "array of const char" instead of
"const char
*". So, it's really an array, and *then* that array undergoes a
conversion to a pointer when the compiler encounters "modem".
Finally then, as I said before, a pointer cannot be assigned to an
array...

It is not the chars that are const --- if they were you couldn't change the
char values. Rather, it is the location in memory of the chars that cannot
be changed. In pointer terms, it is like a const pointer to char, i.e.,
char * const

I think Dave was talking about the literal. String literal is an "array
of const char" and its 'char' values cannot be changed.

--
Best regards,
Andrey Tarasevich