CODEWITHSUNDEEP
carraysstructsizeof
DasIstLars
DasIstLars

Reputation: 1

Using sizeof with struct with fixed length array

Is it good coding practice, to use sizeof(struct) when the struct has a fixed length array?

#define NUM_TRACKS 10
#define NUM_SAMPLES 250
typedef struct _DATA_PROC
{
    uint16_t count[NUM_TRACKS] ;
    int16_t data[NUM_TRACKS][NUM_SAMPLES];
} tDataProcessing ;



tDataProcessing* tDp = malloc(sizeof(tDataProcessing)) ;

Upvotes: 0

Views: 260

Answers (2)

Vlad from Moscow
Vlad from Moscow

Reputation: 310980

For starters in expressions, with rare exceptions including the sizeof operator, arrays are implicitly converted to pointers to their element types.

So for example the array

int16_t data[NUM_TRACKS][NUM_SAMPLES];

is converted in expressions like 

int16_t ( *data )[NUM_SAMPLES];

not as you wrote in a comment like

int16_t** data

However a declaration is not an expression. So the sizeof operator returns the number of bytes to accommodate all data members of the structure without any implicit conversion.

From the C Standard (6.5.3.4 The sizeof and alignof operators)

2 The sizeof operator yields the size (in bytes) of its operand, which may be an expression or the parenthesized name of a type. The size is determined from the type of the operand. The result is an integer. If the type of the operand is a variable length array type, the operand

Here is a demonstrative program

#include <stdio.h>

int main(void) 
{
    enum { N = 10 };
    int a[N];

    printf( "sizeof( a ) = %zu\n", sizeof( a ) );

    struct A
    {
        int a[N];
    };

    printf( "sizeof( struct A ) = %zu\n", sizeof( struct A ) );

    return 0;
}

Its output is

sizeof( a ) = 40
sizeof( struct A ) = 40

Upvotes: 0

Lee Daniel Crocker
Lee Daniel Crocker

Reputation: 13171

I personally prefer to use the pointer variable itself rather than the type:

tDataProcessing *tDp = malloc(sizeof *tDp);

This is sometimes clearer in cases with pointers buried in types and such.

Upvotes: 1

Related Questions