Anybody good at C programming? (1 Viewer)

ADP said:

Luckily not a computer degree. Only computer class I have, and thus far, only problem i have not been able to figure out. I understand arrays mostly. It is the function part of it where I have to have it subtract the two arrays and place in array 3 that is getting me.

Click to expand...

Drunken Poster said:

You don't understand the question so providing the answer won't help you. Consider a different major/career.

Click to expand...

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

ADP said:

What? I provided that answer. I just needed some help in understanding it. And this is not a major/career, it was just one class I took as an option. I already have one degree in meteorology.

You don't understand the question so providing the answer won't help you. Consider a different major/career.

Click to expand...

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

ADP said:

Think i got it....thanks for help.

#include <stdio.h>
#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()
{
int arr1, arr2, arr3;
int i;

printf ("Please enter 10 numbers for the first array: \n");
for (i=0; i<SIZE; i++)
scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");
for (i=0; i<SIZE; i++)
scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");
for (i=0; i<SIZE; i++)
printf("%d ", arr3);

return 0;
}
void subtract(int arr1[], int arr2[], int arr3[])
{
int i;

for (i=0; i<SIZE; i++)
arr3 = arr2 - arr1;
}

Click to expand...

ADP said:

Royally confused by that. lol

Click to expand...

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

Drunken Poster said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>
#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()
{
int arr1, arr2, arr3;
int i;

printf ("Please enter 10 numbers for the first array: \n");
for (i=0; i<SIZE; i++)
scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");
for (i=0; i<SIZE; i++)
scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");
for (i=0; i<SIZE; i++)
printf("%d ", arr3);

return 0;
}
void subtract(int arr1[], int arr2[], int arr3[])
{
int i;

for (i=0; i<SIZE; i++)
arr3 = arr2 - arr1;
}

Click to expand...

Consider detecting the size of both arrays instead of relying on a hardcoded constant.

Click to expand...

ADP said:

Drunken Poster said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>
#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()
{
int arr1, arr2, arr3;
int i;

printf ("Please enter 10 numbers for the first array: \n");
for (i=0; i<SIZE; i++)
scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");
for (i=0; i<SIZE; i++)
scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");
for (i=0; i<SIZE; i++)
printf("%d ", arr3);

return 0;
}
void subtract(int arr1[], int arr2[], int arr3[])
{
int i;

for (i=0; i<SIZE; i++)
arr3 = arr2 - arr1;
}

Click to expand...

Consider detecting the size of both arrays instead of relying on a hardcoded constant.

Click to expand...

Yeh, I am not sure really how to do that, but figured since question said 10 int's specifically I would be ok with the define.

Click to expand...

ADP said:

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

can you show me in my code where you would put the arr3=subtract(arr1, arr2). I understand swapping out the arr3 in the function declaration with results at top and bottom of code, as well as changing out the arr3 in the actuall function itself with results. Just not too sure about the main section. Currently have subtract(arr1, arr2, arr3); Are you saying replace that?

Click to expand...

kutta said:

ADP said:

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

can you show me in my code where you would put the arr3=subtract(arr1, arr2). I understand swapping out the arr3 in the function declaration with results at top and bottom of code, as well as changing out the arr3 in the actuall function itself with results. Just not too sure about the main section. Currently have subtract(arr1, arr2, arr3); Are you saying replace that?

Click to expand...

Actually, you are best off just doing the printing from the subtract function. It's too difficult to explain how to return an array because you would really be returning a pointer, which you probably haven't learned yet.So I would pass in arr1 and arr2, then do the subtract and print in your subtract function.

Click to expand...

"Drunken said:

kutta said:

ADP said:

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

can you show me in my code where you would put the arr3=subtract(arr1, arr2). I understand swapping out the arr3 in the function declaration with results at top and bottom of code, as well as changing out the arr3 in the actuall function itself with results. Just not too sure about the main section. Currently have subtract(arr1, arr2, arr3); Are you saying replace that?

Click to expand...

Actually, you are best off just doing the printing from the subtract function. It's too difficult to explain how to return an array because you would really be returning a pointer, which you probably haven't learned yet.So I would pass in arr1 and arr2, then do the subtract and print in your subtract function.

Click to expand...

Returning a pointer may have been part of the question. Why else would the problem require the results to be printed in the main program instead of the called function?

Click to expand...

kutta said:

"Drunken said:

kutta said:

ADP said:

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

can you show me in my code where you would put the arr3=subtract(arr1, arr2). I understand swapping out the arr3 in the function declaration with results at top and bottom of code, as well as changing out the arr3 in the actuall function itself with results. Just not too sure about the main section. Currently have subtract(arr1, arr2, arr3); Are you saying replace that?

Click to expand...

Actually, you are best off just doing the printing from the subtract function. It's too difficult to explain how to return an array because you would really be returning a pointer, which you probably haven't learned yet.So I would pass in arr1 and arr2, then do the subtract and print in your subtract function.

Click to expand...

Returning a pointer may have been part of the question. Why else would the problem require the results to be printed in the main program instead of the called function?

Click to expand...

