Assign vector<bool> - c++

I'm trying to assign a vector bool as follows:
#include <iostream>
#include <vector>
using namespace std;
struct item {
int value;
vector<bool> pb;
vector<bool> *path = &pb;
};
int main(int argc, char* argv[]) {
vector<item> dp(10);
for (int n = 0; n < 10; n++)
dp[n].pb = vector<bool>(10);
}
It compiles ok but get the following runtime error:
An invalid parameter was passed to a function that considers invalid parameters fatal.
The value of dp and n are
How can I assign a vector bool to an existing variable?

Related

Create a class run at main() get undefined reference to... error [duplicate]

This question already has answers here:
Undefined reference to a static member
(5 answers)
What is an undefined reference/unresolved external symbol error and how do I fix it?
(39 answers)
Closed last year.
arrayADT.h
#include <iostream>
using namespace std;
template <class T>
class arrayADT
{
private:
T *A;
static int size;
static int length;
public:
arrayADT(){
size=10;
A= new T[size];
length=0;
}
void increaseSize(){
T *p;
size=size*2;
p= new T[size];
delete[] A;
A=p;
p=NULL;
}
int getSize(){
return size;
}
~arrayADT();};
example.cpp
#include<iostream>
#include<stdio.h>
#include"arrayADT.h"
using namespace std;
int main(int argc, char const *argv[])
{
arrayADT<int> s;
s.increaseSize();
s.getSize();
return 0;
}
Get error:
undefined reference to `arrayADT::~arrayADT()'
undefined reference to `arrayADT::~arrayADT()'
undefined reference to `arrayADT::length'
undefined reference to `arrayADT::size'
Can anyone help me? Thanks very much!
I found two different problems:
The destructor method must have a scope with {} even if it is empty.
I removed the static keyword as each object will have size and length property.
#include <iostream>
#include <stdio.h>
#include <string>
using namespace std;
template<class T>
class arrayADT
{
private:
T *A;
int size; /* Removed the static keyword used in data members. */
int length; /* Removed the static keyword used in data members. */
public:
arrayADT()
{
size = 10;
A = new T[size];
length = 0;
}
~arrayADT(){} /* The destruction method has been edited. */
void increaseSize()
{
T *p;
size = size * 2;
p = new T[size];
delete[] A;
A = p;
p = NULL;
}
int getSize()
{
return size;
}
};
int main(int argc, char const *argv[])
{
arrayADT<int> s;
s.increaseSize();
cout << s.getSize() << endl;
return 0;
}

How to create and call function with same name as other function?

I am working on my own library and I want to create function max(). I know that function like this exists in C++ and it isn't in namespace std, so erasing using namespace std; won't help. I am creating this function in my namespace like this:
namespace ml
{
template<typename T>T max(T cntr, int size)//I'm getting errors here
{
sort(cntr,0,size-1);//my function which just sorts elements, it's working fine
return cntr[size-1];
}
}
Here is my main function:
#include <iostream>
#include <ctime>
#include "mylib.hpp"
int main()
{
srand(time(NULL));
int* arr, n;
std::cin>>n;
arr = new int [n];
for(int i = 0; i < n; i++)
{
arr[i] = rand()%100;
}
int maximum = ml::max(arr,n);//I'm getting errors here
std::cout<<maximum<<'\n';
return 0;
}
Sorry for grammatical mistakes if i've done so.
If the purpose of the function is to search a C-style array, the signature should be template <typename T> T max(T* cntr, int size). (note the T* as the type of cntr) That way, when it's called with an int*, T is deduced as int, and that's the correct return type.

