Function with input of pointer to pointer to

Question

I saw this question about the the result of sizeof() and i didn't understand the output of this function.

#include <stdio.h>
void checkSizes(char matrix3d[10][20][30])
{
    printf ("%lu\t", sizeof(matrix3d));
    printf ("%lu\t", sizeof(*matrix3d));
    printf ("%lu\t", sizeof(**matrix3d));
    printf ("%lu\t", sizeof(***matrix3d));
}
int main()
{
    char matrix[10][20][30];
    checkSizes(matrix);
}

as i was assuming , when a function get an array to work with , it's getting a pointer to the array, so when i'll print the first line of

printf ("%lu\n", sizeof(matrix3d));

the output will be 8.

BUT isn't the next printf will get me to diffrent pointer to pointer that his size will be also 8 ?

so i thouth this will lead to

8    8    8    1

the real output is

8    600    30   1

matrix3d is not `char***` but `char (*)[20][30]`. Only the outermost dimension decays to a pointer. The rest remain baked into the type. — StoryTeller - Unslander Monica, Sep 01 '16 at 13:13
You can convert size into unsigned long. `printf ("%lu\n", (unsigned long)sizeof(matrix3d));` — Arpit Svt, Sep 01 '16 at 13:16
8 = size of a pointer, 600 (20*30) size of 1st block of 3-d array, 30 is the size of the 1st row of the block, and 1 is sizeof(char) which is one cell of the row — Rohith R, Sep 01 '16 at 13:16
to print the result of `sizeof` use [`%zu`](http://stackoverflow.com/q/940087/995714), not `%lu` — phuclv, Sep 01 '16 at 13:21

Vlad from Moscow · Accepted Answer · 2016-09-01T14:45:52.100

In this statement

printf ("%lu\t", sizeof(*matrix3d));
// it is better to use "%zu\t"

expression *matrix3d has type char[20][30].

Using an array in the sizeof operator as an expression does not convert the expression to the pointer to the first element of the array.

Thus you will get output 600.

However if you would write

sizeof( *matrix3d + 0)

then indeed *matrix3d having type char[20][30] will be converted in the expression *matrix3d + 0 implicitly to pointer of type char ( * )[30] and the corresponding printf statement will output 8,

As for the function parameter and its argument then they are implicitly converted to type char ( * )[20][30]

So for example these function declarations are equivalent

void checkSizes(char matrix3d[10][20][30]);
void checkSizes(char ( *matrix3d )[20][30]);

and declare the same one function.

haccks · Answer 2 · 2016-09-01T14:36:27.050

First you need to know that sizeof is an operator and not a function. Arrays do not convert to pointer to its first element when it is an operand of sizeof or unary & operator.

In function checkSizes, the parameter matrix3d is of type pointer to an array of 20 arrays of 30 chars . Since it is a pointer, the statement

printf ("%lu\t", sizeof(matrix3d));

will print the size of a pointer.
Dereferencing the pointer matrix3d will give its first element. Its first element is an array of 20 arrays of 30 chars. sizeof(*matrix3d) will return size of array char[20][30] which is 600 Bytes. sizeof(*matrix3d) is equivalent to sizeof(matrix3d[0]).

The type of **matrix3d is an array of 30 chars, sizeof(**matrix3d) will print 30 Bytes.
The type of ***matrix3d is achar, sizeof(***matrix3d) will print 1 Byte.

score 0 · Answer 3 · answered Sep 01 '16 at 13:19

The first is, as you said the size of a pointer, so 8; the last is the size of a character, so 1; the second is sizeof(char)*30*20, because you have dereferenced the first "dimension" of the array, so you are looking at what's contained in one of the 10 elements of that dimension. The third is sizeof(char)*30 because you are looking inside an element of the second dimension.

Function with input of pointer to pointer to

3 Answers3