Allocating memory with malloc getting error

Question

int n;
scanf("%d",&n);
int *score;
score=(int *)malloc(sizeof(int)*n);
int i;
for (i=0;i<n;i++)
{
    scanf("%d",sizeof(int)*i+score);
}
printf("ok");

In the above code I get an error but when I comment the last line printf the program runs correctly. what is the problem?? (I want to give n number from user without using arrays)

`scanf("%d",sizeof(int)*i+score);`-->`scanf("%d",&score[i]);` and `score=(int *)malloc(sizeof(int)*n);`-->`score=malloc(sizeof(int)*n);` — user2736738, Dec 25 '17 at 07:41
of `scanf("%d", &score[i]);`. And check return value of scanf. — M.M, Dec 25 '17 at 07:50
@M.M.: That's what was being mentioned and then OP commented a requirement *I shouldn't use [ or ]* — user2736738, Dec 25 '17 at 08:00

score 2 · Accepted Answer · answered Dec 25 '17 at 07:43

Pointer arithmetic of the form score + i is already done in multiples of sizeof(*score). So when you write score + i * sizeof(int) you doubly multiply by the size of the items. You reach way beyond the bounds of the buffer.

Either write it simply as score + i, or if you insist on doing the multiplication yourself, be sure to cast to a character pointer type first:

(int*)((char*)score + i * sizeof(int))

Oh, and don't cast the result of malloc. C doesn't require it, and it's somewhat unidiomatic.

score 0 · Answer 2 · answered Dec 25 '17 at 07:45

0

scanf("%d",sizeof(int)*i+score);

pointer arithmetic uses the pointer type, so here you are moving to sizeof(int)isizeof(int) bytes after score, instead just use scores+i.

answered Dec 25 '17 at 07:45

SoronelHaetir

14,104
1
12
23

Allocating memory with malloc getting error

2 Answers2