I've written a simple test program to try to learn how to use template static member functions in C++. The code compiles, but doesn't work right (prints out some garbage). I guess I'm using the right syntax. I've read this or this and some other stuff but still don't know what I'm doing wrong. The code below:
#include <iostream>
using namespace std;
class Util {
public:
Util();
virtual ~Util();
template <typename T> static void printTab(T tab[]);
};
template <typename T>
void Util::printTab(T tab[]) {
for (unsigned int i=0; i<sizeof(tab)/sizeof(tab[0]); i++) {
cout << tab[0] << " ";
}
cout << endl;
}
int main() {
float tabFloat[5] {1, 2, 3, 4, 5};
unsigned char tabChar[3] {1, 2, 3};
Util::printTab(tabFloat);
Util::printTab(tabChar);
return 0;
}
Any hints appreciated.
You need to pass the size as another template argument :
#include <iostream>
using namespace std;
class Util {
public:
Util();
virtual ~Util();
template <typename T,int N> static void printTab(T (&tab)[N])
{
for (int i=0; i<N; i++) {
cout << tab[i] << " ";
}
cout << endl;
}
};
int main() {
float tabFloat[5] {1, 2, 3, 4, 5};
unsigned char tabChar[3] {1, 2, 3};
Util::printTab(tabFloat);
Util::printTab(tabChar);
}
sizeof(tab) is the size of a T*, it will not return the size of the whole array. You need to pass that in yourself as another argument to the function. See here for an explanation and another potential workaround: When a function has a specific-size array parameter, why is it replaced with a pointer?
Note that the second printTab will not output readable characters. If you want to see something printed out, try with:
unsigned char tabChar[3] {'1', '2', '3'};
How about trying, you need to send the size of the array when calling a function:
#include <iostream>
using namespace std;
class Util {
public:
Util();
virtual ~Util();
template <typename T> static void printTab(T tab[], size_t sz);
};
template <typename T>
void Util::printTab(T tab[], size_t sz) {
for (unsigned int i=0; i<sz; i++) {
cout << tab[i] << " ";
}
cout << endl;
}
int main() {
float tabFloat[5] {1, 2, 3, 4, 5};
unsigned char tabChar[3] {1, 2, 3};
Util::printTab(tabFloat, sizeof(tabFloat)/sizeof(float));
Util::printTab(tabChar, sizeof(tabChar)/sizeof(char));
return 0;
}
I'd pass the number of elements of T as a function argument or use a STD container such as a Vector.
Your for loop is just printing the first element tab[0] not tab[i]
Your initialization of tabFloat and tabChar are missing =
float tabFloat[5] {1, 2, 3, 4, 5};
unsigned char tabChar[3] {1, 2, 3};
(also I'd use 65, 66, 67 instead of 1,2,3 for console readability in your testing)
float tabFloat[5] = {1, 2, 3, 4, 5};
unsigned char tabChar[3] = { 65, 66, 67};
Related
#include <algorithm>
#include <array>
#include <iostream>
int main() {
std::array<int, 10> s{5, 7, 4, 2, 8, 6, 1, 9, 0, 3};
struct {
bool operator()(int a, int b) const
{
return a < b;
}
} customLess;
std::sort(s.begin(), s.end(), customLess);
for (auto a : s) {
std::cout << a << " ";
}
std::cout << '\n';
return 0;
}
I want to sort the array using std : : sort. The code above works fine, but when I try with template array in my main task (GArr<int, 5> arr{5, 2, 3, 4, 1};) the compiler gives the following
error: no match for 'operator-' (operand types are 'Iterator' and
'Iterator')|
How can I fix it?
What can be wrong this piece of code.
I'm trying to call countLessThan3 by instantiating with std::vector.
// 3. Lambdas
template<typename T>
const auto countLessThan3(const T & vec, int value)
{
const auto count = std::count(vec.begin(), vec.end(),
[](int i){ return i < 3;}
);
return count;
}
int main(int argc, char const *argv[])
{
// 3
std::vector<int> vector = {1, 2, 3, 4, 5, 2, 2, 2};
countLessThan3<std::vector<int>>(vector, 3);
return 0;
}
compiled with g++ -std=c++14 1.cpp -o 1 on linux.
