[jjl]

Does this function change the value of any pointer?

int adder(int temp){
	temp++;
	cout << "\n\nIn the 'adder' function we change *num to " << temp << endl;
	return (temp);     
} 

Ill tell you what this function does

int adder(int temp) //creates a "new" variable with a "new memory location named temp, then the parameter you pass in it copied to temp
{
	temp++; //increments the new variable, 
	cout << "\n\nIn the 'adder' function we change *num to " << temp << endl;
	return (temp);  //now returns the new variable   
} 

What you need to do is pass in a pointer and change the value at that memory location rather than creating a new variable


Hers a function that will show the value at that address, maybe that will help you get going

void displayPointer(int *temp) 
{
	cout <<"The value at memory location "<<temp<<" is "<<*temp<<endl;
}

This post has been edited by ImaSexy: 24 July 2010 - 04:06 PM

[Nakor]

ok, now i've got a question on this. I was running this just to see what I could come up with and when I did

void adder(int* temp){
    *temp += 1; // also works with *temp = *temp + 1;
    cout << "\n\nIn the 'adder' function we change *num to " << *temp << endl;    
} 

I got the desired 7 both in the function and back in main after I returned from the function. However, when I did this

void adder(int* temp){
    *temp++;
    cout << "\n\nIn the 'adder' function we change *num to " << *temp << endl;    
} 

the value of temp inside the functions was 134515024 and back in main it was still 6. I had always been told that ++ and +=1 were the same thing but there must be some difference. I'm using gcc on a linux box.

[jjl]

take a look


(*temp)++ is equivelent to *temp = *temp +1;

when you do *(temp++), you are actually incrementing your pointer location

	const int SIZE = 5;
	int myArray[SIZE] = {1,2,3,4,5};
	int *ptr = &myArray[0];
	for(int i=0; i<SIZE; i++)
		cout<<*ptr++<<endl;

This post has been edited by ImaSexy: 24 July 2010 - 05:24 PM

[Nakor]

ah ok, so we need to actually get the value that's being pointed to before we apply the ++.

But we don't need the () if we're doing ++*temp?

[Nakor]

didn't mean to hijack this thread but maybe it'll help them learn a little bit more about it too

[theRedNeck]

ImaSexy, on 24 July 2010 - 03:03 PM, said:

Heres a function that will show the value at that address, maybe that will help you get going

void displayPointer(int *temp) 
{
	cout <<"The value at memory location "<<temp<<" is "<<*temp<<endl;
}

I'm not real big on pointers, but um.... Shouldn't that be *&temp???

cout <<"The value at memory location "<<temp<<" is "<<*temp<<endl;

gives me "invalid type argument of unary" error. Maybe it's just my compiler. (Dev c++)

As far as the apology comment.
Looking at Richard Bucklands lectures on Youtube, he made mention that some super guru programmer once said "Comments in your code are an apology, because it means your code is not clear enough to understand without the comments." Or something to that effect. Basically, good codes have comments. Great codes don't need them.

This post has been edited by theRedNeck: 24 July 2010 - 06:30 PM

[jjl]

Quote

A comment is an apology for not choosing a more clear name, or a more reasonable set of parameters, or for the failure to use explanatory variables and explanatory functions. Apologies for making the code unmaintainable, apologies for not using well-known algorithms, apologies for writing 'clever' code, apologies for not having a good version control system, apologies for not having finished the job of writing the code, or for leaving vulnerabilities or flaws in the code, apologies for hand-optimizing C code in ugly ways. And documentation comments are no better. In fact, I have my doubts about docstrings

[baavgai]

If you're supposed to be changing a value in a function with a pointer, then this is a total fail.

int adder(int temp);
//...
*num=adder(*num);

Since you've done nothing with pointers, the C was a gift.

This is probably what was expected:

#include <iostream>

using namespace std;

void adder(int *n) { *n = *n + 1; }

int main() {
	int n=6;
	
	cout << "value of n - " << n << endl;
	adder(&n);
	cout << "value of n - " << n << endl;
	return 0;
	
}