c - Getting the size of the data of a Pointer

Question

I tried the following code in order to see how to get size of the data of a pointer:

 #include <stdio.h>
 #include <stdlib.h>
 #include <string.h>

 int main(){
  char *test_char_ptr = "This is just a test";
  int *test_int_ptr = (int *) malloc(16*sizeof(int));
  for(int i = 0; i<16; i++){
    test_int_ptr[i] = i;
  }
  printf("%s
",test_char_ptr);
  printf("Test char 1: %d
", (int)sizeof test_char_ptr );
  printf("Test char 2:%d
", (int)sizeof *test_char_ptr );
  printf("Test char 3: %d
", (int)((strlen(test_char_ptr)+1) * sizeof(char)));
  printf("Test int 1:%d
", (int)sizeof test_int_ptr );
  printf("Test int 2:%d
", (int)sizeof *test_int_ptr );
  return EXIT_SUCCESS;
 }

And code's output is(On 32 bit gcc 4.3):

This is just a test
Test char 1: 4
Test char 2:1
Test char 3: 20
Test int 1:4
Test int 2:4

I thought that sizeof(*test_char_ptr) would give me the sizeof the data inside of the *test_char_ptr. But instead it gave me 1 , which i think that is the sizeof char instead of the data. The same is for the test_int_ptr. Long story short, my question is how can i get the sizeof data inside a pointer or dynamically memory allocated array.

Answer 1

That's not something you can do in C without maintaining the information yourself. You created the arrays, so you knew their sizes at one point, you just have to keep track of it on your own. You could create a data structure to help you, or just maintain the array and size information carefully without any data structure at all.

In addition, your code is using strlen() to get the size of the string - make sure you remember that the size returned will not include the terminating null character (''); the in-memory size of a string constant is strlen(string) + 1.