I have two char arrays as below:
char *t1[10];
char(*t2)[10];
when using sizeof to find their size
printf("sizeof(t1): %d\n", sizeof(t1));
printf("sizeof(t2): %d\n", sizeof(t2));
I found that the output is:
sizeof(t1): 80
sizeof(t2): 8
I am quite confused by why I have two different results when using the sizeof operator.
For starters you have to use the conversion specifier zu instead of d when outputting values of the type size_t
printf("sizeof(t1): %zu\n", sizeof(t1));
printf("sizeof(t2): %zu\n", sizeof(t2));
This record
char(*t2)[10];
does not declare an array. It is a declaration of a pointer to the array type char[10].
So sizeof( t1 ) yields the size of an array while sizeof( t2 ) yields the size of a pointer.
Consider for example declarations
char t1[5][10];
char ( *t2 )[10] = t1;
The pointer t2 is initialized by the address of the first element (of the type char[10]) of the array t1. That is it points to the first element of the array t1.
Also consider the following call of printf
printf("sizeof( *t2 ): %zu\n", sizeof( *t2 ));
The output of the call will be 10.
That is dereferencing the pointer you will get one dimensional array of the type char[10].
If you want to get identical outputs then the second declaration should be rewritten like
char *t1[10];
char *( t2)[10];
char(*t2)[10];is a pointer to array of tenchars. So it's size is a size of a pointer.char *t1[10];is an array of pointers. Notchars. So it is ten times as a size of a pointer.char *t1[10]means an item in t1 can be accessed with for example*t1[5](which does [] first and then * so it must be an array of pointers).char (*t2)[10]means an item in t2 can be accessed with for example(*t2)[5](which does * first and then [] so it must be a pointer to an array)