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I am passing a char array by reference but when I return from function and print the array, it displays nothing. What am I doing wrong?
#include <iostream>
using namespace std;
void func(char []);
int main()
{
char a[100];
func(a);
cout << a<<endl;
return 0;
}
void func(char *array)
{
array="Inserting data in array a";
cout << array<<endl;
}
Regards
252
There are two ways. The first is to declare the parameter as a pointer to a const string. The second is to declare the parameter as a reference to an array. Also ypu can write a template function.
Arrays can be passed by reference OR by degrading to a pointer. For example, using char arr[1]; foo(char arr[]). , arr degrades to a pointer; while using char arr[1]; foo(char (&arr)[1]) , arr is passed as a reference. It's notable that the former form is often regarded as ill-formed since the dimension is lost.
Passing arrays to functions in C/C++ are passed by reference. Even though we do not create a reference variable, the compiler passes the pointer to the array, making the original array available for the called function's use. Thus, if the function modifies the array, it will be reflected back to the original array.
A whole array cannot be passed as an argument to a function in C++. You can, however, pass a pointer to an array without an index by specifying the array's name. In C, when we pass an array to a function say fun(), it is always treated as a pointer by fun(). The below example demonstrates the same.
What you can probably do is:
void func( char (& array)[10] ) {
}
Which translates to: pass an array ([..]) of 10 ( [10] ) characters ( char ) by reference ( (& ..) ).
78
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