Changing an array value from a function in C

Question

The program crashes right in the instruction mentioned in the source code (I didn't write all the code because it's too long)

int main()
{
char screen[24][80];

//......every thing is well until this instruction

backgrounds(5,screen);

//......the program doesn't execute the rest of the code

}
//______________________________________________________

//this is a header file

void backgrounds(int choice,char **screen)
{   
    if(choice==5)
    {
        screen[18][18]='-';
        screen[18][19]='-';
        screen[18][20]='-';
    }
}

A 2D array is not the same as a pointer to a pointer. See this — samgak
– samgak, Commented Nov 30, 2016 at 21:53
also having a function in a header file is (generally) a Bad Idea(tm) — pm100
– pm100, Commented Nov 30, 2016 at 23:27
this line: backgrounds(5,screen); is not passing a **screen Suggest the backgrounds() function signature be: void backgrounds( int choice, char screen[][80] ) — user3629249
– user3629249, Commented Dec 1, 2016 at 17:39
@samgak I see, now I understand how it's working thank you ! — heiwiper
– heiwiper, Commented Dec 1, 2016 at 18:04

dbush · Accepted Answer · 2016-11-30 23:38:20Z

3

A char [24][80] cannot be converted to a char **.

When passed to a function, an array decays into a pointer to its first element. This is simple for a 1 dimensional array, but less so for higher dimensions.

In this case, a char [24][80] is an array of char [80]. So passing a variable of this type to a function yields a char (*)[80].

Change your function definition to either this:

void backgrounds(int choice,char (*screen)[80])

Or this:

void backgrounds(int choice,char screen[24][80])

Or you can use a variable length array for maximum flexibility:

void backgrounds(int choice, int x, int y, char screen[x][y])

edited Nov 30, 2016 at 23:38

answered Nov 30, 2016 at 21:53

dbush

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5 Comments

too honest for this site Over a year ago

"This is straightforward for a 1 dimensional array, but not for higher dimensions." - That's wrong. It is very well straight-forward: It decays to a pointer to an array of one less dimension: a 2D array decays to a pointer to a 1D array, a 3D array to a pointer to 2D array, etc. As you wrote: an array decays to a pointer to its element. An element of a (n)D-array is an (n-1)D array. With a 0D array being a single element.

dbush Over a year ago

@Olaf "straightforward" probably wasn't the best choice of words. Changed to "simple".

too honest for this site Over a year ago

I'm still not happy with this. Maybe I'm too deep into this, but to me it as both, straight-forward and obvious: an array always decays to a pointer to the first element. There is no exception how many dimensions. Actually the confusion is that ppl don't stop after this decaing, but think it recurses to the next dimension and so on. But that's not obvious to me. Anyway, I don't intend to DV for this, so feel free to ignore my comment :-)

pm100 Over a year ago

I would just say 'do this' and have the direct [24][80] version

too honest for this site Over a year ago

@pm100: Agreed with a slight modification: No magic numbers. Either a VLA with dimensions passed (most flexible) or macros with the constants.

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