help with first and second problem

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help with first and second problem

 
-1
  #1
Nov 10th, 2004
hi guys i ask for help but know one responded, so i figured i will ask again.





this is the first problem and why i can't print out a rectangle, triangle and diamond. i need know what am i doing wrong with both of these questions.

someone please help me.

#include <iostream>
#include "stdafx.h"
void DoRectangle();

int main()
{

void DoRectangle();
int height, width, i;
char symbol;
char response;
int shape;


cout<< " Do you wish to create a shape?:";
cin>> response;
cout<< " Enter 'Y' for yes, and 'N' for no:";

cout<<" enter a shape to create:";
cin>> shape;


cout<<" Enter rectangle width:";
cin>> width;
cout<< "Enter rectangle height:";
cin>> height;

cout<<" enter any character to be displayed:";
cin>> symbol;

for( i=1, i<=height, i++)
{
cout<<i;
}




return (0);
}








for the second problem this is what i have. what am i doing wrong

this is an ARRAY problem.

#include "stdafx.h"
#include <iostream>
using namespace std;

int main()
{

int num;
int avg;
int num;
int testscore1, testscore2, testscore3;
const int SIZE =3;
char terminate;

cout<<"enter testscore1:";
cin>> num;

cout<<" enter testscore2:";
cin>> num[82.2];

cout<<"enter testscore3:";
cin>> num[97.8];

avg = (testscore1 + testscore2 + testscore3)/3;

cout<< "the average is"<< avg << "\n";

cout<<"Enter any character to terminate:";
cin>> terminate;



return 0;
}
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Re: help with first and second problem

 
0
  #2
Nov 11th, 2004
>know one responded, so i figured i will ask again
You figured wrong. Asking again is rude. Asking again in a new thread is even worse.

>why i can't print out a rectangle, triangle and diamond
The evidence suggests that you're lazy. Here's a hint: Use nested loops so that you have control over columns as well as rows.

>for the second problem
num is not an array, and array subscripts have to be integral unless you can come up with a definition of arrays where floating-point indices make sense and can sell it to ISO for standardization.
I'm here to prove you wrong.
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