Ther are some problems:
The code uses std::count, but a lambda is passed, so it should be std::count_if instead.
value is passed to the function as a parameter, but it is not used and there is an hard coded 3 in the lambda.
Other minor issues are fixed in the snippet below
#include <iostream>
#include <vector>
#include <algorithm>
template<typename T>
auto countLessThan(const T & vec, typename T::value_type value)
{
return std::count_if(vec.begin(), vec.end(),
[value](typename T::value_type i){ return i < value;}
);
}
int main()
{
std::vector<int> vector = {1, 2, 3, 4, 5, 2, 2, 2};
std::cout << countLessThan(vector, 3) << '\n';
return 0;
}
I am trying to write a function which would Print Data on the console. The function is to be templated as it should accept different types of Data.
The code is as shown below:
template<typename DataType>
void PrintData(DataType *X)
{
for (DataType Data : X)
{
cout << Data << "\t";
}
cout << endl;
}
int main()
{
int nArray[7] = { 7, 5, 4, 3, 9, 8, 6 };
double dArray[5] = { 4.3, 2.5, -0.9, 100.2, 3.0 };
PrintData(nArray);
PrintData(dArray);
system("pause");
return EXIT_SUCCESS;
}
I get an error that variable Data is undeclared in the templated function PrintData.
error C2065: 'Data' : undeclared identifier
error C3312: no callable 'begin' function found for type 'double *'
error C3312: no callable 'begin' function found for type 'int *'
error C3312: no callable 'end' function found for type 'double *'
error C3312: no callable 'end' function found for type 'int *'
Any help would be appreciated.
Thanks
Assuming you have included the iostream header file and the using namespace std; directive. Then your problems are:
You should not use DataType *. Your code makes X a pointer, which is different from array. Use DataType const& or DataType&& instead.
You have to include the iterator header file which provides the begin and end function for C-style array.
The following code works for me.
#include <iostream>
#include <iterator>
using namespace std;
template<typename DataType>
void PrintData(DataType const& X)
{
for (auto Data : X)
{
cout << Data << "\t";
}
cout << endl;
}
int main()
{
int nArray[7] = { 7, 5, 4, 3, 9, 8, 6 };
double dArray[5] = { 4.3, 2.5, -0.9, 100.2, 3.0 };
PrintData(nArray);
PrintData(dArray);
return EXIT_SUCCESS;
}
As commented by Igor Tandetnik, you may use template<struct DataType, size_t N> if you want to refer the size of the array.
Update:
With template<struct DataType> and DataType *X, DataType is deduced as int and double, and X is a pointer which is not a container.
With template<struct DataType, size_t N> and DataType (&X)[N], DataType is deduced as int and double, and X is an array which can be used with range-based for loop.
With template<struct DataType> and DataType&& X or DataType const& X, DataType is deduced as int[7] or double[5], and X is an array as well.
The problem is that you're passing the name of the array (e.g., nArray), which is just a pointer to the first element, basically. However, new-style for loops expect something on which you can call begin and end - a range.
The following changes make it work by using a vector instead of an array. Note that there are very few differences otherwise, and, in general, very few reason to use C-style arrays in contemporary C++.
