C dynamic allocating memory for array within a struct

Question

First I defined this struct:

typedef struct{ int **_data; int _num_of_lines; int *_lines_len; } Lines;

my goal is to receive _num_of_lines as input from user. which will be used to define number of line of the 2D array data whereas the array _lines_len repreasants length of each line in data

I'm trying to malloc memory for _lines_len, but I always get back that the size of the array is 2, and I don't understand why...

int main(int argc, char *argv[]) { Lines linesStruct; printf("enter num of lines:\n "); scanf("%d",&linesStruct._num_of_lines); printf("Num of lines is = %d \n", linesStruct._num_of_lines); linesStruct._lines_len = (int*) malloc(linesStruct._num_of_lines * sizeof(int)); int len = (int) ((sizeof(linesStruct._lines_len))/(sizeof(int))); printf("number of lines len = %d \n", len);

sizeof on a pointer returns the size of the pointer, not the buffer that follows it. — Kninnug
– Kninnug, Commented Oct 28, 2013 at 17:39
sizeof(linesStruct._lines_len) returns the size of memory for an int *, not the length of your array. — lurker
– lurker, Commented Oct 28, 2013 at 17:40
hmmm I don't quite get it then, how do I figure out what is the length of the array after the allocation? — sony jimbo
– sony jimbo, Commented Oct 28, 2013 at 17:43

Paul Evans · Accepted Answer · 2013-10-28 17:40:17Z

sizeof(linesStruct._lines_len) return the size of the pointer (i.e. two words). There's really no way to statically determine the array size at compile time. But you've got that stored in _num_of_lines anyway.

Community · Accepted Answer · 2017-05-23 12:28:06Z

linesStruct._num_of_lines already tells you the array size.

Variable len will always have the value of 1, because the size of integer pointer is the same as integer.

int len = (int) ((sizeof(linesStruct._lines_len))/(sizeof(int)));

Plus, You don't cast the malloc() result.

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