Basic Char and Pointer questions

maskedMan · 11-11-2009, 10:03 AM

Hello,

I'm approaching C from a background consisting of about 6 years worth of ActionScript, including AS3, so certain concepts like pointers and memory allocation are new to me. What's particularly confusing to me is the way C treates Strings as char arrays. Can anyone shed some light on why the following code yields the following results?

Code:

#include <cstdlib>
#include <iostream>

using namespace std;

// String Max Length.
#define FOOLEN 11

char* aString(char *str);


int main(int argc, char *argv[])
{
    int safety;
    char foo[FOOLEN] = "Foo";
    
    
    /* 
       this assignment appears to have the same result as
       the following uncommented assignment...
    */
    
    // char *cp = foo;
    
    char *cp;
    *cp = foo[0];
    
    cout << foo << endl << *cp << endl << aString(foo) << "\n\n";
    
    strncpy(cp, aString("What is the meaning of this?"), FOOLEN - 1);
    
    safety = FOOLEN - 1;
    
    
    
    // But traversing the resulting char array tells a different story.
    
    while(*cp && safety){
       cout << cp << endl; 
       cp++;   
       safety--;
    }
    
    cout << safety;
    
    system("PAUSE");
    return EXIT_SUCCESS;
}

char* aString(char *str){
   // I am unclear on why returning *str results in a compiler error.
   return str;   
}

As far as I can tell from syntax:

char *cp = foo;

and

char *cp;
*cp = foo[0];

Should have the same result, but it clearly does not.

Furthermore, what if in the aString function I wish to return a modified version of the char array passed to it? I could instantiate a new char array and return that value, but it only contains the first character from the passed argument. The goal would be to create a duplicate of the char array passed to the function rather than to directly modify the char array, in the event that you want to preserve the original data.

Dino · 11-11-2009, 10:22 AM

foo represents the pointer to the char array.
foo[0] represents the first element in the array, not the address of the first element.

When you code

Code:

char *cp = foo;

you are assigning the char array pointer (foo) to the char pointer. Correct.

When do code this:

Code:

*cp = foo[0];

you are assigning the first element (the actual value contained in foo[0]) to a pointer that has not been initialized. Wrong.

If you wanted the address of the first element in foo to be assigned to thechar pointer, you would code

Code:

cp = &foo[0];

using the & to get the address of the first element.

bithub · 11-11-2009, 10:31 AM

Quote:

Originally Posted by Dino

If you wanted the address of the first element in foo to be assigned to thechar pointer, you would code

Code:

*cp = &foo[0];

using the & to get the address of the first element.

That's not right. You can't assign a pointer value to a char. Get rid of the '*'.

Dino · 11-11-2009, 10:33 AM

Oops - missed that. I know that. Thanks for the correction.

11-11-2009, 10:03 AM	#1
maskedMan Registered User Join Date: Nov 2009 Posts: 1	Basic Char and Pointer questions Hello, I'm approaching C from a background consisting of about 6 years worth of ActionScript, including AS3, so certain concepts like pointers and memory allocation are new to me. What's particularly confusing to me is the way C treates Strings as char arrays. Can anyone shed some light on why the following code yields the following results? Code: #include <cstdlib> #include <iostream> using namespace std; // String Max Length. #define FOOLEN 11 char* aString(char str); int main(int argc, char argv[]) { int safety; char foo[FOOLEN] = "Foo"; /* this assignment appears to have the same result as the following uncommented assignment... / // char cp = foo; char cp; cp = foo[0]; cout << foo << endl << cp << endl << aString(foo) << "\n\n"; strncpy(cp, aString("What is the meaning of this?"), FOOLEN - 1); safety = FOOLEN - 1; // But traversing the resulting char array tells a different story. while(cp && safety){ cout << cp << endl; cp++; safety--; } cout << safety; system("PAUSE"); return EXIT_SUCCESS; } char* aString(char str){ // I am unclear on why returning str results in a compiler error. return str; } As far as I can tell from syntax: char cp = foo; and char cp; *cp = foo[0]; Should have the same result, but it clearly does not. Furthermore, what if in the aString function I wish to return a modified version of the char array passed to it? I could instantiate a new char array and return that value, but it only contains the first character from the passed argument. The goal would be to create a duplicate of the char array passed to the function rather than to directly modify the char array, in the event that you want to preserve the original data.
maskedMan is offline

11-11-2009, 10:22 AM	#2
Dino Jack of many languages Join Date: Nov 2007 Location: Katy, Texas Posts: 2,071	foo represents the pointer to the char array. foo[0] represents the first element in the array, not the address of the first element. When you code Code: char cp = foo; you are assigning the char array pointer (foo) to the char pointer. Correct. When do code this: Code: cp = foo[0]; you are assigning the first element (the actual value contained in foo[0]) to a pointer that has not been initialized. Wrong. If you wanted the address of the first element in foo to be assigned to thechar pointer, you would code Code: cp = &foo[0]; using the & to get the address of the first element. __________________ Mac and Windows cross platform programmer. Ruby lover. Last edited by Dino; 11-11-2009 at 10:35 AM. Reason: fix code and comments.
Dino is offline

11-11-2009, 10:33 AM	#4
Dino Jack of many languages Join Date: Nov 2007 Location: Katy, Texas Posts: 2,071	Oops - missed that. I know that. Thanks for the correction. __________________ Mac and Windows cross platform programmer. Ruby lover.
Dino is offline

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