Is a char array more efficient than a char pointer in C?

Question

I'm trying to understand the under-the-hood difference between these two char declarations:

char* str1;
char str2[10];

I understand that using char* gives a pointer to the first character in str1, while char[10] results in an array with a length of 10 bytes (because char is a single byte... I recall that in some encodings it can be more but let's just assume one byte to keep it simple).

My question is when actually assigning values to them the data obviously has to be stored somewhere and in the case of char[10] we're telling the compiler upfront to allocate ten bytes, whereas in the case of char* we're really just saying allocate a pointer to a single byte. But what happens if the string we assign to str1 is more than a single byte, how is that allocated? How much more work is needed to appropriately allocate that? Plus, what happens if we want to reassign str1 to be something longer than what was previously assigned, how is that allocated?

Because of the uncertainty from the compiler's point of view when dealing with char pointers, is it more efficient to use a char array when I either know the length ahead of time or want to limit the length to start with?

Answer 1

When discussing performance in general, allocation, access time and copy time separate things. You seem mostly concerned about allocation.

But there are lots of misconceptions here. Arrays are used for storing. Pointers are used to point at things stored elsewhere. You cannot store any data in a pointer, you can only store an address to data allocated elsewhere.

So comparing pointers or arrays is pretty much nonsense, because they are separate things. Similar to "should I live in my house at a street address or should I live in the sign stating my street address".

I understand that using char* gives a pointer to the first character in str1

No, it gives a pointer to a single character which is allocated somewhere else. Though it doesn't point anywhere meaningful until you assign an address to it. In case of arrays, it will typically get set to point at the first character of the array.

I recall that in some encodings it can be more

No, a character is per definition always 1 byte. Some exotic systems might have 16 bits per bytes or such though. This is of no concern unless you program exotic DSPs and the like. As for other character encodings, there's wchar_t which is a different topic entirely.

whereas in the case of char* we're really just saying allocate a pointer to a single byte

No, we tell it to allocate room for the pointer itself. Which is typically of a size between 2 to 8 bytes depending on address bus width of the specific system.

But what happens if the string we assign to str1 is more than a single byte, how is that allocated?

However you like. You can assign it to a read-only string literal, or a static storage duration variable, or a local automatic storage variable, or dynamically allocated variables. The pointer itself doesn't know or care.

How much more work is needed to appropriately allocate that?

It depends on what you want to allocate.

Because of the uncertainty from the compiler's point of view when dealing with char pointers

What uncertainty is that? Pointers are pointers and the compiler don't treat them much differently than other variables.

is it more efficient to use a char array when I either know the length ahead of time or want to limit the length to start with?

You need to use an array, because data cannot be stored in thin air. Again, data cannot be stored "in pointers".