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 stackoverflow.com/questions/330793/…

Shmil The Cat
– Shmil The Cat

2014-04-19 15:20:29 +00:00
Commented Apr 19, 2014 at 15:20 — Shmil The Cat
– Shmil The Cat, Commented Apr 19, 2014 at 15:20

this · Accepted Answer · 2014-04-19 15:30:00Z

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, 2014 at 15:30

this

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5 Comments

this Over a year ago

@LeeDuhem The brackets are unnecessary.

this Over a year ago

@LeeDuhem Yes; No; Operator precedence makes them unnecessary.

Lee Duhem Over a year ago

Yes, you are right, those parentheses are not necessary. I made another mistakes when I test it.

user2166351 Over a year ago

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.

this Over a year ago

You would have an array of integer pointers, without the brackets.

Community · Accepted Answer · 2017-05-23 12:28:34Z

0

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));

edited May 23, 2017 at 12:28

CommunityBot

11 silver badge

answered Apr 19, 2014 at 15:18

MBlanc

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Comments

Mohsen Kamrani · Accepted Answer · 2014-04-19 17:12:35Z

0

error due to

int pulses[][2];

you need to define size !! try it once.

edited Apr 19, 2014 at 17:12

Mohsen Kamrani

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answered Apr 19, 2014 at 15:20

Aman Kaushal

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1 Comment

Aman Kaushal Over a year ago

do int pulses[31][2]; and than write a function to copy upButtonPulses[31][2] to int pulses[31][2]; in struct irPulseSet

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