problem in code

You do read: y1-Jan, y2-Jan, y3-Jan, y1-Feb, ...
Is your input: y1-Jan, y1-Feb, y1-Mar, y1-Apr, ...

Sep 5, 2014 at 11:30am

Yeah the input is y1 - Jan, y1 - Feb....etc

Sep 5, 2014 at 12:12pm

mutexe (2372)

Yeah the input is y1 - Jan, y1 - Feb....etc
are you sure? Because i disagree.

In wordy terms your for loop is telling me this: "For January, tell me your sales for year 1, year 2, year 3. Now tell me for February your sales for year 1, year 2, year 3". etc..

Last edited on Sep 5, 2014 at 12:34pm

Sep 5, 2014 at 4:19pm

That is the point. The data is in one order, but the code expects different order.

Code can change:

const size_t Years {3};
int yearlytotals[Years] {};
for ( size_t year = 0; year < Years; ++year ) {
  // accumulate months of year 'year+1' to yearlytotals[year]
}

Sep 5, 2014 at 4:45pm

@mutexe: I wasn't really sure what I was doing...I thought I was taking inputs for each year separately, instead of each month over three years. Thanks for pointing that out.

Should the 2D sales array be sales[12][3] for the program to work according to what I was thinking?

Sep 5, 2014 at 5:17pm

store them directly in the sales array, it might work

Sep 5, 2014 at 5:19pm

No. &(sales[0][0]) + row*Cols + col == &( sales[row][col] )

1
2

std::cout << "Enter sales for " << months[i] << " of year " << year+1 << '\n';
std::cin >> sales[year][i];

Sep 5, 2014 at 5:21pm

This short code in C(not C++) but you can convert it into C++ might help you understand how to get the right input.

#include <stdio.h>

int main()
{
	int i, j, sales[3][12];
	unsigned int total = 0;
	
	for( i=0; i<3; i++ )
	{
		for( j=0; j<12; j++ )
		{
			printf("Enter sale %d for month %d: ", i, j);
			scanf("%d", &sales[i][j]);
			total += sales[i][j];
		}
	}
	
	printf("%d", total);
}

Edit & run on cpp.sh

In this case, total is the amount of sales at the end of 3 years.

Last edited on Sep 5, 2014 at 5:22pm

Sep 5, 2014 at 6:04pm

How to get the sales total per year?

Sep 5, 2014 at 7:15pm

this is a slight workaround to the solution. kinda hacky but it does the trick.

#include <stdio.h>
#include <string.h>

int main()
{
	int i, j, sales[3][12];
	unsigned int yearlyTotal[3];
	unsigned int total = 0;
	
	memset(yearlyTotal, 0, sizeof yearlyTotal);
	
	for( i=0; i<3; i++ )
	{
		for( j=0; j<12; j++ )
		{
			printf("Enter sale %d for month %d: ", i, j);
			scanf("%d", &sales[i][j]);
			total += sales[i][j];
			yearlyTotal[i] += sales[i][j];  // get yearly total for each year
		}
		printf("Yearly total %d: %d\n", i, yearlyTotal[i]); // print it
		yearlyTotal[i] = 0;  // reset value to 0
		printf("Yearly total %d reset to %d\n", i, yearlyTotal[i]);
	}
	
	printf("%d", total);
}

Edit & run on cpp.sh

my coding skills suck. apologies. this is the best i could do. :)

Sep 5, 2014 at 8:22pm

You don't actually need any arrays in that last form.

Sep 5, 2014 at 9:30pm

if that's directed to me, I was showing the OP that you can create a loop within a loop with fewer variables hence saving space and code.

Sep 6, 2014 at 7:23am

Here's my loop, please tell me what I am doing wrong:

int sales[3][12];
int tot_sales = 0;
int year1 = 0;
int year2 = 0;
int year3 = 0;

for (int i = 0; i < 3; ++i) {
for (int j = 0; j < 12; ++j) {
cin >> sales[i][j];

year1 += sales[i][j];

cin >> sales[i][j];
year2 += sales[i][j];

cin >> sales[i][j];
year3 += sales[i][j];

tot_sales = year1 + year2 + year3;
}
}

I don't seem to be getting the yearly total.

Sep 6, 2014 at 12:38pm

The outer loop executes the inner loop 3 times.
The inner loop executes the "read input" 12 times.
The "read input" reads 3 values.
3 * 12 * 3 == 108 However, three years have only 36 months.

Lets demonstrate:

for (int i = 0; i < 3; ++i) {
  for (int j = 0; j < 12; ++j) {
    cout << "I want sale of month " << j+1 << " of year " << i+1 << " and I will add it to total for year 1\n";
    cout << "I want sale of month " << j+1 << " of year " << i+1 << " and I will add it to total for year 2\n";
    cout << "I want sale of month " << j+1 << " of year " << i+1 << " and I will add it to total for year 3\n";
  }
}

Run that loop. It does not read any values, but the output should help you to understand what your loop does.

Sep 6, 2014 at 7:22pm

@keskiverto: Thanks for pointing that out!

I finally solved my problem. But, I have a problem. Here's my final code:

// sales2.cpp -- sales per month for 3 years using 2D array
#include <iostream>

using namespace std;
int main()
{
string months[12] =
{
"January", "February", "March", "April",
"May", "June", "July", "August",
"September", "October", "November", "December"
};

int sales[3][12];
int tot_sales = 0;
int year1 = 0;
int year2 = 0;
int year3 = 0;

int i, j;
for (i = 0; i < 3; ++i) {
for (j = 0; j < 12; ++j) {
cin >> sales[i][j];
}
}

for (i = 0; i < 3; ++i) {
//year1 += sales[0][j];
//year2 += sales[1][j];
//year3 += sales[2][j];
for (j = 0; j < 12; ++j) {
year1 += sales[0][j];
year2 += sales[1][j];
year3 += sales[2][j];

tot_sales += sales[i][j];
}
}

//cout << "Sale on year 1: " << year_tot << endl;

cout << "Sale on year1: " << year1 / 3 << endl;
cout << "Sale on year2: " << year2 / 3 << endl;
cout << "Sale on year3: " << year3 / 3 << endl;

cout << "Total sales for 3 years: " << tot_sales << endl;

return 0;
}

Here's the output:
1 2 3 4 5 0 0 0 0 0 0 0
0 0 0 0 0 1 1 1 1 2 3 4
1 2 3 4 1 2 3 4 1 2 3 4
Sale on year1: 15
Sale on year2: 13
Sale on year3: 30
Total sales for 3 years: 58

Before, I noticed that the output was repeated 3 times for each year.
What I mean is for Sale on year1 it showed 45, for sale on year2 it showed 39, and for sale on year3 it showed 90.
So, I just divided the final answer by 3 in the cout statements.
My question is, is there any way I can get the loop to get the correct value, without having to divide the final values by 3.

Thanks for all the help previously. Any help for the current problem would be appreciated.

Sep 6, 2014 at 7:52pm