If i declare a function as a void function. But for testing purpose if i use a return statement in the function definition. i have tested and found that the function does not return and executes the entire function. How does the function not return even if a return statement is available? Does the compiler removes this return statement or how it is?
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I'm writing some functions pertaining to binary trees. I've used recursion once before while learning quicksort but am still quite new and unfamiliar with it. And this is my first time touching a binary tree. So my question: In my addnode function, will the return root statement at the end ever return a value other than the value passed to the function?
Code:
#include <stdlib.h>
struct tnode
{
int data;
struct tnode * left;
struct tnode * right;
}
[code]....
From the page: [URL] ....
#include <iostream>
using namespace std;
int n;
int& test();
[Code] ....
Explanation
In program above, the return type of function test() is int&. Hence this function returns by reference. The return statement is return n; but unlike return by value. This statement doesn't return value of n, instead it returns variable n itself.
Then the variable n is assigned to the left side of code test() = 5; and value of n is displayed.
I don't quite understand the bold sentence. Shouldn't value of n and variable n be the same?
I've been trying to get my program to call void functions with an if statement, but when i run my program and try to call one of the functions "worst case, best case, or random case" it doesn't get called. It just prompts the original menu.
#include<iostream>
#include<fstream>
using namespace std;
void bubbleSort();
void selectionSort();
[Code] .....
I've encountered a slight logical error in my code
/*
Lab06_pensionplans.cpp
Purpose :
- Create a simple financial application to calculate retirement plans
*/
#include <iostream>
#include <cstdlib>
using namespace std;
void displayMenu() {
system("cls");
[Code] ....
Look at case 2, which the user supposed to key in a new input, the problem is the value will never got into main function, I don't know what should I modify with the function.
Figured out I need to change the
void changeData(int startingAge, int numOfYears,
double lumpSumAmount, double yearlyAmount, double interestRate )
into
void changeData(int &startingAge, int &numOfYears,
double &lumpSumAmount, double &yearlyAmount, double &interestRate )
i need to return a struct pointer dynamically allocated inside a function call void function() which is done using 'out parameters' in following code
struct my_struct {
int x;
} void my_function( my_struct** result ) {
my_struct* x = new my_struct{ 10 };
//...
*result = x;
}
Now i have a doubt, so if i want to print the return value from struct pointer, should i need to print it in the void function() or in the caller the function...
I'm beginners in C Programming, i have a question about Structure in C. i know in Function it is possible for
1. Sending the value of argument
2. Sending the address of the argument
3. Returning values from a function by return statement
Does in Structure it is possible for
1. returning structure from a function using by return statement
I am supposed to update the value temp without using a return statement, or a global variable. I have never ran across a problem like this before and I am totally stuck. I think I'm missing something really simple and need a hint. Here is the code that was provided for "fixing".
#include "stdafx.h"
#include <iostream>
#include <cstdlib>
using namespace std;
[Code]....
I'm trying to pass 2 arrays into a void funtion, and return values to one function.
this is the the program I'm working with, after I'm done I have to split it into 3 files, a header, a main, and a separate cpp file for the functions to live in.
#include <iostream>
using namespace std;
void processArrary(int numberCount[], int Numbers[], int intnumberSize, int numberCountSize);
int main() {
int Scores[26] = {76, 89, 150, 135, 200, 76, 12, 100, 150, 28, 178, 189, 167, 200, 175, 150, 87, 99, 129, 149, 176, 200, 87, 35, 157, 189};
int numberCount[8] = { 0 };
[code]...
The goal of this program is to separate and count the groups of numbers then output the amount of numbers in each group. Near as I can tell, everthing should work, but I'm getting all zeros to be displayed in each group.
How to make if else code below into SWITCH STATEMENT?
cout << "Enter the temperature: ";
cin >> temp;
cout << "Enter the pressure: ";
cin >> pressure;
cout <<endl;
[Code]....
difference between return 0 and return -1 .and exit(0) and exit(1)
View 1 Replies View RelatedThe error is coming up in line 13(srand() issue I think) of the following code:
#include <cstdlib>
#include <iostream>
#include <cmath>
#include <ctime>
using namespace std;
int main(int argc, char *argv[]){
[Code] ....
I'm sure it's something really simple that I'm overlooking.
Using a template in the assignment, I don't know what I did wrong?
Code:
#include <iostream>
#include <string>
using namespace std;
template<class T>
void mpgcalc(T& Miles, T& Gallons)
[Code] .....
I have a void that needs to end a program but a break and return 0 both won't work. Instead I have it cout (1/0). It works but is there an alternative?
