C++ initial value of reference to non-const must be an lvalue

问题

I'm trying to send value into function using reference pointer but it gave me a completely non-obvious error to me

#include "stdafx.h"
#include <iostream>

using namespace std;

void test(float *&x){

    *x = 1000;
}

int main(){
    float nKByte = 100.0;
    test(&nKByte);
    cout << nKByte << " megabytes" << endl;
    cin.get();
}

Error : initial value of reference to non-const must be an lvalue

I have no idea what I must do to repair above code, can someone give me some ideas on how to fix that code? thanks :)

回答1:

When you pass a pointer by a non-const reference, you are telling the compiler that you are going to modify that pointer's value. Your code does not do that, but the compiler thinks that it does, or plans to do it in the future.

To fix this error, either declare x constant

// This tells the compiler that you are not planning to modify the pointer
// passed by reference
void test(float * const &x){
    *x = 1000;
}

or make a variable to which you assign a pointer to nKByte before calling test:

float nKByte = 100.0;
// If "test()" decides to modify `x`, the modification will be reflected in nKBytePtr
float *nKBytePtr = &nKByte;
test(nKBytePtr);

回答2:

The &nKByte creates a temporary value, which cannot be bound to a reference to non-const.

You could change void test(float *&x) to void test(float * const &x) or you could just drop the pointer altogether and use void test(float &x); /*...*/ test(nKByte);.

回答3:

When you call test with &nKByte, the address-of operator creates a temporary value, and you can't normally have references to temporary values because they are, well, temporary.

Either do not use a reference for the argument, or better yet don't use a pointer.

来源：https://stackoverflow.com/questions/17771406/c-initial-value-of-reference-to-non-const-must-be-an-lvalue

标签

c++

pointers

reference