C ++从int *到int的无效转换

Question

I have the following C++ code:

#include <iostream>

using namespace std;
template<class T>
T f(T x, T y)
{
    return x+y;
}
int f(int x, int y)
{
    return x-y;
}
int main()
{
    int *a=new int(3), b(23);
    cout<<f(a,b);
    return 0;
}

I get these errors: invalid conversion from 'int*' to 'int'
initializing argument 1 of 'int f(int, int)' . What does it mean ?

Thanks

Answer 1

You need to dereference a in your call to f . This is because a is a pointer to an int , while b is a bare int . Dereferencing a pointer to an int returns an int . This results in the function f being called with two ints.

cout<<f(*a,b);

Answer 2

Please tell us what exectly you want to achieve ??

You need to use * operator to get value from 'a' pointer.

f(*a,b) //this invocation should work.

regardless of the fact that this code compiles without errors, it is ugly

Answer 3

int f(int x, int y)
{
    return x-y;
}

In your above function, first parameter expects an integer argument, and not an address pointing to an integer value. So what you need to make sure is that the data types of your FUNCTION PARAMETERS must match the data types of your FUNCTION ARGUMENTS. Otherwise, such compilation errors will always pop up at compilation time.

So in your case, there are following solutions to this problem:

1. change your function parameter int x to int* x Or
2. change your f(a,b) to f(*a,b) so that value of a is sent not address Or
3. declare a as a simple integer like you did for b. Then while calling f function, send address of variable a. Make sure for this option, function parameter int x should be changed to int* x