Does array declaration mean consecutive memory allocation?

Question

int main() {
 char a[5];
 a[0] = 0;
 return a[0];
}

In this snippet, does char a[5];, which is an array declaration without any initialization, guarantee allocation of 5 bytes of consecutive memory?

Or since only 1 element gets initialized later, compiler is free to use a register for that 1 element?

I'm assuming that reading any other index of this array beside 0 is undefined behaviour.

0___________ · Accepted Answer · 2021-06-13 23:02:45Z

5

It guarantees that the observable behaviour of the program will be the same as if it has allocated this memory.

The compiler is free to optimize out any objects as in your trivial example:

main:
        xor     eax, eax
        ret

https://godbolt.org/z/KE5PzvTKq

answered Jun 13, 2021 at 23:02

0___________

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15 Comments

Dan Over a year ago

Thank you. I also wanted to confirm, reading any other index here beside 0 is considered undefined behaviour correct? unless the entire array in 0 initialized in the beginning like char a[5] = {0}; ?

0___________ Over a year ago

@Dan yes ---------

Eric Postpischil Over a year ago

No, reading uninitialized elements of the array will produce “indeterminate” values. There is a rule in the C standard that reading uninitialized values may have undefined behavior, but it applies only to objects that could have been declared with register (and that have automatic storage duration). Array elements can never be declared with register, so the rule does not apply.

Dan Over a year ago

@EricPostpischil is this a C and C++ rule or only C? links with more details on this?

Dan Over a year ago

This link says they are undefined in C++: stackoverflow.com/questions/49696810/…

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