#include <iostream>
#include <vector>
using namespace std;
template<typename DataType>
void PrintData(const DataType &X)
{
for (const auto &Data : X)
{
cout << Data << "\t";
}
cout << endl;
}
int main()
{
vector<int> nArray = { 7, 5, 4, 3, 9, 8, 6 };
vector<double> dArray = { 4.3, 2.5, -0.9, 100.2, 3.0 };
PrintData(nArray);
PrintData(dArray);
system("pause");
return EXIT_SUCCESS;
You have some problems:
1) DataType must be a container. So, try to use:
std::vector<int> nArray = { 7, 5, 4, 3, 9, 8, 6 };
std::vector<double> dArray = { 4.3, 2.5, -0.9, 100.2, 3.0 };
2) You are not going to change the container. It's better to pass container by const reference:
template<typename DataType>
void PrintData(const DataType & X);
3) Change the loop like this:
for (const auto & value : X) {
std::cout << value << "\t";
}
Code example:
#include <iostream>
#include <vector>
template<typename DataType>
void PrintData(const DataType & X) {
for (const auto & value : X) {
std::cout << value << "\t";
}
std::cout << std::endl;
}
int main() {
std::vector<int> nArray = { 7, 5, 4, 3, 9, 8, 6 };
std::vector<double> dArray = { 4.3, 2.5, -0.9, 100.2, 3.0 };
PrintData(nArray);
PrintData(dArray);
return 0;
}
As Igor suggested, If at all you wish to use DataType * then you need to do something like this
#include <iostream>
#include <iterator>
using namespace std;
template <typename DataType, size_t N>
void PrintData(DataType (&X)[N])
{
for (auto i : X)
cout << i << "\t";
cout << endl;
}
int main()
{
int nArray[7] = { 7, 5, 4, 3, 9, 8, 6 };
double dArray[5] = { 4.3, 2.5, -0.9, 100.2, 3.0 };
PrintData(nArray);
PrintData(dArray);
return EXIT_SUCCESS;
}
Output
7 5 4 3 9 8 6
4.3 2.5 -0.9 100.2 3
Explanation:
If you see void PrintData(int* nArray); & void PrintData(int (&nArray)[7] );
are similar declarations, except that seconds one tells where array ends.
Template function
template <typename DataType, size_t N>
void PrintData(DataType (&X)[N])
is deduced as'
void PrintData(int (&nArray)[7] )
You could also write
void PrintData(int (&nArray)[7] )
{
for (auto i : nArray)
cout << i << "\t";
cout << endl;
}
I wrote a c++ program as fllow(3.43.cpp):
#include <iostream>
using std::cout;
using std::endl;
void version_1(int **arr) {
for (const int (&p)[4] : arr) {
for (int q : p) {
cout << q << " ";
}
cout << endl;
}
}
int main() {
int arr[3][4] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12};
version_1(arr);
return 0;
}
Then I compile it by using: gcc my.cpp -std=c++11, there is an error I can not deal with.
Info:
3.43.cpp:6:30: error: no matching function for call to ‘begin(int**&)’
for (const int (&p)[4] : arr) {
^
3.43.cpp:6:30: note: candidates are:
In file included from /usr/include/c++/4.8.2/bits/basic_string.h:42:0,
from /usr/include/c++/4.8.2/string:52,
from /usr/include/c++/4.8.2/bits/locale_classes.h:40,
from /usr/include/c++/4.8.2/bits/ios_base.h:41,
from /usr/include/c++/4.8.2/ios:42,
from /usr/include/c++/4.8.2/ostream:38,
from /usr/include/c++/4.8.2/iostream:39,
from 3.43.cpp:1:
/usr/include/c++/4.8.2/initializer_list:89:5: note: template<class _Tp> constexpr const _Tp* std::begin(std::initializer_list<_Tp>)
begin(initializer_list<_Tp> __ils) noexcept
I search it in google, but not find similar answer.
Since arr is just a pointer, there's no way to deduce how big it is. But, since you are actually passing in a real array, you can just template your function on its dimensions so you take the actual array by reference rather than having it decay to a pointer:
template <size_t X, size_t Y>
void foo(const int (&arr)[X][Y])
{
std::cout << "Dimensions are: " << X << "x" << Y << std::endl;
for (const int (&row)[Y] : arr) {
for (int val : row) {
std::cout << val << ' ';
}
std::cout << std::endl;
}
}
int main() {
int arr[3][4] = {{1, 2, 3, 4}, {5, 6, 7, 8}, {9, 10, 11, 12}};
foo(arr);
}
std::begin() and std::end() won't work for pointers. They only work for arrays. If you want to deduce the size of your array you'll need to pass it as a reference your function:
#include <cstddef>
template <std::size_t A, std::size_t B>
void version_1(int (&arr)[B][A]) {
for (const int (&p)[A] : arr) {
for (int q : p) {
cout << q << " ";
}
cout << '\n';
}
}
Pointers are not the same as arrays. To be able to use range based for, your container must support std::begin and std::end. Standard C arrays can be used, but not pointers.
simple call by reference
void foo(int* A)
{
// ...
