How does sizeof() work for char* (Pointer variables)?

Question

I have a C code:

char s1[20];
char *s = "fyc";
printf("%d %d\n", sizeof(s1), sizeof(s));
return 0;

It returns

20 8

I'm wondering how does the 8 come from, thanks!

Solution 1

sizeof(s1) is the size of the array in memory, in your case 20 chars each being 1 byte equals 20.

sizeof(s) is the size of the pointer. On different machines it can be a different size. On mine it's 4.

To test different type sizes on your machine you can just pass the type instead of a variable like so printf("%zu %zu\n", sizeof(char*), sizeof(char[20]));.

It will print 4 and 20 respectively on a 32bit machine.

Solution 2

sizeof(char *) is the size of the pointer, so normally 4 for 32-bit machine, and 8 for 64-bit machine.

sizeof an array, on the other hand, outputs the size of the array, in this case, 20*sizeof(char) = 20

One more thing, you should use %zu for size_t type in printf format.

printf("%zu %zu\n", sizeof(s1), sizeof(s));

Solution 3

The sizeof operator returns the size of a type. The operand of sizeof can either be the parenthesized name of a type or an expression but in any case, the size is determined from the type of the operand only.

sizeof s1 is thus stricly equivalent to sizeof (char[20]) and returns 20.

sizeof s is stricly equivalent to sizeof (char*) and returns the size of a pointer to char (64 bits in your case).

If you want the length of the C-string pointed by s, you could use strlen(s).

Solution 4

8 is the size of a pointer, an address. On 64 bit machine it has 8 bytes.

Solution 5

If you are on a 64 bit computer, the memory addresses are 64 bit, therefore a 64 bit (8 bytes x 8 bits per byte) numeric value must be used to represent the numeric pointer variable (char*).

In other words, sizeof() works the same way for pointers as for standard variables. You just need to take into account the target platform when using pointers.

View more solutions

Comments

[Chandra Shekhar]

Just had a doubt what if we are to find sizeof(*s) ? it is giving me 1 byte but it should have been 3 bytes for "fyc". Kindly clarify this doubt.

[Suri]

@ChandraShekhar That's because char* s is pointing to the first character of "fyc". So if you printf *s, you get the first character of "fyc" which is 'f'. Individual character weighs 1 byte. That's why you got 1 byte.