Oh. Good point. I didn't even read the whole problem statement, just read the code.OK. So with all that said, the easiest thing to do is what you posted earlier. Go ahead and pass all three arrays in and modify arr3 in subtract. It's not the best practice but it works and should be acceptable.

Click to expand...

Run It Up said:

Ashamed to say this took me an hour to remember how to do and it still looks like crap.

Java:

Spoiler

public static void main(String[] args) { Scanner input = new Scanner(System.in); int[] array1 = new int[10]; int[] array2 = new int[10]; for (int i=0; i<array1.length; i++) { System.out.println("Enter the numbers for each element of the first array"); array1 = input.nextInt(); } for (int i=0; i<array2.length; i++) { System.out.println("Enter the numbers for each element of the second array"); array2 = input.nextInt(); } int[] array3 = subtract(array1, array2); System.out.println("Array 1: \n-------------\n"); for (int number : array1) { System.out.println("Number = " + number); } System.out.println("\nArray 2: \n-------------\n"); for (int number : array2) { System.out.println("Number = " + number); } System.out.println("\nArray 3: \n-------------\n"); for (int number : array3) { System.out.println("Number = " + number); } } public static int[] subtract(int arrayOne[], int arrayTwo[]) { int[] arrayThree = new int[10]; for (int i=0; i < arrayTwo.length; i++) { arrayThree = arrayTwo - arrayOne; } return arrayThree; }

Click to expand...

ADP said:

No more coding crap for me ever...

Click to expand...

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

belljr said:

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

standard c allows output and input argumentsshould use a pointer though

Click to expand...

hofer said:

belljr said:

kutta said:

ADP said:

Think i got it....thanks for help.

#include <stdio.h>

#define SIZE 10

void subtract (int arr1[], int arr2[], int arr3[]);

int main()

{

int arr1, arr2, arr3;

int i;

printf ("Please enter 10 numbers for the first array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr1);

printf ("\nPlease enter 10 numbers for the second array: \n");

for (i=0; i<SIZE; i++)

scanf("%d", &arr2);

subtract(arr1, arr2, arr3);

printf ("\n\nResultant third array is: ");

for (i=0; i<SIZE; i++)

printf("%d ", arr3);

return 0;

}

void subtract(int arr1[], int arr2[], int arr3[])

{

int i;

for (i=0; i<SIZE; i++)

arr3 = arr2 - arr1;

}

Click to expand...

I wouldn't in pass arr3 into the subtract function. Declare a "results" array in subtract() and use that instead. Then make your subtract function return an array and return results, and in main, set arr3=subtract(arr1, arr2).ETA: it's bad form to pass something to a function and allow that function to manipulate it without really knowing how it is being manipulated. You are manipulating arr3 in subtract but you declared it in main, and you are trying to use it again in main. And actually, you are really just kind of getting lucky if that is working.

Click to expand...

standard c allows output and input argumentsshould use a pointer though

Click to expand...

I'm mostly a c# programmer, and I'm probably mistaken, but I'm pretty sure that your code will not work as intended because array 3 is passed by value (the values altered in subtract() in the array will not persist in main()). hence the need to use a pointer.

if this is a basic programmer class, are you sure it isn't c++?

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ADP said:

I have a problem I have been racking my brain with it. I am sure this would take all of 5 mins for a knowledgeable programmer; however, that I am not.

Question:

Create two arrays of integers, each holding ten elements of data. Create a third array of integers for a result array. Your main program will take the two arrays of integers and pass them to the function subtract(). Inside the function, subtract the first array from the second array and put the difference into the third array. Print the values of all three arrays back in the main program.

Any help would be much appreciated.

Click to expand...

int* subtract( int*  x,  int*  y)
{
    int* retval = new int[10];// use malloc here for straight c
    for (int i = 0; i < 10; ++i)
        retval = x - y;
    return retval;
}

BowieMercs said:

I haven't written anything in C in a loooong time... & I don't have a C compiler installed anywhere but i'd do something like this ... well it's almost C... its C++ :

You can't return an int array directly in C,  but you can return an int* which is pretty much the same thing under the hood.

    int a[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10};
    int b[] = { 10, 20, 30, 40, 50, 60, 70, 80, 90, 100 };

    int *c = subtract(a, b);
    for (int i = 0; i < 10; ++i)
        printf("a: %d - b:%d = c:%d \n", a, b, c);

 

Since you said pass the two arrays to the subtract function, I'm going to assume where subtract is something like this:

int* subtract( int*  x,  int*  y)
{
    int* retval = new int[10];// use malloc here for straight c
    for (int i = 0; i < 10; ++i)
        retval = x - y;
    return retval;
}

Click to expand...

NutterButter said:

I'm sure he still struggling with this over 2 years later

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BowieMercs said:

LOL. I spent the weekend replacing fence posts instead of watching hoops & drinking beer. Im sore, tired & apparently unable to read the timestamp on the OP.

They shoot horses, don't they?  

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Man of Constant Sorrow said:

Depends...what's your 1/4 mile time Bowie?

You could be a stable pony...you have a calm and reassuring demeanor...that's a big plus!

Also, you could go the stud route...how's your pedigree?  Are ya fixed?

Either way, the Churchill Downs stables are just across the street from me...we have room for ya buddy. 

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