C++. help in making this code shorter.

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hi. can someone help me with the syntax in my code. im still beginning to learn about C++.

C++

#include<iostream>
#include<math.h>
#define _USE_MATH_DEFINES // for C++
#include <cmath>
using namespace std;

int main(){
	
	int  iTerm;
	float  iSum=0.00,iDenom=1.00,iError;
	float iPi =  M_PI;
	
    cout<<"|Term|Value|Error|"  <<endl;

    for(iTerm=1;iTerm<=100;iTerm++){
	
		if(iTerm%2==1){
			iSum=iSum + (4/iDenom);
			cout<<"|"<<iTerm << "|"<<iSum;
			iError = iSum - iPi;  
			cout<<"|"<< iError <<endl;
			cout<<"\n";
			iDenom = iDenom + 2;
        }
        else{
		    iSum =iSum - (4/iDenom);
		    cout<<"|"<<iTerm<<"|"<<iSum;
		    iError = iSum - iPi; 
		    cout<<"|"<< iError <<endl;
		    cout<<"\n";
		    iDenom = iDenom + 2;
        }
    }
}

can someone help me shorten the the syntax inside if and else. in C there is a way to shorten this.

additional help:
how can I make iError into printing 2 decimal places? in C its %.2f

thank you

Posted 3-Feb-15 4:24am

Reuben Cabrera

Updated 3-Feb-15 4:29am

Tomas Takac

v2

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3 solutions

Solution 3

C++

for(iTerm=1;iTerm<=100;iTerm++)
{
	iSum += (8*(iTerm % 2)-4)/iDenom;
	cout<<"|"<<iTerm << "|"<<iSum;
	iError = iSum - iPi;  
	cout<<"|"<< iError << endl << endl;
	iDenom = iDenom + 2;
 }

Posted 3-Feb-15 23:37pm

CPallini

Comments

Reuben Cabrera 4-Feb-15 8:22am

thank you :)

CPallini 4-Feb-15 12:16pm

You are welcome.

TheRealSteveJudge 4-Feb-15 8:41am

This is even shorter than my suggestion. :-(
5*

CPallini 4-Feb-15 12:18pm

Thank you (your asnwer had my 5 as well). :-)

TheRealSteveJudge 5-Feb-15 2:48am

Thank you, too!

Member 10641779 4-Feb-15 22:57pm

High 5 :)

CPallini 5-Feb-15 2:54am

Thank you.

Reuben Cabrera 5-Feb-15 5:01am

nice code but can you explain how you came up with this line?
iSum += (8*(iTerm % 2)-4)/iDenom;

where did 8 come from

CPallini 5-Feb-15 8:07am

(iTerm % 2) result can be 1 or 0. We have to transform 1 into 4 and 0 into -4. So:
8*1-4 = 4
8*0-4 = -4

Solution 2

Try

C++

for (iTerm = 1; iTerm <= 100; iTerm++){
 
	if (iTerm % 2)
		iSum += (4 / iDenom);
	
	else
		iSum -= (4 / iDenom);
	
	cout << "|" << iTerm << "|" << iSum;
	iError = iSum - iPi;
	cout << "|" << iError << endl;
	cout << "\n";
	iDenom = iDenom + 2;
}

Posted 3-Feb-15 23:19pm

Member 10641779

Comments

Reuben Cabrera 4-Feb-15 8:22am

thank you :)

Member 10641779 4-Feb-15 22:52pm

you are welcome :)

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TheRealSteveJudge · Accepted Answer · 2015-02-03T04:42:00

If you have a look at your code
then you will see that both branches of your if/else statement
contain mostly the same.
The only difference is "iSum=iSum + (4/iDenom);" vs. "iSum =iSum - (4/iDenom);"

So your code should look like this:

C++

#include<iostream>
#include<math.h>
#define _USE_MATH_DEFINES // for C++
#include <cmath>
using namespace std;

int main(){

	int  iTerm;
	float  iSum = 0.00, iDenom = 1.00, iError;
	float iPi = M_PI;

	cout << "|Term|Value|Error|" << endl;

	for (iTerm = 1; iTerm <= 100; iTerm++){

		if (iTerm % 2 == 1){
			iSum = iSum + (4 / iDenom);
		}
		else{
			iSum = iSum - (4 / iDenom);
		}

		cout << "|" << iTerm << "|" << iSum;
		iError = iSum - iPi;
		cout << "|" << iError << endl;
		cout << "\n";
		iDenom = iDenom + 2;
	}
}

You should always follow the DRY principle:
http://en.wikipedia.org/wiki/Don%27t_repeat_yourself[^]

For string formatting please have a look at:
http://www.tutorialspoint.com/c_standard_library/c_function_sprintf.htm[^]