this runs and everything just fine, however i cant seem to figure out why the program isnt telling to user to use a valid time for hours and minutes when it's entered over the standard time limits (i.e. cant go over 23 hours and over 59 minutes)
#include <iostream>
using namespace std;
//functions for converting 24hr to 12hr notation
//parameters use int and void as a data type
[code]....
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.
I'm writing a program that converts seconds to hours and minutes - That works fine! But if I write: 3700 seconds It answers That equals to: 1 hours, 1 minutes and 40 seconds.
I now want my program to write 1 hour, 1 minute and 40 seconds.
And if there is no hours at all it should only write 400 seconds equals 6 minutes and 40 seconds.
To sum up: The program shall write hour/hours minute/minutes second/seconds correct.
I've made one succesfull program, but it was written with 27 "if-statements" and I know there is a better way.
Code:
#include <stdio.h>
int
main(void) {
[Code]...
how to make the program know when to write hour or hours?
I want to add a count down timer to my testing software 1hr and 30min
Its a 32bit console application, I want the countdown timer outputted above each question on the right side, timer only needs to output the time left at the top of each question so a live outputted timer is not important.
Questions are not random generated.
Also how to output "out of time" if the timer reaches 0 and for the test to end, I think branching is involved not too sure.
I already have, std::srand( std::time(0) ) ; at the start of the test to randomly generate the answer order.
I need to write a program that converts timestamp to hour and days.
View 2 Replies View RelatedI have to program something that would convert miles per hour to seconds per mile.
I'm having an issue with something that I coded. Lines 21-23 are not showing up when I compile the program. What I'm trying to do is convert seconds into minutes.
What I have so far is
#include <iostream>
using namespace std;
int main() {
double miles_per_hour, minutes_per_hour, seconds_per_hour, seconds, minutes;
cout << "Enter the miles per hour: ";
cin >> miles_per_hour;
[Code] ....
I need to make a recursive function(s) which prints the hour glass pattern. This is my code so far:
void HourGlass(int n) {
if (n==1){
cout<<"*"<<endl;
cout<<"*"<<endl;
} else {
stars(n-2);
HourGlass(n-2);
[Code] ....
Problem is, it doesn't print the spaces in front of the pattern. This is what it prints:
*****
***
*
*
***
*****
and this is what it is supposed to print:
*****
***
*
*
***
*****
Do I make a separate function for the spaces? How do I go about it?
I have to write a program for class in which users will be inputting:
start time = startTime
number of minutes = callDuration
so first off, teacher wants us to input time as a floating point number such as 09.00
How do I validate the time so that users cant input i.e. 09.61 for every hour? In other words, so the user cant input minutes 60-99 for every hour.
then, when calculating:
endTime = startTime + callDuration
how would I make endTime display a correct time in such a situation: endTime = 09.45 + 00.30
so that it displays 10.15 not 09.75
Write a program in C++ that calculates the amount to be paid to an employee based on the hours worked and rate per hour. A user will enter hours worked and rate per hour for an employee. Your program should have a function payCheck that calculates and returns the amount to be paid. The formula for calculating the salary amount is as follows: for the first 40 hours, the rate is the given rate (the rate that a user has entered); for hours over 40, the rate is 1.5 times the given rate.
I had write my own coding, but it keeps error. it is okay. But i still don't think i could share here? Still running... I will update my coding for you alls to check for any syntax or logic error. The problem is i need to calculate if the employee work for more than 40 hours which 1.5 times.
Code:
#include <stdlib.h>
#include <stddef.h>
#include <string.h>
#include <stdio.h>
#include <iostream.h>
using namespace std;
int main(void) {
char *empname, *test, *final_output;
[Code] .....
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int displayMenu ();
void addRecords ();
void zip ();
void city ();
[Code] ...
I want to convert time from est to utc. Is there any function for this.
View 3 Replies View RelatedI am looking for a math/big num library, that allows me to convert 32/64/80 bot float numbers to string and vice versa.
Precision & accuracy is of importance here, and since this is an IEEE standard, i have high hopes that there are libraries for this out there, which would save me the hassle of trying to implement this myself...
I am having trouble in creating a program (named "up.c") that should do the following:
- if you run the command ("up") with no arguments, it should read input from stdin and display it on stdout, converting lowercase letters to uppercase.
- if you run the command with one or more files (as arguments to the "up" command), your program should read input from each file and display it on stdout, again converting lowercase letters to uppercase.
I'm trying to create a single program that can do both of these, and handle errors gracefully. I've found that some codes work to convert the letters, but they seem overly simple and aren't giving me what I'm looking for in my program.
why does the following code output "0.000000" instead of "1.000000"?
Code:
#include <stdio.h>
int main(void) {
int x=3, y=2;
printf("3/2 is %f
", x/y);
return 0;
}
the code was compiled and run using gcc 4.4.7 and glibc 2.17 on linux kernal 2.6.32 running on a PC with an intel i5-2500k cpu(sandy bridge)
I have this project, How do I ever Convert a .Pdf file to .txt file using C++.
