Size of pointer variable in a structure

Question

int main()
{
    struct
    {
        char *name_pointer;
        char all[13];
        int foo;
    } record;

    printf("%d\n",sizeof(record.all));
    printf("%d\n",sizeof(record.foo));
    printf("%d\n",sizeof(record));

    return 0;
}

I want the size of the pointer vatiable "*name_pointer" in the structure....

read: What does sizeof(&arr) return? to know differences between name_pointer and all. — Grijesh Chauhan
– Grijesh Chauhan, Commented Aug 1, 2013 at 11:57
you want sizeof(record.name_pointer) or strlen(record.name_pointer) ? — Sander De Dycker
– Sander De Dycker, Commented Aug 1, 2013 at 11:57

chux · Accepted Answer · 2013-08-01 12:36:58Z

7

To get the size of the pointer use

printf("%d\n", (int) sizeof(record.name_pointer));

Your might get 2, 4, 6, 8, etc.

To get the size of the data that is pointed to by the pointer (a char) use

printf("%d\n", (int) sizeof(*record.name_pointer));

Your should get 1.

To get the string length of the string pointed to by the pointer, assuming record.name_pointer points to legitimate data, use

printf("%d\n", (int) strlen(record.name_pointer));

BTW As @alk says and why the (int) casting above, a suitable conversion specifier to use with sizeof() includes the 'z' prefix. The result of sizeof() and strlen() is of type size_t. Although size_t and int are often the same, there are many systems where they are of different sizes. And since sizeof() is an "unsigned integer type" (C11 6.5.3.4), I recommend

printf("%zu\n", sizeof(...

edited Aug 1, 2013 at 12:36

answered Aug 1, 2013 at 12:06

chux

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5 Comments

Manoj Awasthi Over a year ago

in the second one - data being pointed is a string - not the first character. so this is wrong.

alk Over a year ago

Nitpick: As strlen() returns size_t the suitable conversion specifier should be prefixed by the appropriate length modifier z to become zd.

alk Over a year ago

@ManojAwasthi: name_pointer is defined as char *, so it points to exactly one character. So everything is fine.

chux Over a year ago

@alk Yes - I concur - thought of that but was uncertain to add that improvement to a basic question.

chux Over a year ago

@Manoj Awasthi record.name_pointer is a pointer to a char, thus the sizeof(*record.name_pointer) will be 1. As a string in C is a variable length array of characters and the string value is passed about by the address of the first char, whether record.name_pointer is used to point to only 1 char, a string, or a fixed char array is indistinguishable from OP's code snippet.

someone · Accepted Answer · 2013-08-01 12:02:01Z

0

Pointer variable will always(32 bit system architecture) have size 4. if you have 64 bit system architecture it is 8.

answered Aug 1, 2013 at 12:02

someone

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2 Comments

Grijesh Chauhan Over a year ago

remove always add in your system

alk Over a year ago

Not on 64 bit systems. It's 8 there. And there are others ...!

Manoj Awasthi · Accepted Answer · 2013-08-01 12:09:15Z

0

I believe it is a GOOD idea to use data types in sizeof rather than variables for native data types. so use:

sizeof(char *)

answered Aug 1, 2013 at 12:09

Manoj Awasthi

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1 Comment

sje397 Over a year ago

Really? I prefer to use the variable, so code usually remains correct if you change the variable type.

Hemjal · Accepted Answer · 2020-02-15 13:36:46Z

-1

use the following code:

printf("%d\n", (int) sizeof(record->name_pointer)/4);

answered Feb 15, 2020 at 13:36

Hemjal

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2 Comments

JerodG Over a year ago

It would be more helpful to provide an explanation along with a working solution.

Blastfurnace Over a year ago

This won't work. record isn't a pointer so you can't dereference it with ->. In addition, why are you dividing by 4?

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