A question between Array and Pointer

Program 1:
#include <iostream.h>

int main()
{
char *nz;
nz="hello";

cout << nz << "\n";
return 0;
}

Program 2:
#include <iostream.h>

int main()
{
char nz[6];
nz="hello";

cout << nz << "\n";
return 0;
}

Isnt that,they are the same?howcome the first one run without any error,but the second one got an error?

Quote:

---

Originally posted by Unregistered
Program 1:
#include <iostream.h>

int main()
{
char *nz;
nz="hello";

cout << nz << "\n";
return 0;
}

Program 2:
#include <iostream.h>

int main()
{
char nz[6];
nz="hello";

cout << nz << "\n";
return 0;
}

Isnt that,they are the same?howcome the first one run without any error,but the second one got an error?

---

With the second you are trying to assign a r-value to an l-value. To do what you want to do you have to use the function strcpy( );

strcpy( nz, "Hello" );

In the first program, nz is a string initialized as pointer. It's legal to assign "hello" (nameless) string to nz. Therefore your first program works perfectly.

However, in the second program, nz is a C string. And in order to assign "hello" to nz string, you must put it this way:

nz[ ] = "hello" .

otherwise your program won't work.

Why Cant i separate

nz[]="hello";

into

char nz[6];
nz="hello";

Both are wrong.

I just told you. You can't assign an r-value to an l-value. If you need further explanation lookup those terms on google.com

Program:
#include <iostream.h>

int main()
{
char *nz;
nz="hello";

cout << nz << "\n";
return 0;
}


char *nz;
nz="hello";

isnt this also assign r-value to l-value?

Quote:

---

Why Cant i separate

nz[]="hello";

into

char nz[6];
nz="hello";

---

I didn't mention that in your 2nd example, you need to initialize your string this way

char nz[6]="hello"; or char nz[]="hello";

You can't just use nz[] = "hello";

Quote:

---

char *nz;
nz="hello";

isnt this also assign r-value to l-value?

---

Nope, that's assigning an r-value to an r-value.

Wut Does l-value and r-value means?

and

why initialize it :char nz[6]="hello"; or char nz[]="hello";

but not char nz[6];
nz="hello"; <---- is that coz,r-value cant be assigned to an l-value?

Quote:

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Originally posted by Unregistered
Wut Does l-value and r-value means?

---

http://msdn.microsoft.com/library/de...xpressions.asp

Quote:

---

and

why initialize it :char nz[6]="hello"; or char nz[]="hello";

---

I'm not sure why either, I only remember that's how the c-string is initialized. char nz[6] ="Hello"; is a string that can hold 5 characters, if you initialize to a string with more than 5 character, you'll get error messages. char nz[ ] = "hello" can hold as a string with many characters (undeterminated).

Quote:

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but not char nz[6];
nz="hello"; <---- is that coz,r-value cant be assigned to an l-value?

---

You can't assign any value with this syntax nz="hello", because nz is a c-string (an array of character termintated by a null). And you have to use strcpy() function.

l-value can not be Array

but howcome this below is correct?

char array[]="hello";

Code:

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char *nz = "hello";

---

Isn't exactly cool either. You should get in the habbit of allocating memory.

Code:

---

char *nz = new char[5];
strcpy(nz, "hello");

---

or

Code:

---

char *nz = (char *)malloc(5);
strcpy(nz, "hello");

---

Your compiler can interpret char *nz = "hello"; to mean one of many things. "hello" could be a static array of chars and nz is pointing to it (this one will run error free) or it could think "hello" is a temporary array of chars and nz points to it (this one will give you a good old seg fault). The fundamental difference between an array and a pointer is that an array has a finite size and a predictable location whereas a pointer is not. Before someone posts how wrong I am, I'll go ahead and point out that there are exceptions however a newbie would best understand what I just said.

Quote:

---

Originally posted by Unregistered
l-value can not be Array

but howcome this below is correct?

char array[]="hello";

---

Just an arays can be initialized, so alos can strings. The string can be intialized as the following exmaple:

Code:

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char str[ ] = {'s', 't', 'r', 'i', 'n', 'g', '\0'};  // notice that this string is ends up with a null character

---

However, C concedes that strings are a special kind of character array by providing a shortcut:

PHP Code:

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char str[ ] = "string";  // notice that we don't need to insert the null character 


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I hope that helps. :)

Quote:

---

Originally posted by master5001


Isn't exactly cool either. You should get in the habbit of allocating memory.

Code:

---

char *nz = new char[5];
strcpy(nz, "hello");

---

or

Code:

---

char *nz = (char *)malloc(5);
strcpy(nz, "hello");

---

Your compiler can interpret char *nz = "hello"; to mean one of many things. "hello" could be a static array of chars and nz is pointing to it (this one will run error free) or it could think "hello" is a temporary array of chars and nz points to it (this one will give you a good old seg fault). The fundamental difference between an array and a pointer is that an array has a finite size and a predictable location whereas a pointer is not. Before someone posts how wrong I am, I'll go ahead and point out that there are exceptions however a newbie would best understand what I just said.

---

Even these examples are not correct.

The reason you can not separate it in to two lines is nz is constant, the first line is initialization and the second is assignment.

can you do the following:

const int i = 5;
i = 55;

same story.

what do I know?

Quote:

---

Originally posted by master5001

Code:

---

char *nz = "hello";

---

Isn't exactly cool either. You should get in the habbit of allocating memory.

Code:

---

char *nz = new char[5];
strcpy(nz, "hello");

---

or

Code:

---

char *nz = (char *)malloc(5);
strcpy(nz, "hello");

---

Your compiler can interpret char *nz = "hello"; to mean one of many things. "hello" could be a static array of chars and nz is pointing to it (this one will run error free) or it could think "hello" is a temporary array of chars and nz points to it (this one will give you a good old seg fault). The fundamental difference between an array and a pointer is that an array has a finite size and a predictable location whereas a pointer is not. Before someone posts how wrong I am, I'll go ahead and point out that there are exceptions however a newbie would best understand what I just said.

---

yours is wrong too, you need to malloc() 6 chars for null terminator at the end

char nz[x]; initializes an ARRAY of x * sizeof(char). that means: if x == 10, 10 bytes are
allocated.
Thus:
char nz[10]; //declare
nz[0] = value; //valid!
nz[9] = value; //valid!
nz[10] = value; //BAD IDEA! you would access memory outside the array nz;

char *nz; just declares a POINTER! A pointer can point to ANY address. BUT IT ONLY POINTS TO IT! no memory is allocated here (except the four bytes (on 32bit systems) to store the address where the pointer points to)
Thus:
char *nz; //declare pointer;
nz[0] = value; //BAD IDEA! you would assign a value to a random location - because the pointer is NOT INITIALIZED!
SO WE INITIALIZE IT:
nz = "hello"; means: nz points to a location in memory, where the string "hello" is stored. so you can now access each letter of hello by using nz[0] up to nz[6] (nz[6] would be the NULL-character)