#include <iostream>
#include <thread>
#include <Windows.h>
#include <limits>
using namespace std;
double clicks=0;
double result;
bool gameon=false;
[Code] .....
how can i use the void pointers? i understand that can recive an adress variable(&). but can recive a value?
Code:
int a=5;
void *d;
b=&a;
b=100;//???
why i can't do these?
Why is it not okay to return void? Most compilers will probably let you (gcc does) but it gives you a warning that you aren't supposed to. Most languages allow you to return void.
Something like
Code:
void log(const std::string & txt){ std::cout << txt << std::endl; }
//C++ way to do it
void bar(int i){
[Code].....
I am trying to add data to a queue with the following simplified code:
Code:
typedef struct Queue {
void * data;
int head;
int tail;
int elementSize;
My question is, how do I move the queue->data pointer to the correct memory location in order to copy given data to head? The code above inside memcpy gives med the error: "expression must be a pointer to a complete object type".
Do I need an extra pointer to be able to navigate between the queue's head and tail, and keep queue->data as a reference to the first byte of the allocated memory, or is it possible with only queue->data?
Edit. Just noticed I have mixed up head and tail. The enqueued data should probably go to the Queue's tail and not the head. However, the problem is still the same.
I could understand void pointers I created the following program:
Code:
#include <stdio.h>
#include <string.h>
int main(void) {
char word[] = "Zero";
int number = 0;
void *ptr = NULL;
[Code] .....
The program works fine, however i really want to fully understand what is going on with the dereferencing of the void pointer, for example: With the following code:
Code:
ptr = &number;
*((int *)ptr) = 1;
Why can't you just do:
Code:
ptr = &number;
*(int *)ptr = 1;
And again with this code, (i'm guessing it's becuase its a pointer to a pointer?):
Code:
ptr = &word;
strcpy(ptr,"One");
Code:
int main() {
List* newList= lst_new();
names* nama;
char* data;
int x=1;
[Code] ....
I cant seem to be able to print a string.. the functions lst_next() lst_first() return void*.
Write a C++ program consisting of main plus two other functions which will do the following:
Take an integer input from the keyboard.
Send the integer to a function which will output the integer to the screen.
Send the integer to a second function which will tell the user that the integer is an odd value.
Do not tell the user anything if the integer is an even value.
Repeat this process until the user enters something which is not an integer; use input validation to check for validity.
Any not valid input should terminate the program.
I was reading about void as function argument, but I did not fully understand it's meaning in C.
In C++
void foo(void) {}
and
void foo() {}
are the same. It means no arguments for foo function. But in C it's different. First function means the same as in C++, but second means
In C, an empty parameter list means that the number and type of the function arguments are unknown. But if it is unknown you can't use this arguments if user specifies same. Because here are no variables to store them. So doesn't result are the some? You do not get any arguments. O do I can get this arguments from some hidden variable?
For example.
void foo() {
printf("%d", var);
}
foo(5);
It is very unclear for me. Do this apply to main function too?
int main(void)
int main()
or can I use arguments given to int main() like given to int main(int argc, char* argv[])
My assignment is to write a program using VOID FUNCTIONS WITH AN ARGUMENT.
*I need one non-void function with an argument to generate the first 15 numbers greater than 500, another non-void function with an argument to generate the first 15 perfect squares that are greater than 500. Last, they need to be in columns next to each other.* also i cant use x,y, coordinates to align them. i must create a for loop with the
These are some notes from examples in the class. i just don't know how to do it with non void functions with an argument.
#include "stdafx.h"
#include <iostream>
#include <iomanip>
using namespace std;
void ClearTheScreen();
void NormalTermination();
[Code] ....
I want to have a function that has a pointer to an int/double/string so I thought I'd use a void pointer like so:
int someFnc(int a, int b, const void *key){
// take care of converting key into appropriate type in here
}
And when I want to use this function I'd like to be able to do something like this:
main{
...
int myKey;
someFnc(1,2,myKey);
]
But I get a compiler error telling me:
invalid conversion from 'int' to 'const void' -[fpermissive]
Do I need to convert myKey into a void pointer before passing it as an argument?
Why does passing myKey like this work?
someFnc(1,2,&myKey);
How to declare variable for all void() as I have another void s in my C++ program. I want to have a variable that can use for all the void and not only in a simple void.Is it possible?
View 17 Replies View Relatedint (*cInts)(void*,void*);
cInts= &compareInts;
int x=(cInts)(2,5); //This wont work. I tried a bunch of other things
printf(x);
I need this break in my main function, but I'm not allowed to put it in void since void is not located in a loop. How can I solve this?
Code:
#include <iostream>#include <string>
using namespace std;
void login(string x, string y);
string username;
string password;
string x;
string y;
[Code] ....
