pointers help

[jsbeckton]

I've been working on this C programming assignment all week and am getting nowhere, its due monday so I'm running out of time. Any help would be greatly appreciated, thank you.

CIT 145 Pointer Assignment

Construct a program that declares a size 10 array of doubles and an array of pointers to doubles.

In a for loop, using pointer arithmetic to increment, initialize the integer array from 1.0 to 10.0 and in same loop initialize the pointer array to point to the array elements holding the data.

Create a print function that accepts the pointer array and its size as an argument. The function should, using square bracket notation, print out the contents of the pointer array and the values in the double array.

An possible example of output:

0x001f20 1.0
0x001f28 2.0

etc.

[Dave_Sinkula]

And the code you have so far...?

[jsbeckton]

Im almost too embarrassed to post it, I think I was closer o few days ago than it is now.

#include <stdio.h>
#define SIZE 10

int main()
{
	int ptr;
	double values[10];	
    void array(double *ptr[10]);
    int i;

   

	for (ptr = 0; ptr < 10; ptr++ ) {
		values[ptr]=1;
		
	}
	for (i = 1; i< 11; i++) {
	
		printf("%p%3d\n", &values[i], i);

   


	}

}

[jsbeckton]

I'm afraid that looking at this code will lower IQ's!

[Dave_Sinkula]

If this is to have meaning...

#define SIZE 10

...then you may need to use it.

C doesn't have much magic, so...

void array(double *ptr[10]);

... is going a little overboard. An array of pointers would be:

double *ptr[10];

This doesn't change much:

values[ptr]=1;

does it?

This looks troublesome:

for (i = 1; i< 11; i++)

If you can't explain "magic" numbers such as one and eleven, maybe it's not quite correct.

It may not be as far off as you think, though.

[jsbeckton]

when I change to

double *ptr[10];

I get an error message:
C:\Documents and Settings\Joshua\Desktop\CIT-145\trial.c(9) : error C2040: 'ptr' : 'double *[10]' differs in levels of indirection from 'double '

[Dave_Sinkula]

when I change to

double *ptr[10];

I get an error message:
C:\Documents and Settings\Joshua\Desktop\CIT-145\trial.c(9) : error C2040: 'ptr' : 'double *[10]' differs in levels of indirection from 'double '

Where? Post the associated code or else your note may not have any meaning.

[jsbeckton]

here it is after change, thank you for your help by the way...

#include <stdio.h>
#define SIZE 10

int main()
{
	double ptr;
	double values[10];	
    double *ptr[10];
    int i;

   

	for (ptr = 0; ptr < 10; ptr++ ) {
		values[ptr]=i;
		
	}
	for (i = 1; i< 11; i++) {
	
		printf("%p%3d\n", &values[i], i);

   


	}
 

}

[Dave_Sinkula]

How many variables with the name ptr do you have? And where is the print function?

[edit]If in pointer notation values[i] is *(values + i)...

[jsbeckton]

I thought that I needed to declare ptr as a double and then use it as a array of pointers to doubles.

[Dave_Sinkula]

Construct a program that declares a size 10 array of doubles and an array of pointers to doubles.

It might not be an assignement I'd create, but I'll work with what's there.

[jsbeckton]

in my first for statement I was trying to initilize the pointer array to point to the array elements holding the data.

[jsbeckton]

in the second I was trying to print out the contents of the pointer array and the values in the double array

[jsbeckton]

I was a bit thrown by that too, but i'm lost anyways

[Dave_Sinkula]

The second loop is supposed to be a function, so I pointed you that way.

The first loop is supposed to use pointer notation, so I was trying to steer you that way. This is way too much of a clue, but I need another beer and a cigarette:

      *(values + i) = i + 1;
      *(ptr    + i) = values + i;