Why am I getting error in initializing array using pointers in C++ (C++14)?

Question

I have problem initializing array in C++14.

int *arr={1,2,3,4};     // Works in C but does not work in C++

I can't understand the error message it shows

prog.cpp: In function 'int main()':
prog.cpp:6:26: error: scalar object 'arr' requires one element in initializer
     int *arr = {1, 22, 5, 20};   //NOT WORKING IN C++

What is the exact meaning of initializer in the error message?

Link of code of C++ which is not working http://ideone.com/684V5t

Link of Same code in C which works http://ideone.com/RAL3te

If anyone can explain why this is happening it would be nice. Thank you.

"Works in C" -- With loads of warnings telling you that it is wrong. — Spikatrix
– Spikatrix, Commented Aug 26, 2015 at 10:00
gcc gives me this feedback to your C-Code: warning: initialization makes pointer from integer without a cast [enabled by default] int *arr={1,2,3,4};. And this right. You can't put a integer-array initialisation on a int pointer. — Alex44
– Alex44, Commented Aug 26, 2015 at 10:09
int *arr = (int[]){1, 2, 3, 4} will work in C, but not in C++. Without proper tagging, I don't know if that's an answer. — Quentin
– Quentin, Commented Aug 26, 2015 at 10:12

Jabberwocky · Accepted Answer · 2015-08-26 11:36:25Z

4

This doesn't work in C++ because arr is not a variable of type class and therefore there is no size member:

#include <iostream>
using namespace std;

int main() {
    // your code goes here
    int arr[] = {1, 22, 5, 20};         //NOT WORKING IN C++ < yes it does
    cout << arr.size();          // << this does not "work" in C++
    return 0;
}

BTW the error message:

prog.cpp:7:12: error: request for member 'size' in 'arr', which is
               of non-class type 'int [4]' cout<<arr.size();

is quite explicit.

And this should not compile (or you should get at least warnings), neither in C neither in C++:

int *arr={1,2,3,4};

You want this (C and C++):

int arr[] = {1, 22, 5, 20};
printf("%d", sizeof(arr)/sizeof(*arr));

Output will be 4.

In C this may compile depending on the compiler, but you should get warnings:

#include <stdio.h>

int main() {
    // your code goes here
    int *arr={1,2,3,4};
    printf("%d\n", sizeof(arr));
    printf("%d\n", sizeof(*arr));
    printf("%p\n", arr);        
    return 0;
}

The output is :

4
4
0x1

because sizeof(arr) is 4 and sizeof(*arr) which is the same as sizeof(int) because arris an int pointer (assuming a 32 bit system).

The arrpointer will contain 1 because of nthe 1st 1 in the initializer list.

edited Aug 26, 2015 at 11:36

answered Aug 26, 2015 at 10:35

Jabberwocky

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2 Comments

sian Over a year ago

Sorry, I accidently changed the code of ideone link. Here is the actual problem - <ideone.com/684V5t> If you can explain the error message and why am I getting it.

Jabberwocky Over a year ago

@SHWETANK, because it's just plain wrong. In that example arr is a pointer and it's nonsense to initialoze it with more than one value. See also the comments of your questions.

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