C array of pointers and malloc

Question

I'm trying to get data from std stored into an array, using pointers. The main declares d as int *d; and the function is called using x = getdata(&d); When I step through it with gdb it dies at the first iteration of the for loop, when it tries *d[i]=val;

int getdata(int **d)
{ 
    int count,val,i,j=0; 

    scanf("%d", &count);

    d = malloc(sizeof *d * count);
    for( i = 0; i < count-1; i++) {

    scanf("%d",val);

    *d[i]=val;  

    }

    for ( i = 0; i < count; i++)
            printf("Number %d\n",*d[i]);

    return count;

}

scanf("%d", &val); BTW: a serious compiler should warn you about this. — wildplasser
– wildplasser, Commented Jan 18, 2013 at 0:39

AnT stands with Russia · Accepted Answer · 2013-01-18 00:52:45Z

2

The memory should be allocated as follows

*d = malloc(count * sizeof **d);

The values should be accessed as follows

(*d)[i] = val;

It is also not clear why you allocate count elements and only initialize count - 1 elements in the input cycle (and later print all count elements in output cycle).

edited Jan 18, 2013 at 0:52

answered Jan 18, 2013 at 0:35

AnT stands with Russia

323k44 gold badges548 silver badges793 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

AdamTopi Over a year ago

thanks, it was this and me forgetting the & in scanf("%d",&val) !

Pavan Yalamanchili · Accepted Answer · 2013-01-18 00:28:20Z

2

*d = malloc(count * sizeof(int));

then

(*d)[i] = val

answered Jan 18, 2013 at 0:28

Pavan Yalamanchili

12.1k2 gold badges38 silver badges56 bronze badges

Comments

Mel Nicholson · Accepted Answer · 2013-01-18 00:40:28Z

2

What you have is a pointer to an array, not an array of pointers.

1) malloc returns a void* so you need to assign the result to *d instead of d

2) the data size you want is of an int, not an int* (using *d gets you an int* where **d is an int)

*d = malloc(sizeof(**d) * count);

3) Indexing the array requires slightly different syntax

(*d)[i] = val;

printf("%d\n", (*d)[i]);

4) The second scanf needs a pointer, not an int

scanf("%d", &val);

edited Jan 18, 2013 at 0:40

answered Jan 18, 2013 at 0:27

Mel Nicholson

3,22416 silver badges25 bronze badges

5 Comments

wildplasser Over a year ago

*d = malloc(count * sizeof **d ); is more robust than sizeof type .

Mel Nicholson Over a year ago

I agree with this. Edited

wildplasser Over a year ago

you don't need the parentheses with sizeof expression (that's why I put "count" in front)

AdamTopi Over a year ago

I'm still getting a segmentation fault

Mel Nicholson Over a year ago

I missed the &val on your second scanf.

Collectives™ on Stack Overflow

C array of pointers and malloc

3 Answers 3

1 Comment

Comments

5 Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

5 Comments

Your Answer

Sign up or log in

Post as a guest

Related