CODEWITHSUNDEEP
c
Saiteja Pabisetti
Saiteja Pabisetti

Reputation: 109

sizeof operator in c on pointers of same type giving different results in C

I know that a pointer has a size similar to the device address architecture. But I am unable to get why same function and arguments are giving different results. Please look at the code and let me know what's the cause.

#include <stdio.h>
#include <stdint.h>

int main()
{
    uint8_t a[22]={1,2,3,4,5};
    uint8_t* p[2]={a};
    printf("%d\n%d\n",&a,p[0]);
    printf("%d,%d",sizeof(a),sizeof(p));

    return 0;
}

606584768 606584768 22,16

Upvotes: 0

Views: 78

Answers (4)

user13158930
user13158930

Reputation:

sizeof(a) give array size which is 22 but sizeof(b) give the size of array of pointer which consists of two pointers (2*8=16)

Upvotes: 0

sara hamdy
sara hamdy

Reputation: 412

Because there is a different between a and b firstly let examine a 

uint8_t a[22]={1,2,3,4,5};

this line of code means a is array consists of 22 elements each element has 8 bits (one byte) so sizeof(a) is 22*1=22 secondly with respect to b 

uint8_t* p[2]={a};

this line of code means b is array of pointers consists of 2 elements type of pointers which have the same size with respect to design architecture so so sizeof(p) is (2 pointer * 8 )=16

Upvotes: 1

John Bode
John Bode

Reputation: 123568

An expression of type "N-element array of T" will "decay" to an expression of type "pointer to T" except when the expression is the operand of the sizeof or unary & operators.

The expression a has type "22-element array of uint8_t", so sizeof a will give you the same result as sizeof (uint8_t) * 22. The expression p has type "2-element array of pointer to uint8_t (uint8_t *)", so sizeof p is the same as sizeof (uint8_t *) * 2.

To summarize:

a           == some_address;
sizeof a    == 22 ( sizeof (uint8_t) should be 1)

p           == some address;
p[0]        == address of the first element of a
sizeof p    == 2 * sizeof (uint8_t *) (assuming pointer size is 8, this will be 16)
sizeof p[0] == sizeof (uint8_t *)

Upvotes: 1

SoronelHaetir
SoronelHaetir

Reputation: 15172

An array and pointer are not of the same type, an array will decay to a pointer but is not itself a pointer.

sizeof(a) gives the actual size of an uint_8[22], while sizeof(p) gives the actual size of an uint8_t *[2];

Upvotes: 3

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