Cannot assign array to struct property

Question

I'm just learning C, and I'm having problems with assigning an array to a property (pulses).

I have a struct:

typedef struct irPulseSet
{
    int pulseCount;
    int pulses[][2];
} irPulseSet;

I create a new variable with the irPulseSet type that I created above, and define an array:

irPulseSet upButton;

upButton.pulseCount = 31;
int upButtonPulses[31][2] = 
{
    { 0 , 120 },
    { 440 , 360 },
    { 440 , 340 },
    { 440 , 1120 },
    { 420 , 380 },
    { 420 , 360 },
    { 400 , 1140 },
    { 420 , 1120 },
    { 420 , 380 },
    { 420 , 1140 },
    { 420 , 1120 },
    { 440 , 340 },
    { 440 , 360 },
    { 440 , 1120 },
    { 440 , 1120 },
    { 420 , 1120 },
    { 400 , 1140 },
    { 420 , 360 },
    { 440 , 340 },
    { 440 , 360 },
    { 440 , 1140 },
    { 440 , 360 },
    { 440 , 340 },
    { 440 , 380 },
    { 420 , 360 },
    { 440 , 1120 },
    { 440 , 1120 },
    { 440 , 1120 },
    { 440 , 27400 },
    { 7160 , 1500 },
    { 0 , 0 }
};

I then assign that array to a property in the irPulseSet struct.

upButton.pulses = upButtonPulses;

But when I compile, I get the error:

invalid use of flexible array member

What am I doing wrong here?

Read this answer http://stackoverflow.com/questions/330793/how-to-initialize-a-struct-in-ansi-c — Shmil The Cat, Apr 19 '14 at 15:20

score 2 · Accepted Answer · answered Apr 19 '14 at 15:30

2

You have to change the type of the pulses member in the struct to a pointer to a 2d array, before you had a flexible array member which you have allocate dynamically.

typedef struct irPulseSet
{
    int pulseCount;
    int (*pulses)[2];  //pointer to a 2d array

} irPulseSet;

And to set the member you do the same:

upButton.pulses = upButtonPulses;

Or a more clever way to initialize the struct

irPulseSet upButton = { 31 , upButtonPulses } ;

answered Apr 19 '14 at 15:30

this

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@LeeDuhem The brackets are unnecessary. – this Apr 19 '14 at 15:41
@LeeDuhem Yes; No; Operator precedence makes them unnecessary. – this Apr 19 '14 at 15:44
Yes, you are right, those parentheses are not necessary. I made another mistakes when I test it. – Lee Duhem Apr 19 '14 at 15:50
Thanks for this! Just one other thing: what are the brackets around the `*pulses` for? I know the * is used when declaring a pointer, but not sure about the brackets. – user2166351 Apr 19 '14 at 18:56
You would have an array of integer pointers, without the brackets. – this Apr 19 '14 at 18:59

score 0 · Answer 2 · edited May 23 '17 at 12:28

What am I doing wrong here?

You are making an assignment (=) to an array-type. See self.'s answer points in the right direction if you want both the structure and the array to point to the same place in memory. However, if you want a copy of the data, read on.

invalid use of flexible array member

The reason you're getting this error, is that to use a flexible array member, you must allocate the extra space for the array, for example when you malloc'd it. Eg.

irPulseSet upButton = malloc(sizeof(irPulseSet) + sizeof(upButtonPulses));
memcpy(upButton->pulses, upButtonPulses, sizeof(upButtonPulses));

score 0 · Answer 3 · edited Apr 19 '14 at 17:12

0

error due to

int pulses[][2];

you need to define size !! try it once.

edited Apr 19 '14 at 17:12

Mohsen Kamrani

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answered Apr 19 '14 at 15:20

Aman Kaushal

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do int pulses[31][2]; and than write a function to copy upButtonPulses[31][2] to int pulses[31][2]; in struct irPulseSet – Aman Kaushal Apr 19 '14 at 15:26

Cannot assign array to struct property

3 Answers3