C++ vector of class objects and dynamic memory allocation

This is a simplified version of what I want to do, it crushes when push_back is called, specifically when the destructor is called. If I delete the body of the destructor it works, but I want to be sure that it is deleted. Instead of calloc/free I tried new/delete with the same result. What do I do wrong here?
#include <cstdlib>
#include <vector>
using namespace std;
class Buffer
{
public:
float *DATA;
Buffer(int length) {DATA = (float*) calloc (length,sizeof(float));};
virtual ~Buffer(){free (DATA);};
private:
};
int main(int argc, char** argv)
{
vector <Buffer> v;
for (int i =0; i<10; i++)
v.push_back(Buffer(1000));
return 0;
}
Here's a working code: https://godbolt.org/z/ex9oMG.
#include <cstdlib>
#include <vector>
using namespace std;
class Buffer
{
public:
float *DATA;
Buffer(int length) {DATA = (float*) calloc (length,sizeof(float));};
Buffer(const Buffer &buffer) = delete;
Buffer(Buffer&& buffer) {
DATA = buffer.DATA;
buffer.DATA = nullptr;
}
~Buffer(){
if (DATA) free(DATA);
};
private:
};
int main(int argc, char** argv)
{
vector <Buffer> v;
for (int i =0; i<10; i++)
v.push_back(Buffer(1000));
return 0;
}
You need to define a move constructor here, primarily because v.push_back(Buffer(1000)) requires a move op otherwise the deletion of the original copy would free the resource.
I've explicitly deleted the copy ctor because when handling such resources - copy should not be allowed.

Using functions within functions

Below is a piece of code I was trying to run where I have function I want to run (dN) within my main function which returns a value of type complex<double>.
#include <iostream>
#include <complex>
#include <cmath>
using namespace std;
const complex<double> Im1(0.0,1.0); //imaginary number definition
class Functions{
public:
complex<double> dN(complex<double> **N, int k, int i, complex<double> kN, double T1){
complex<double> OUT = Im1*(N[k][i]+kN)/(T1);
return OUT;
};
};
int main(int argc, const char * argv[]) {
//...more code here
complex<int> **NM = new complex<int>*[1000]; //1000x500 array
//run loop to initialize
for (int i = 0; i < 1000; ++i)
{
NM[i] = new complex<int>[500];
}
complex<double> dN_OUT = Functions::dN(**NM,1,20,0.,20.);
return 0;
};
Although this code does not run and returns the error: Call to non-static member function without an object argument
Based on my understanding, C++ does not allow nested functions and my code above would not work because I call a separate function in my main function. Although (based on the link) it does appear one can implement "local classes" by defining a function within a struct which would have to be inside the main function. Although when I try doing that:
int main(int argc, const char * argv[]) {
complex<int> **NM = new complex<int>*[1000]; //1000x500 array
//run loop to initialize
for (int i = 0; i < 1000; ++i)
{
NM[i] = new complex<int>[500];
}
struct Functions{
complex<double> dN(complex<double> **N, int k, int i, complex<double> kN, double T1){
complex<double> OUT = Im1*(N[k][i]+kN)/(T1);
return OUT;
};
};
complex<double> dN_OUT = Functions::dN(**NM,1,20,0.,20.);
return 0;
};
the error persists. Ultimately, I am simply wanting to use the function dN that returns output of type complex<double> within my main function, but am unsure of the best/operational way to implement this.
I believe you misunderstand what a nested function is.
A nested function would look like this:
int main()
{
void nested() {} // not allowed in C++
}
The solution to your problem is in the error message provided by your compiler:
Call to non-static member function without an object argument
Take a look at the following:
// Example 1
struct Functions {
void func() {}
};
int main()
{
// to call Functions::func() you would need to have an object
// of type Functions because Functions::func() is not a static function
Functions f;
f.func();
}
// Example 2
// by making func() static you can call it without an object:
struct Functions {
static void func() {}
};
int main()
{
Functions::func(); // OK
}
Ultimately, I am simply wanting to use the function dN that returns output of type complex within my main function, but am unsure of the best/operational way to implement this.
Use a free function, like main is, unless dN has a specific reason to be part of a class:
complex<double> dN(complex<double> **N, int k, int i, complex<double> kN, double T1)
{
...
}
int main(int argc, const char * argv[]) {
...
//like this
complex<double> dN_OUT = dN(NM,1,20,0.,20.);
//not like this
//complex<double> dN_OUT = dN(**NM,1,20,0.,20.);
}
Option 1:
You can do this with class like below
#include <iostream>
#include <complex>
#include <cmath>
using namespace std;
const complex<double> Im1 (0.0, 1.0); //imaginary number definition
class Functions {
public:
complex<double> dN (complex<double> **N, int k, int i, complex<double> kN, double T1)
{
complex<double> OUT = Im1*(N[k][i] + kN) / (T1);
return OUT;
};
};
int main (int argc, const char * argv[]) {
//...more code here
complex<double> **NM = new complex<double>*[1000]; //1000x500 array
//run loop to initialize
for (int i = 0; i < 1000; ++i)
{
NM[i] = new complex<double>[500];
}
Functions fun; //create class instance
//call the function NOTE the changes here i.e not correct passing **NM
complex<double> dN_OUT = fun.dN (NM, 1, 20, 0., 20.);
return 0;
};
Option 2 (mentioned by others direct calling with changes instead of **NM you should use NM.
complex<double> dN(complex<double> **N, int k, int i, complex<double> kN, double T1){
...
}
int main(int argc, const char * argv[]) {
...
complex<double> dN_OUT = dN(NM,1,20,0.,20.);
}