}
void main()
{
int A[] = {1,1,1,1,1,1,1,1,1};
foo(A);
}
not sure why but it is lessening the size of the array and losing/leaking information on the array....
You're passing a pointer to the first element of an array. Create your function with the prototype
void foo(int* A, int size);
You will still be able to access A[0...size-1] like normal.
You should totally drop C-style arrays and use std::array instead. Just compare this (which is the solution to your problem):
void foo(int* A, std::size_t size) {
// ...
}
int main() {
int A[] = {1,1,1,1,1,1,1,1,1};
foo(A, (sizeof(A) / sizeof(int)));
}
to:
template<std::size_t Size>
void foo(const std::array<int, Size>& array) {
// ...
}
int main() {
std::array<int, 9> A {{ 1, 1, 1, 1, 1, 1, 1, 1, 1 }};
foo(A);
}
Ain't it beautiful? Or just take a look at how gorgeous it is with std::vector:
void foo(const std::vector<int>& vector) {
// vector.size() is the size
}
int main() {
std::vector<int> A = { 1, 1, 1, 1, 1, 1, 1, 1, 1 };
foo(A);
}
And if you really want foo to be a generic algorithm, I'll just blow your mind with iterators:
template<class Iterator>
void foo(Iterator begin, Iterator end) {
// ...
}
int main() {
std::array<int, 9> A {{ 1, 1, 1, 1, 1, 1, 1, 1, 1 }};
std::vector<int> B = { 1, 1, 1, 1, 1, 1, 1, 1, 1 };
foo(A.begin(), A.end()); // not a single problem
foo(B.begin(), B.end()); // was given that day
}
C++ has an amazing (arguably) standard library and an amazing type system (if you don't overlook C legacy "features", like void*): use them.
not sure why but it is lessening the size of the array and losing/leaking information on the array....
In foo(), sizeof(A) == 8 not because it is "leaking" information, but because 8 is the size of the pointer of type int*. This is true regardless of how many integers A was initialized with in main().
This may shine some light on what is happening:
#include<iostream>
using namespace std;
void foo(int* A)
{
cout << "foo: " << sizeof(A) << endl; // 8
}
void bar(int A[])
{
cout << "bar: " << sizeof(A) << endl; // still 8
}
int main()
{
int A[] = {1,1,1,1,1,1,1,1,1};
cout << "main: " << sizeof(A) << endl; // 36 (=4*9)
foo(A);
bar(A);
return 0;
}
Output:
main: 36
foo: 8
bar: 8
In main, sizeof "knows" the size of A[] - it is sizeof(int) * length = 4 * 9 = 36. This information is lost when A[] is cast to a pointer A* in foo.
What if we pass in A as bar(int A[]) instead. Will that retain the array length? No! In that case, sizeof(A) is still 8, the size of the pointer. Only in main does the compiler retain the information of array size of A.
If you want your functions to know the size of the array, use the std::vector<int> template, or pass in the size separately.
Here's another discussion on this: When a function has a specific-size array parameter, why is it replaced with a pointer?
The correct way to pass array by reference is
void foo(int (&A) [9])
{
// sizeof(A) == sizeof(int) * 9
}
the generic way is so:
template <std::size_t N>
void foo(int (&A) [N])
{
// sizeof(A) == sizeof(int) * N
}
You may use std::array (C++11 required) which has a syntax more intuitive
template <std::size_t N>
void foo(std::array<int, N> &A)
{
// A.size() == N
}