dev-cpp-users

[Dev-C++] char Array - DevC++ beginner...

[Aurea J. L. <jan...@ho...>]

<html><div style='background-color:'><DIV>
<P>I am trying to create an array type char in Dev++ like this:</P>
<P><EM><FONT size=2>#include &lt;iostream.h&gt;<BR>#include &lt;stdlib.h&gt;<BR>#include &lt;string&gt;</FONT></EM></P>
<P><EM><FONT size=2>int main()<BR>{<BR>char Name[];</FONT></EM></P>
<P><EM><FONT size=2>Name[1]="J";</FONT></EM></P>
<P><EM><FONT size=2>cout &lt;&lt; sizeof(Name);</FONT></EM></P>
<P><EM><FONT size=2>system("PAUSE");<BR>return 0;<BR>}</FONT></EM></P>
<P>and DevC++ is telling me that <EM><FONT size=2>the array size is missing </FONT></EM>and that <FONT size=2><EM>assig to "char" from "const char" lacks a cast</EM>.</FONT></P>
<P>Anyone could please help me with what&nbsp;I am doing wrong?</P>
<P>Thanks.<BR></P>
<P>Janine</P></DIV></div><br clear=all><hr>Add photos to your messages with  <a href="http://g.msn.com/8HMEENUS/2749??PS=">MSN 8. </a> Get 2 months FREE*.</html>

RE: [Dev-C++] char Array - DevC++ beginner...

[Abhijit S. <mu...@gm...>]

Greets, Janine.

> char Name[];
> Name[1]="J";

When creating an array, you need to specify the size. An exception is when
you're specifying the array elements alongside (or when you're allocating
memory dynamically, but that would be called a pointer, not an array).

Also, a char is represented using single-quotes. So "J" is wrong. 'J' is
right. When you use double-quotes, the compiler sees it as an array of
chars, terminated by a null chracter (\0).
So,
	"foo"
is the same as
	{'f', 'o', 'o', '\0' }
or
	{'f', 'o', 'o', 0 }

In the above program, use

char Name[25]; // Some arbitrary size, otherwise, do char Name[] = "foo";
Name[1] = "J";

and it should work. One more thing - arrays begin at zero. I suspect that
here, you're trying to access the first element, in which case, it would be
Name[0].

HTH.
________________________________________

Abhijit Shylanath
http://mudeth.tripod.com/

RE: [Dev-C++] char Array - DevC++ beginner...

[<a_g...@ya...>]

another thing. it should be 
Name[1]='J';
and not
Name[1]="J";
cause "J" is a string by itself. It should be 'J' which is treated as a character constant.

Abhijit Shylanath <mu...@gm...> wrote:

Greets, Janine.

> char Name[];
> Name[1]="J";

When creating an array, you need to specify the size. An exception is when
you're specifying the array elements alongside (or when you're allocating
memory dynamically, but that would be called a pointer, not an array).

Also, a char is represented using single-quotes. So "J" is wrong. 'J' is
right. When you use double-quotes, the compiler sees it as an array of
chars, terminated by a null chracter (\0).
So,
"foo"
is the same as
{'f', 'o', 'o', '\0' }
or
{'f', 'o', 'o', 0 }

In the above program, use

char Name[25]; // Some arbitrary size, otherwise, do char Name[] = "foo";
Name[1] = "J";

and it should work. One more thing - arrays begin at zero. I suspect that
here, you're trying to access the first element, in which case, it would be
Name[0].

HTH.
________________________________________

Abhijit Shylanath
http://mudeth.tripod.com/



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