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

Question

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

#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?

Can you just drop the pointer, use a plain reference instead? — Micha Wiedenmann, Jul 21 '13 at 10:37
haha, now because im following tutorial at learncpp.com and that site teach me how to use reference pointer in function, so i want to try it too.. :p btw thanks @Micha Wiedenmann — Mohd Shahril, Jul 21 '13 at 10:52
Use a reference to a pointer only when you want to modify a pointer outside the function. — Neil Kirk, Jul 25 '14 at 12:27
Or, you could just overload your test function to accept rvalues like this: ```void test(float *&&x)```. — Nanashi No Gombe, Jan 13 '20 at 14:50

score 73 · Accepted Answer · edited Jul 25 '14 at 12:22

73

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);

edited Jul 25 '14 at 12:22

Rachit Agrawal

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answered Jul 21 '13 at 10:39

Sergey Kalinichenko

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your example with explanation is awesome, so i understand it now `temporary value`, also, your solution `float *nKBytePtr = &nKByte;` also different with another answers, so i have(must!) choose your solution as an answers.. thanks again :D – Mohd Shahril Jul 21 '13 at 10:51

score 14 · Answer 2 · answered Jul 21 '13 at 10:38

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);.

score 9 · Answer 3 · answered Jul 21 '13 at 10:36

9

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.

answered Jul 21 '13 at 10:36

Some programmer dude

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float nKByte = 100.0; float *ptr = &nKByte; test(ptr); – thomas Jul 21 '13 at 10:38
@thomas Yes that would work, but in this case there's no need to use a reference to a pointer. – Some programmer dude Jul 21 '13 at 10:40
thanks for the answers, i have learn new thing that &nKByte create a temporary value and i don't know it until now..and thanks again :) – Mohd Shahril Jul 21 '13 at 10:48

score 0 · Answer 4 · answered Jan 04 '22 at 15:26

0

Simply replace test(&nKByte); with test(nKByte); in main function

answered Jan 04 '22 at 15:26

Anurag Maurya

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score 0 · Answer 5 · answered Jan 10 '22 at 12:23

Simply, initializers of references are expected to be objects not their addresses, but if you need to assign a reference to otherwise (like your code above), declare this reference as const. Why const? because if not, the compiler does think that you need to, later on, modify the pointer itself not what it's pointing to.

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

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