Does sizeof evaluate at compile-time or runtime?

问题

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++?

回答1:

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).

回答2:

Almost always compile time. But the following examples might be of interest to you:

char c[100];
sizeof(c); // 100

char* d = malloc(100);
sizeof(d); //probably 4 or 8.  tells you the size of the pointer!

BaseClass* b = new DerivedClass();
sizeof(b); //probably 4 or 8 as above.

void foo(char[100] x) {
    sizeof(x); //probably 4 or 8.  I hate this.  Don't use this style for this reason.
}

struct Foo {
    char a[100];
    char b[200];
};

sizeof(struct Foo); //probably 300.  Technically architecture dependent but it will be
//the # of bytes the compiler needs to make a Foo.

struct Foo foo;
sizeof(foo); //same as sizeof(struct Foo)

struct Foo* fooP;
sizeof(fooP); //probably 4 or 8

class ForwardDeclaredClass;

ForwardDeclaredClass* p;
sizeof(p); //4 or 8
ForwardDeclaredClass fdc; //compile time error.  Compiler
//doesn't know how many bytes to allocate

sizeof(ForwardDeclaredClass); //compile time error, same reason

回答3:

Compile time , because it's calculate size at compile time."Compile time" is when you build your code - when the compiler converts your source code into IL.

来源：https://stackoverflow.com/questions/11179592/does-sizeof-evaluate-at-compile-time-or-runtime

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c++

operators