将unsigned int *'转换为'unsigned int *＆

Question

I have a function which gets 'unsigned int *& argument The parameter I want to transfer in my code is located in the std::vector<unsigned int> data So,what I do is : I transfer the following parameter &data[0] But get the compilation error:

unsigned int *' to 'unsigned int *&

What can be a work around? Thanks,

Answer 1

&data[0] is an rvalue and cannot be bound to non-const reference.

You can make it work this way:

unsigned int *ptr = &data[0];
func(ptr);

But possibly it's better to just change the signature of your function to

void foo(unsigned int &val); //or any other return type

There is a sense of passing a reference to a pointer in case you want to make a pointer point somewhere else. But I don't see a reason to do so in your case

Answer 2

The expression &data[0] yields indeed an r-value, which your function cannot accept.

A simple work-around if you don't want to alter your function (make sure you understand the reasons it requires a reference):

unsigned int* ptr = &data[0];
func(ptr);

Answer 3

&data[0] is a rvalue, but the function parameter type is non-constant reference to a pointer. A non-const reference can't be bound to an rvalue.

It's hard to suggest a good workaround until OP provides more context for his problem.

Answer 4

Write

unsigned int *argument = &data[0];
call_function(argument);

Be aware that the function may be expecting to change the value of argument , so it may expect you to then write:

data.assign(argument, argument + new_length);