C++ – Does sizeof Evaluate at Compile-Time or Runtime?

c++operators

I am confused about the evaluation time of sizeof operator.
When does the sizeof operator get evaluated?

Does its evaluation time (compile-time or runtime) depend on the language (C? C++?)?

Can we use sizeof in case of objects created at runtime in C++?

Best Answer

In almost all cases, sizeof is evaluated based on static type information (at compile-time, basically).

One exception (the only one, I think) is in the case of C99's variable-length arrays (VLAs).

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C++ sizeof Operator – How is sizeof Implemented?

sizeof is not a real operator in C++. It is merely special syntax which inserts a constant equal to the size of the argument. sizeof doesn't need or have any runtime support.

Edit: do you want to know how to determine the size of a class/structure looking at its definition? The rules for this are part of the ABI, and compilers merely implement them. Basically the rules consist of

size and alignment definitions for primitive types;
structure, size and alignment of the various pointers;
rules for packing fields in structures;
rules about virtual table-related stuff (more esoteric).

However, ABIs are platform- and often vendor-specific, i.e. on x86 and (say) IA64 the size of A below will be different because IA64 does not permit unaligned data access.

struct A
{
    char i ;
    int  j ;
} ;

assert (sizeof (A) == 5)  ; // x86, MSVC #pragma pack(1)
assert (sizeof (A) == 8)  ; // x86, MSVC default
assert (sizeof (A) == 16) ; // IA64

C++ sizeof – Evaluated at Compilation or Runtime?

sizeof is a compile time operator.

Best Answer

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C++ sizeof Operator – How is sizeof Implemented?

C++ sizeof – Evaluated at Compilation or Runtime?

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