array's

[las]

Hello, Im trying to figure out the difference between int[] and char[]

#include <iostream>
#include <stdlib.h>

int main()
{
int i[10] = {1};
char c[10] = {'1'};

cout << "&int \t" << &i << endl;
cout << "int \t " << i << endl;
cout << "int[0]\t" << i[0] << endl;
cout << endl;
cout << "&char \t" << &c << endl;
cout << "char\t " << c << endl;
cout << "char[0]\t" << c[0] << endl;
cout << endl;
system("PAUSE");
return 0;
}

I get this output :

&int 0x254fda8
int 0x254fda8
int[0] 1

&char 0x254fd98
char 1
char[0] 1

I belive long, float, double etc work the same way as int does in my above
example. How is char[] different for these other variables ? Any links to
related tutorials would be great aswell.

Cheers :)

[Stewart Gordon]

las wrote:

Hello, Im trying to figure out the difference between int[] and char[]
int[] is an array of integers.
char[] is an array of characters.
#include <iostream>
#include <stdlib.h>

int main()
{
int i[10] = {1};
char c[10] = {'1'};

cout << "&int \t" << &i << endl; <snip> I get this output :
That's interesting. On mine, it doesn't even compile.
&int 0x254fda8
int 0x254fda8
int[0] 1

&char 0x254fd98
char 1
char[0] 1

I belive long, float, double etc work the same way as int does in my above
example. How is char[] different for these other variables ?

<snip>

It isn't, apart from the syntactic sugar and standard libraries enabling
char arrays to be used as strings.

When you send an int, long, float or double to a stream, the output
defaults to the numerical value in decimal.

OTOH, when a char is sent, the output is the character itself, as
distinct from the numerical value, which is the ASCII (or other platform
character set) code of the character.

When a pointer is sent, the default is the memory address in
hexadecimal. The name of an array by itself denotes a pointer. The
exception is a pointer to char, IWC the char array is output as a
null-terminated string.

As for why &char and char gave different values, &c and c are the same
memory address but are of different types. They are char[10]* (pointer
to array of 10 chars) and char* respectively. This difference in
referent size means that incrementing a char* will increment by one
char, whereas incrementing a char[10]* will increment by a block of 10
chars.

Stewart.

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[John Harrison]

"las" <la******@online.no> wrote in message
news:cE*******************@news4.e.nsc.no...

Hello, Im trying to figure out the difference between int[] and char[]

#include <iostream>
#include <stdlib.h>

int main()
{
int i[10] = {1};
char c[10] = {'1'};

cout << "&int \t" << &i << endl;
cout << "int \t " << i << endl;
cout << "int[0]\t" << i[0] << endl;
cout << endl;
cout << "&char \t" << &c << endl;
cout << "char\t " << c << endl;
cout << "char[0]\t" << c[0] << endl;
cout << endl;
system("PAUSE");
return 0;
}

I get this output :

&int 0x254fda8
int 0x254fda8
int[0] 1

&char 0x254fd98
char 1
char[0] 1

I belive long, float, double etc work the same way as int does in my above
example. How is char[] different for these other variables ? Any links to
related tutorials would be great aswell.

Cheers :)

Such a simple question, such a complex answer.

First your code is wrong. The only way to guarantee the behaviour you see is
to do this

char c[10] = {'1', '\0'};

This makes c a 'null terminated string' ('\0' is the null). In C++ arrays of
chars are often used as null terminated strings. operator<< knows this an
given an array of char will attempt (rightly or wrongly) to interpret it as
a null terminated string and print out all of the string. Gvien any other
sort of array operator<< will print out the address of the first element of
the array, that's what you see with your int array.

You might want to experiment with other values for c, e.g.

char c[10] = {'h', 'e', 'l', 'l', 'o', '\0'};

john

[Default User]

John Harrison wrote:

First your code is wrong. The only way to guarantee the behaviour you see is
to do this

char c[10] = {'1', '\0'};

This makes c a 'null terminated string' ('\0' is the null).

Nope nope nope. If you init any values in an array, the remainder of the
elements are initialized to the type-specific 0 value. So the addition
of the null-terminator here is superfluous.


Brian Rodenborn

[John Harrison]

"Default User" <fi********@company.com> wrote in message
news:3F***************@company.com...

John Harrison wrote:

First your code is wrong. The only way to guarantee the behaviour you see is to do this

char c[10] = {'1', '\0'};

This makes c a 'null terminated string' ('\0' is the null).

Nope nope nope. If you init any values in an array, the remainder of the
elements are initialized to the type-specific 0 value. So the addition
of the null-terminator here is superfluous.

OK, I stand corrected.

john