The issue arises with case 3 where it tells produces an error when I attempt to compile and says "lvalue required for left operand of assignment error". How to fix this so that I can properly run the program.
I am getting this error when compiling my program with quincy:
Error: I value required as left operand of assignment
The program is meant to calculate how much parking costs based on the amount of hours in a park and what type of vehicle it is. the error is coming from my function definitions which i have just started to add in.
Code: float calcCarCost (char vehicletype, int time, float car) { if ((time > MINTIME) && (time <= 3)) calcCarCost =( CAR * time ); }
The error is on line 72 which is: calcCarCost =( Car * time);
I should probably point out CAR is already defined as a constant with a numerical value given and time is previously asked to be input in when the program runs.
I am having problems compiling this program. line 29 causes the error "left operand must be l-value".
// chap5proj.cpp : Defines the entry point for the console application. // # include <stdafx.h> # include <iostream> using namespace std; int main() { double mph, time, disPerHour, milesTrav;
I actually am wanting to practice on a 2 dimension char array. so, I am using the below program and getting the error - Lvalue Required For Function Main
i want to modify value of whole array by passing it to a function and make each value of array multiplied by 3 ,return the value to main and print it using pointer.
error : invalid Lvalue is the error
Code:
#include<stdio.h> main() { int i,arr[10]; for (i=0;i<=9;i++) { printf("Enter value of arr[%d] : ",i); scanf("%d",&arr[i]);
I am working on a double linked list and inside of my function to insert a node, I am getting an error of "Incompatible types in assignment". Here is my function code. Line 55 is where I am receiving the error.
My understanding is that the rhs of line 1 construct only a temporary object. getMe() then return the reference of this temp object and bind it to thg (as a lvalue reference). After line one, the temp object is really destroyed (hence the first output line). At this point thg is really binding to a destroyed, invalid object. But somehow the 2nd line still prints the correct value of 10 is because the memory storage is not yet corrupted (still holding the previous value). Is this correct?
I get Error this error. Did I miss something or is that some kind of bug?
Code: error C2679: binary '<<' : no operator found which takes a right-hand operand of type 'const std::string' (or there is no acceptable conversion)301
Code: 2IntelliSense: no operator "<<" matches these operands operand types are: std::ostream << const std::string307
while (getline(inStream, line)) { while (inStream >> Student.getId() >> Student.FNAME >> Student.MINIT >> Student.LNAME >> Student.GENDER >> Student.UNITS >> Student.getGpa()) { while (Student.getId() != id) { outStream << line << endl; } } }
This is what I have right now. It shouldn't be a problem, but for some reason I am getting an error trying to >> Student.getGpa()
Error1error C2679: binary '>>' : no operator found which takes a right-hand operand of type 'double' (or there is no acceptable conversion)c:location1301Project 5
I will post more code if needed, but... I just don't know. I have a TON of code so I would rather not if I don't have to.
Info:Program that stores information about reports .the above function searches the report according to its title. list is the name of the structure that stores the records.
Why i'm using strstr:
for eg. there is a report titled 'report on tigers'
I want the report information to be output if someone searches for 'tiger'
Output:displays all the entries i have made till now
Im using the remquo function in the cmath library as follows:
int quotient; double a = remquo (10.3, 4.5, "ient);
This gives the correct remainder (a = 1.3) and quotient (quotient = 2).
Infact about 50% of the answers are right when I play around, however, trying something like:
int quotient; double a = remquo (2.25, 1.5, "ient);
yields an incorrect quotient of 2 and remainder of 0.
I do think this has something to do with float arithmetic. I recall tinkering with the float number 0.500 and that the CPU actually saves it as 0.50000000000000231. However if my suspicion of float arithmetic as the suspect is correct, I do not understand why a tenth decimal would make such a drastic difference as changing the quotient result.
Write a program in C that will allow a user to enter an amount of money, then display the least number of coins and the value of each coin required to make up that amount. The twist to this program is the introduction of a new coin - the jonnie (value = $1.26). Therefore, the coins available for your calculations are: Twonies, Loonies, Jonnies, Quarters, Dimes, Nickels and Pennies. You can assume the amount of money user enters will be less than $100.00.
User enters $4.53. Your output should be:
The least number of coins required to make up $4.53 is 4. The coins needed are:
You enter decimal number into the program and what base you want. The integer part of the decimal is being handled fine, but the decimal is not.
For example, I enter 15.6847 and base 10, which means I'm going from base 10 to base 10. It spits out 68469999999999 for the decimal part. (Do not worry about the first block of numbers. The second block seperated from the first by a space is where the decimal will appear in order.)
#include <iostream> #include <string> #include <math.h> using namespace std; int baseConverter(int, int, int *, int *);
From my understand the cast (reg8 *) applies to the result of the bitwise OR. But what is the left most asterisk doing?Is it just dereferencing the casted pointer?
I am creating and implementing a left and a right rotation to balance a bst into an avl tree. I have made and tried 5 different codes that are commented in the functions left_rotate() and right_rotate() but none have run correctly. Sometimes the program works, sometimes there is a segmentation fault and sometimes not all inserted numbers are shown.
avl.c
Code: #include<stdio.h>#include<stdlib.h> #include<time.h> #include "avl.h" #define N 10 void swap(int *a, int *b){