what is empty function size?

Question

integer is 4 byte

double is 8 byte

What is the size of an empty function?

void test(){} //-> size????
void test1(){ int a, int b, double c } //-> size????
void test2(){ test() }  -> size??????

When I run the program, the result is the same

void test(){} 
void test1() {int a} 

void main()
{
     cout<<sizeof(&test) <<endl;
     cout<<sizeof(&test1) <<endl;
}

please solve my questions

You may want to post the output of that program, instead of only saying that the result is "the same". — Andreas Wenzel, Jan 07 '22 at 06:18
Check this question and its answers: https://stackoverflow.com/questions/4156585/how-to-get-the-length-of-a-function-in-bytes/22047976#22047976 — Paulo1205, Jan 07 '22 at 06:24
*"integer is 4 byte double is 8 byte"* -- these sizes are not guaranteed across all platforms. How did you come up with these sizes? (Did they involve using an `&`, like your attempt to get the size of a function?) — JaMiT, Jan 07 '22 at 06:46
Hi welcome to StackOverflow. While there already are decent answers to this question, please fix the code in your question so that it at least can compile. That way the question will still be readable to others when they read it in years. — Tommy Andersen, Jan 07 '22 at 06:51
You can always try to get an idea of what the size could be, using the compiler, but you should not have to: https://godbolt.org/z/nEGWjMaar — Tommy Andersen, Jan 07 '22 at 06:53
What meaning would you expect by _size of a function_. Number of generated instructions? Please, specify. — Daniel Langr, Jan 07 '22 at 07:32

Andreas Wenzel · Accepted Answer · 2022-01-07T06:55:09.283

ISO C++ does not have a notion of the size of a function.

How the compiler creates the machine-level instructions for the individual functions and merges them into an entire program is not specified by the ISO standard. Every individual platform can do this its own way. Therefore, it would not make sense for the ISO standard to attempt to define what the size of a function is.

As a consequence, you cannot use the sizeof operator to determine the size of a function.

When you write the expression &test, you get a pointer that points to the function test. On 32-bit platforms, pointers are usually 32 bits (i.e. 4 bytes), and on 64-bit platforms, they are usually 64 bits (i.e. 8 bytes).

That is why the expressions

sizeof(&test)

and

sizeof(&test1)

will both evaluate to either 4 or 8, depending on your platform.

Depending on how code is generated on your platform, you may be able to find a meaningful definition of the size of a function, and find the size of a function in a platform-specific manner. However, most platforms perform compiler optimizations which allow functions to be, for example, inlined, so that they may not even exist as a separate entity. In such a case, it is probably hard to find a meaningful definition of the size of a function.

score 2 · Answer 2 · answered Jan 07 '22 at 06:34

There is no concept of size of a function in C++. When you do sizeof(&test), you just get the size of a function pointer, which can only dependent on the function type and it is likely that all function pointer types will have the same size.

If you want to know the size of a function in terms of the number of instructions in it or the number of bytes these instructions occupy in memory, you cannot use core C++ or its standard library to achieve that directly. Instead you can look at the compiled object files with external tools, see e.g. this question, or potentially with platform-specific libraries or compiler intrinsics/extensions.

The size of function pointer types are unspecified by the standard. Practically, as you say, all functions pointers will probably have the same size for a particular implementation (e.g. compiler). But that's because of what makes sense to compiler developers - there is no requirement in the standard (which means each implementation can do whatever makes sense for that implementation). — Peter, Jan 07 '22 at 14:20

what is empty function size?

2 Answers2