View 5 Replies View RelatedI'm writing a small pice of code that takes numbers and convert it to string. So that's what I got
int main() {
char letter;
char letter1;
char letter2;
cin >> letter;
cin >>letter1;
int sum = letter * 128 + letter1;
[Code] ....
Then it asks the user to enter a letter, say 'w' , which it's value is 119
Then enter another letter, say 'o' which value is 111
then it do the equation, which is take the value of the first letter multiple it by 128 then add it to the value of the second letter.
How can I make it does this process in a For loop, or any kind of loop??
Then I have the second part which is the other way around. That is when the user enter a numbers then the code convert these numbers to a string.
251394404 - "d"
1964018 - "rd"
15343 - "ord"
119 - "word"
so what's happening here is the it divide each time by 128. But I can't do that in a code.
I have a question on conversion between char & string. I have cut & pasted the part of the code from my C++ code and my function "decryptPwd" uses C style code with "char" etc.
I need to pass a string (mypwd) somehow to this function after conversion & then compare it to another string (newmypwd).
I tried - "decryptPwd(mypwd.c_str())==newmypwd.c_str()" but it did not work.
..
#include <string>
..
char* decryptPwd(char hash[64]);
main () {
string mypwd;
string newmypwd;
if (decryptPwd(mypwd)==newmypwd)
[Code] ...
Converting an IP address to a binary number? As in the entire thing 123.45.555.49 to it's binary equivalent.
View 14 Replies View RelatedI am trying to insert values from a form into my SQL database. Two of the fields in the SQL database are of type DateTime.
When trying to insert I am receiving the error "Conversion failed when converting date and/or time from character string."
The relevant code is as follows:
DateTime saveNow = DateTime.Now;
string sqlFormattedDate = saveNow.ToString("yyyy-MM-dd HH:mm:ss");
conn.Open();
[Code] ....
The table layout:
I am beginner in developing C # and I will like to develop an application that a converte. csv file (table) in an xml file
View 1 Replies View Relatedi have used single argument conversion function which gets called in below scenario.
Code:
#include<iostream>
using namespace std;
class Demo
{
[Code]....
It is giving error "test3.cpp: In function `int main()':test3.cpp:18: syntax error before numeric constant"
But it should work as Demo d=100; works
I'm trying to make OBJ converter. The file structure of which I am trying to convert is an XML structure. Here's an example. I currently have a lua script that does this for me. But it always crashes on big objects. It's for a game called "ROBLOX" The only information I need to convert is.
<int name="BrickColor">194</int> <-- Brick Color
<CoordinateFrame name="CFrame"> <-- Position
<X>0</X>
<Y>0.600000024</Y>
<Z>0</Z>
And
Vector3 name="size"> <-- Part size.
X>1000</X>
Y>1.20000005</Y>
Z>1000</Z>
Here's the lua script. [URL] .....
This currently grabs all of the ClassNames in the place and converts them to OBJ wavefront gemotry.
So I tried debugging this program almost 10 times, each time finding a problem then making code to fix it, I used to get wrong values but then now after I changed the calculation from an infinity while loop to a for loop (because after some changes I realized that now I know the number of entries, meaning the start and the end of the loop) the program now displays no values at all. I checked it thoroughly and until the calculation process its functioning exactly the way it should but then the calculation breaks hell loose . Also the for loop that raises 2 to the power needed works fine too so it must be something with the other processes included in the calculation....
This used to be a function of a multiple value types conversion program, I isolated it for easier analysis as a lone standing program.
Code:
#include<iostream>
using namespace std;
int main() {
int d[10],e[10],anse=1,r,limit;
short int counterd=0,i,j;
float bind=0;
[Code] ....
I am Getting following errors I don't know why.I have mentioned the lines (in comments) where these errors are occurring.
__________________________________________________
warning C4018: '<' : signed/unsigned mismatch
error C2064: term does not evaluate to a function taking 0 arguments
error C2064: term does not evaluate to a function taking 0 arguments
__________________________________________________ _
Code:
/////////////////////////////////////////////////
//libraries
/////////////////////////////////////////////////
#include <iostream>
#include <string>
using namespace std;
[Code] .....
Code:
/*Currency Conv Control Module*/
#include <stdio.h>
#include "currencyconv.h" /* defines constants, declares functions */
int main(void) {
float us_dollars;
float currency;
int user_input;
[Code] ....
This is $$ converter program I created that is split up over 3 files. It takes the users input in US Dollars and converts it to the currency selected. When executing the program I am encountering a small problem. It actually isn't converting anything and returns 0.00 dollars.
I would like for it to convert the amount entered into the currency selected. I would also like to add in the print statement the type of currency the user has selected.
Code:
#include<stdio.h>
#include<conio.h>
#include<string.h>
#include<ctype.h>
#define MAX 50
[Code] ....
[URL] .....
My output should look like this ..
Enter an expression in infix form: (9+6)/(2*4)^15
Specify output expression type,1) Prefix 2)Postfix
The Prefix expression is: /+96^*2415
----------------------------------------------
Enter an expression in infix form: (9+6)/(2*4)^15
Specify output expression type,1) Prefix 2)Postfix
The Postfix expression is:96+24*15^/