C++ :: Stop Scientific Notation

Jul 15, 2013

double number = 10000000;
int range;//the length of the string result
string result;//holds the number in a string
ostringstream convert; //stream used for the conversion
convert << number;
result = convert.str();

range = result.length();

I'm trying to convert a double to a string and when the number goes to ten million it goes to scientific notation and it shows it in the string. How do I stop it from do that?

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Mar 12, 2014

I just wrote code that is a program for a relativity calculator. However many of my outputs (because the values tend to be large) end up in scientific notation. Although useful, its not great for the laymen, or nice looking.

How can I change it so that output is not in scientific notation? here is the code:

// This program/converter is designed to find the desired 'real' values using Einstein's theory of relativity

#include<iostream>
#include<math.h>
#include<stdlib.h>

[Code]....

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Dec 10, 2013

Write a program to convert the time from 24-hour notation to 12-hour notation and vice versa. Your program must be menu driven, giving the user the choice of converting the time between the two notations. Furthermore, your program must contain at least the following function: a function to convert the time from 24-hour notation to 12-hour notation, a function to convert the time from 12-hour notation to 24-hour notation, a function to display the choices, function(s) to get the input, and function(s) to display the results. (For 12-hour time notation, your program must display AM or PM.)

Answer:

#include <iostream>
#include <iomanip>
#include <cmath>

[Code]....

It is showing error because may be I was not able to put that if statement inside any function. find out the error sand complete the program with corrected code.

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Mar 31, 2014

Can we change the floating point number format from scientific format to below example format ?

FROM TO
==========================
2.06374E-03 ---> 206370-8
-4.30311E-01 ---> -.430310
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==========================

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Aug 4, 2013

Is there a simple notation to check if a value is within a plus or minus range?

E.g.
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//Read value again -calling this value C delay(50); //Read value again -calling this value D delay(50);
//Read value again -calling this value E
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I can think of a few round about ways of doing this but is there any simple "equals to plus or minus" notation? (what I actually want to do is to check a lot more values than this and it will get very complicated with any of my solutions)....

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My program suppose calculate postfix expression.

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I need to use vector to compute the value. Here is what I think. First, I save leftmost from the string. If it is a number, I push. If that is a operation, I pop twice and push the result. By doing it until the string is emptied, the vector will only contain the final answer. And here is my code

#include <iostream>
#include <vector>
#include <string>
using namespace std;
int main(){
vector<int> stack;
string input;

[Code] .....

When I put 1 1 + or 2 2 +, it showing me a correct answer, but when I put the above example which is 5 4 + 3 10 * +, it shows 30 instead of 39.

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Sep 14, 2014

I'm trying to make an infix to postfix notation calculator. The difficult thing is th stack class is custom. It's not to hard to understand, I don't know if the fact that it is that way will not allow me to receive support. The difference is that the pop functions is as such:

stack<char> conversion;
char temp;
conversion.pop(temp);//It receives a parameter and puts the popped element in there.
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not only that... but these are boolean functions. they return true if the operation was completed. So they can be used as conditions.

#include<iostream>
#include<string>
#include<stdio.h>
#include"stack.h"
using namespace std;
int main(void) {
Stack<char> conversion;
string infix,inter,temps;

[Code] .....

The error is that i am mismanaging parenthesis handling and i can't seem to grasp where and how.

An example input is:(35+7)-(9-2)
that input gives me:35 7 + 9 2 ) -
but another input such as :(35+7)/7
outputs as: 35 7 + 7 /. Totally fine.

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Please check the attachment.

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been working on this code til my eyes bled, can not get second operand to work, if i enter 1 2 3 + - it will only return 5 from the addition operand.

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#include <math.h>

[Code].....

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#include <stdio.h>
#include <math.h>
int main(int argc, const char * argv[]) {
int n;
register int i=1;
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[Code] ....

For some reason I can't edit printf("%f",/>/>; when I post it as the topic so ignore that part cuz Ik its supposed to be written as printf("%f",/>;

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Code:
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using namespace std;
class person {
protected :
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[Code] ....

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Would also like to know why the argument call needed &head instead of just head.

remove = deleteNode(&head,found); opposed to remove = deleteNode(head,found);

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[Code].....

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Code:

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[code]...

If the user enters an invalid number, then it works fine, but if a character is entered, the while loop in main never breaks.

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...
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[Code]....

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[Code] .....

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Code:
#include <stdio.h>
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/*

You are to write a program that will allow a user to enter infinite numbers (greater than zero)(One number at a time). You must have a way for the user to stop entering values.Once the user stops, you will immediately display the following:

The lowest number entered is:
The highest number entered is:
The number of values entered is:
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*/

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");
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[Code]....

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void MainLoop()
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[Code].....

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