#include "vehicle.h"
...
void exchange(vehicle *&v1, vehicle *&v2) {
vehicle *tmp = v2;
v2=v1;
v1=tmp;
}
Is it right?
How about: void exchange(vehicle *v1, vehicle *v2){...}
What is the difference between *&v1 and *v1 ?
I have a function:
const void insertStuff(const void *key, const int value){
// I want to convert the void pointer into one
// of three types of pointers(int, string, or double)
switch(value){
case 0:
int *intPtr = key;
[Code] .....
But this causes an error of: "crosses initialization of int*intPtr"
What's the correct way of implementing this?
i really don't know why has a error in my code, that pass a pointer of pointer (name of a matrix with 2 dimensions). Here is the source code of a simple example where appears segmentation fault when execute (but compiles normal):
#include <stdio.h>
#define LINHAS 3
#define COLUNAS 5
float a[LINHAS][COLUNAS];
void zeros(float **p,float m, float n){
int i,j;
for(i=0;i<m;i++)
[Code]...
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*.
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);
int (*cInts)(void*,void*);
cInts= &compareInts;
int x=(cInts)(2,5); //This wont work. I tried a bunch of other things
printf(x);
Why does the following code compile and execute without any error? I mean, the function compareid should get 2 arguments so why does the compiler not complaining, is it because of the type of arguments?
Code:
#include <stdio.h>
int compareid(void* info, int value); // ansi declaration
int compareid(void* info, int value)
[Code] .....
I have a the following in a header file.
Code:
struct SortedList{
void * data;
struct SortedList * next;
struct SortedList * previous;
int (*compareFunc)(void *, void *);
void (*destructor)(void *);
[Code] ....
In a .c file, I implemented the SLCreate function.
Code:
SortedListPtr SLCreate(CompareFuncT cf, DestructFuncT df){
struct SortedList item;
item.data = NULL;
item.next = (struct SortedList *) malloc(sizeof(struct SortedList));
[Code] ....
In main.c, I try to do the following:
Code:
SortedListPtr list = SLCreate(&compareInts, &destroy);
A bunch other code that does not alter list or it's contents at all.
struct SortedList item = (*list);
void * data = item.data;
if (data != NULL) {
printf(Why did data become not null???
"); }
How come my variable data became not null anymore when I haven't altered it at all....
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 having a problem which I'm not able to resovle. I try to dereference a void pointer but I always get a C2440 error. It says: 'static_cast':void* cannot be converted in wqueue<T>. I tried different cast ways but I always get the same error. As far as I found out I should get the error if I try to dereference without cast but in my case I cast before and still get that error.
void *srumbler (void *arg) {
wqueue<workclas*> m_queue= static_cast<wqueue<workclass*>>(arg);
return NULL;
}
The according type wqueue in the header file:
template <typename T> class wqueue {
list<T> m_queue;
pthread_mutex_t m_mutex;
pthread_cond_t m_condv;
[Code] .....
How can I add the variable adress to a void pointer inside of a class?
class variant2 {
private:
void *Vvariant=NULL;
public:
template<typename b>
variant & operator = (b *adress)
[Code] ....
if possible i want avoid the '&' when i assign the variable address.(variant2 f=varname;//like you see i don't use the '&')
for the moment i just need put the address to Variant pointer. but i receive several errors .
I declared all functions in header file, such as:
bool readCase();
bool meshing();
bool readMesh();
bool calculateFlowfield();
bool readFlowfield();
bool calculateEvaporation();
And then I define them in separated .cpp files, each .cpp file include the header, but I got multiple definition error, why?
Even the int main() function, which only decalred and defined once got this error, why?
Can I a have one pointer with two reference in it. Here's what I've got.
Code:
char* c;
char x='x' , y='y';
c = &x;
c = &y; -- or --
Code: char* c[2];
char x='x' , y='y';
c[0] = &x;
c[1] = &y;
If it's possible I want to apply it to make AST.
I would like to get the this pointer by call by reference. Is this possible? I hoped to get it with this code, but it doesn't work:
Code:
"cpp"]class DemoClass {
public:
DemoClass();
int x;
void setParam(const DemoClass ¶m){
param=this;
[Code] ....
I get always the error code "C2678". But I don't understand how I should change my code to avoid this.
I am used to the restrict keyword to hint the compiler that no overlap is going to happen in the values passed to the function.
void foo( int * restrict a, char * restrict b)
I understand that I can pass by reference in c++.
void foo( int &a, char &b)
Using it, will it automatically restrict it or not ? It is very important for performance reasons (no checks at each iterations/steps)...
//program to form a header file
/* using pass by reference in pointer */
#include <iostream>
#include<math.h>
[Code].....
I am unable to get "reference to a pointer".
Here:
char*p = "hello";
char*&k = p;
while creating reference to a pointer why we have to write *&k , why not &*k?
I've read about the difference between the two, what one can do that the other can't syntactially, but I'm still very confused by the concept of references in C++. I'm trying to understand what is the difference in terms of undelying implementation.
Take, for example, the following code:
void foo(int *bar) {
++*bar;
}
int main(void) {
int n = 0;
foo(&n);
return 0;
} Translated into: Code: __Z3fooPi:
[Code] ....
Considering the above assembly code, and the memory that was used in both cases, I'm tempted to say that references was added to C++ to make the synax looks prettier... Where would I be wrong?
What is the difference in pass by pointer and pass by reference? As per my understanding, there is no difference much.If a function accepts pointer, then NULL check can be performed.other than this i'm not able to see any big difference..
View 3 Replies View RelatedI have a struct which has an array inside of it:
struct someStruct{
int structArray[999];}
Now when I want to access that array, I have the following:
ptrSomeStruct->structArray[someIndex];
But now I want to pass structArray to this function by reference so that it can adjust the array as needed and I can continue to use the array back in my caller function:
void adjustArray(void *& someArray){}
How do I accomplish this?
I just want to know if there is any real difference between the two below, if yes, when would i use one over the other? I would thought the "&" is pointless in below function, as far as the data is concerned.., the only things is with "&", if the pointer address value is changed in Test function, it will affect the caller's copy of data. Both function should behave the same if data is changed.
Code:
Between
void Test(QSharedPointer<Data> data)
{
}
and
void Test(QSharedPointer<Data> & data)
{
}
what is the advantage of using reference over constant pointer?
Can they be used interchangeably?
i'm still unclear between the difference between using pointer and a reference
I understood the concept of pointers in c in the class i took last year
and that was to change the actual value stored in the memory address Code:
void change_a(int a*){
a=6;
}
int main(){
int a=5;
change_a(&a);
}
but in c++ I've been using references in all my assignments because I don't know how to correctly use pointers in c++ I may have missed a class but I'm on spring break and would like to clear things up
so in c++
in my assignments I would call it like this Code:
void change_a(int &a){
a=6;
}
int main(){
int a=5;
change_a(a);
}
so does this change the value in the address or does it make another copy of a in my c++ code and stores 6 in that copy
I have to write an example in which you use a function having as argument a pointer passed by reference in C++. Can you provide an example like this:
funz.h : void funz( int *&a );
funz.cpp : ? (1)
main.cpp:
#include "funz.h"
#include <iostream>
[Code]...
as I write in (1) and (2) ?
I don't understand how my code not run.
#include "stdafx.h"
#include<iostream>
using namespace std;
struct student{
char name[30];
char birthday[20];
char homeness[50];
float math;
[Code] ....