usage on c++ function pointer

I'm a newbie to C++, learning pointer of function recently, a little confused by usage of pointer of function;
I practiced the following code:
#include <iostream>
#include <sstream>
using namespace std;
int subtraction(int a,int b){
return a-b;
}
int main(int argc, const char * argv[])
{
int (*minus)(int,int)=subtraction;
cout<<minus(5,4);
return 0;
}
it works well;
so,I try a little variation:
#include <iostream>
#include <sstream>
using namespace std;
int subtraction(int a,int b){
return a-b;
}
int main(int argc, const char * argv[])
{
int *minus(int,int)=subtraction;//only here different!
cout<<minus(5,4);
return 0;
}
I practiced it in Xcode on Mac,it give me Error:
Illegal initializer (only variables can be initialized)
but I think compiler can recognized the two is same,why must have a pair of parenthesizes?
In your original code
int (*minus)(int,int)=subtraction;
declares minus as a function pointer that takes parameter int, int and returns int.
In your second code
int *minus(int,int)=subtraction;
declares minus as a function that takes parameter int, int and returns a pointer int *.
You can use a function name(which is automatically converted to a function pointer) to initialize a function pointer, but you can't initialize a function.
This is a matter of operator precedence. The function call operator () has a higher precedence than the dereference operator *. So you must use parentheses to specify the correct order of evaluation.
int *minus(int, int)
means: First call a function named minus, then dereference the return value (int* in this case).
int (*minus)(int, int)
means: First dereference "minus", which returns a function, and then call that function.
You have tagged your code C++ and using iostream so I can safely assume you are looking for a C++ solution.
In such scenario, its best to use class template std::function instead of the function pointer syntax that is prone to error.
#include <iostream>
#include <sstream>
#include <functional>
int subtraction(int a,int b){
return a-b;
}
int main(int argc, const char * argv[])
{
std::function<int(int,int)> minus = subtraction;
//int (*minus)(int,int)=subtraction;
std::cout<<minus(5,4);
return 0;
}
Alternatively, if you would still want to continue with pointer to function, typedefs are recommended
#include <iostream>
int subtraction(int a,int b){
return a-b;
}
typedef int (*MINUS)(int,int);
int main(int argc, const char * argv[])
{
MINUS minus = subtraction;
//int (*minus)(int,int)=subtraction;
std::cout<<minus(5,4);
return 0;
}
And finally, another widely used option is to use functors.
#include <iostream>
struct MINUS
{
int operator()(int a,int b){
return a-b;
}
};
int main(int argc, const char * argv[])
{
//int (*minus)(int,int)=subtraction;
MINUS minus;
std::cout<<minus(5,4);
return 0;
}