Array of pointers + memory Location

[Slain]

I have more of a conceptual question now. Let us say I do this:-

char *str[10]; --create an array of pointers
str[1]= "John";

I thought this would automatically put John at some memory space and
point str[1] to it.
My colleague argues that in flat bed memory this should not be done. A
new should be used to allocate the memory and then point to it. Of
course he didnt do the best job explaining, hence my question to you
guys, if you can shed some more light on this.

Thanks in anticipation

[Nick Keighley]

Re: Array of pointers + memory Location

On 5 Mar, 14:38, Slain <Slai...@gmail. comwrote:

I have more of a conceptual question now. Let us say I do this:-
>
char *str[10]; --create an array of pointers
str[1]= "John";
>
I thought this would automatically put John at some memory space and
point str[1] to it.

yes

My colleague argues that in flat bed memory

in *what*?

this should not be done.

why not?

A
new should be used to allocate the memory and then point to it.

why? Why waste the cost of a new (and they are quite expensive)
when it isn't necessary?

Of
course he didnt do the best job explaining, hence my question to you
guys, if you can shed some more light on this.

I've little sympathy with people who give apparently motiveless
technical advice. If he knows the reason why doesn't he tell you?


--
Nick Keighley

[Victor Bazarov]

Re: Array of pointers + memory Location

Nick Keighley wrote:

On 5 Mar, 14:38, Slain <Slai...@gmail. comwrote:
>

>I have more of a conceptual question now. Let us say I do this:-
>>
>char *str[10]; --create an array of pointers
>str[1]= "John";
>>
>I thought this would automatically put John at some memory space and
>point str[1] to it.

>
yes

Except that it shouldn't compile since "John" is an array of const
char and 'str[1]' is a pointer to non-const char. There is no
conversion without a cast. The intent probably was to declare 'str'
as 'char const* str[10];'

[..]

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

[Slain]

Re: Array of pointers + memory Location

On Mar 5, 9:57 am, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:

Nick Keighley wrote:

On 5 Mar, 14:38, Slain <Slai...@gmail. comwrote:

>

I have more of a conceptual question now. Let us say I do this:-

>

char *str[10]; --create an array of pointers
str[1]= "John";

>

I thought this would automatically put John at some memory space and
point str[1] to it.

>

yes

>
Except that it shouldn't compile since "John" is an array of const
char and 'str[1]' is a pointer to non-const char. There is no
conversion without a cast. The intent probably was to declare 'str'
as 'char const* str[10];'
>

[..]

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

It does compile as long as it is inside main or any function. So may
be it does a type conversion.

[Slain]

Re: Array of pointers + memory Location

On Mar 5, 9:57 am, "Victor Bazarov" <v.Abaza...@com Acast.netwrote:

Nick Keighley wrote:

On 5 Mar, 14:38, Slain <Slai...@gmail. comwrote:

>

I have more of a conceptual question now. Let us say I do this:-

>

char *str[10]; --create an array of pointers
str[1]= "John";

>

I thought this would automatically put John at some memory space and
point str[1] to it.

>

yes

>
Except that it shouldn't compile since "John" is an array of const
char and 'str[1]' is a pointer to non-const char. There is no
conversion without a cast. The intent probably was to declare 'str'
as 'char const* str[10];'
>

[..]

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

It does compile as long as it is inside main or any function. So may
be it does a type conversion.

[Jeff Schwab]

Re: Array of pointers + memory Location

Victor Bazarov wrote:

Nick Keighley wrote:

>On 5 Mar, 14:38, Slain <Slai...@gmail. comwrote:
>>

>>I have more of a conceptual question now. Let us say I do this:-
>>>
>>char *str[10]; --create an array of pointers
>>str[1]= "John";
>>>
>>I thought this would automatically put John at some memory space and
>>point str[1] to it.

>yes

Agreed, "John" now lives in static storage.

Except that it shouldn't compile since "John" is an array of const
char and 'str[1]' is a pointer to non-const char. There is no
conversion without a cast.

There is such a conversion, although it is deprecated. From
[conv.array]: "This conversion is considered only when there is an
explicit appropriate pointer target type, and not when there is a
general need to convert from an lvalue to an rvalue."

The intent probably was to declare 'str'
as 'char const* str[10];'

I believe any attempt to modify the string is undefined behavior, so (as
you suggest) the OP is surely better off with char const*. Even better,
IMO, would be replacing the raw array of raw pointers with a vector (or
tr1 array) of std::string.

[Micah Cowan]

Re: Array of pointers + memory Location

"Victor Bazarov" <v.Abazarov@com Acast.netwrites :

Nick Keighley wrote:

>On 5 Mar, 14:38, Slain <Slai...@gmail. comwrote:
>>

>>I have more of a conceptual question now. Let us say I do this:-
>>>
>>char *str[10]; --create an array of pointers
>>str[1]= "John";
>>>
>>I thought this would automatically put John at some memory space and
>>point str[1] to it.

>>
>yes

>
Except that it shouldn't compile since "John" is an array of const
char and 'str[1]' is a pointer to non-const char. There is no
conversion without a cast. The intent probably was to declare 'str'
as 'char const* str[10];'

No cast is required, for compatibility with C (it's a, deprecated,
standard conversion, and only applies to string literals).

No diagnostic is required, but decent implementations will issue one
anyway.

--
Micah J. Cowan
Programmer, musician, typesetting enthusiast, gamer...

http://micah.cowan.name/

[Pete Becker]

Re: Array of pointers + memory Location

On 2008-03-06 07:07:36 -0500, utab <umut.tabak@gma il.comsaid:

char * p1=0,p2=0; // **

This is equivalent to

char *p1 = 0;
char p2 = 0;

--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

[utab]

Re: Array of pointers + memory Location

On Mar 6, 1:59 pm, Pete Becker <p...@versatile coding.comwrote :

On 2008-03-06 07:07:36 -0500, utab <umut.ta...@gma il.comsaid:
>

char * p1=0,p2=0; // **

>
This is equivalent to
>
char *p1 = 0;
char p2 = 0;
>
--
Pete
Roundhouse Consulting, Ltd. (www.versatilecoding.com) Author of "The
Standard C++ Library Extensions: a Tutorial and Reference
(www.petebecker.com/tr1book)

Ah yes, I missed that. Thanks

[Triple-DES]

Re: Array of pointers + memory Location

On 6 Mar, 13:07, utab <umut.ta...@gma il.comwrote:

[snip]

>
And the last question is that I have an array of pointers to char, how
can I get the start address of each char array that are pointed by the
pointers in the array? This is a bit related to my 1st question where
I thought I can get it with &(str[0][0]), but apparently there is sth
I am missing.

If you have an array of pointers to char, then the start address of
each char array is the pointer itself (at least unformally). So
&(str[0][0]) == str[0]

DP