问题
Passing an object by reference is an easier, faster and safer way to pass an address to it. But for most compilers, it's all the same: references are really pointers.
Now what about basic types like int? Passing an address to an int and using it inside a function would be slower than passing it by copy, because the pointer needs to be dereferenced before use.
How do modern compiler handle, this?
int foo(const int & i)
{
cout << i; // Do whatever read-only with i.
}
May I trust them to compile this into this?
int foo(const int i)
{
cout << i;
}
By the way, in some cases it could even be faster to pass both i and &i, then use i for reading, and *i for writing.
int foo(const int i, int * ptr_i)
{
cout << i; // no dereferencement, therefore faster (?)
// many more read-only operations with i.
*ptr_i = 123;
}
回答1:
May I trust them to compile this into this?
Yes You can.[The Yes here means differently, Please read Edit section, Which clarify's]
int foo(const int & i)
Tells the compiler that i is an reference to type constant integer.
The compiler may perform optimizations but they are only allowed to perform optimizations following the As-If Rule. So you can be assured that for your program the behavior of the above will be as good as(the const qualifier will be respected):
int foo(const int i)
As-If Rule:
The C++ standard allows a compiler to perform any optimization, as long as the resulting executable exhibits the same observable behaviour as if all the requirements of the standard have been fulfilled.
For Standerdese fans:
C++03 1.9 "Program execution:
conforming implementations are required to emulate (only) the observable behavior of the abstract machine.
And the Foot-Note says:
This provision is sometimes called the “as-if” rule, because an implementation is free to disregard any requirement of this International Standard as long as the result is as if the requirement had been obeyed, as far as can be determined from the observable behavior of the program. For instance, an actual implementation need not evaluate part of an expression if it can deduce that its value is not used and that no side effects affecting the observable behavior of the program are produced.
EDIT:
Since there is some confusion about the answer let me clarify:
Optimizations cannot be enforced on the compiler.So How compiler interprets it depends on the compiler.The important thing is the observable behavior of the program will not change.
回答2:
It shouldn't compile it into that because it might not be correct. Consider:
int foo(const int &i, int *p)
{
*p = 42;
cout << i; // prints 42
return 0;
}
int main()
{
int x = 5;
foo(x, &x);
return 0;
}
versus
int foo(const int i, int *p)
{
*p = 42;
cout << i; // prints 5
return 0;
}
int main()
{
int x = 5;
foo(x, &x);
return 0;
}
How does the compiler know that this won't happen? It would have to somehow be able to analyze that it is impossible to access that variable to change it, e.g. (1) someone having a pointer, (2) it might be a global variable, (3) from another thread. Given the unsafe nature of C, with pointer arithmetic and all, even guaranteeing that the function won't be able to get a pointer to the variable might be impossible.
回答3:
Visual Studio 2010 (Express) does, in the simple cases I've tested at least. Anyone to test gcc?
I've tested the following:
1. Passing only i:
int vars[] = {1,2,3,12,3,23,1,213,231,1,21,12,213,21321,213,123213,213123};
int ok1(const int i){
return sqrtl(vars[i]);
}
int ok2(const int & i){
return sqrtl(vars[i]);
}
void main() {
int i;
std::cin >> i;
//i = ok1(i);
i = ok2(i);
std::cout << i;
}
The ASM:
i = ok1(i);
000D1014 mov ecx,dword ptr [i]
000D1017 fild dword ptr vars (0D3018h)[ecx*4]
000D101E call _CIsqrt (0D1830h)
000D1023 call _ftol2_sse (0D1840h)
i = ok2(i);
013A1014 mov ecx,dword ptr [i]
013A1017 fild dword ptr vars (13A3018h)[ecx*4]
013A101E call _CIsqrt (13A1830h)
013A1023 call _ftol2_sse (13A1840h)
Well, the ASMs are identical, no doubt the optimization was performed.
2. Passing i and &i:
Let's consider @newacct 's anser here.
int vars[] = {1,2,3,12,3,23,1,213,231,1,21,12,213,21321,213,123213,213123};
int ok1(const int i, int * pi) {
*pi = 2;
return sqrtl(vars[i]);
}
int ok2(const int & i, int * pi) {
*pi = 2;
return sqrtl(vars[i]);
}
void main() {
int i;
int * pi = &i;
std::cin >> i;
i = ok1(i, pi);
//i = ok2(i, pi);
std::cout << i;
}
The ASM:
i = ok1(i, pi);
00891014 mov ecx,dword ptr [i]
00891017 fild dword ptr vars (893018h)[ecx*4] // access vars[i]
0089101E call _CIsqrt (891830h)
00891023 call _ftol2_sse (891840h)
i = ok2(i, pi);
011B1014 fild dword ptr [vars+8 (11B3020h)] // access vars[2]
011B101A call _CIsqrt (11B1830h)
011B101F call _ftol2_sse (11B1840h)
In ok1 I can't see it writing 2 into pi. Probably it understands that the memory location will be overwritten by the result of the function anyway, so the writing is useless.
With ok2, the compiler is as smart-ass as I expected. It understands that i and pi point to the same place, so it uses a hardcoded 2 directly.
Notes:
- I've compiled twice for both test, once uncommenting only
ok1, once uncommenting onlyok2. Compiling both at the same time leads to more complex optimizations between the two functions, which end up all inlined and mixed up - I've added a lookup in the array
varsbecause simple calls tosqrtlwere simplified into basic ADD- and MUL-like operations without the actual call - Compiled in Release
- Yielded the expected results, of course
回答4:
gcc does not appear to do this optimization with -O3 (gcc version 4.7.2). Using Gabriel's code, note how ok2 loads a dereferenced address before indexing into vars while ok1 does not.
ok1:
.cfi_startproc
subq $40, %rsp
.cfi_def_cfa_offset 48
movslq %edi, %rdi
fildl vars(,%rdi,4)
fld %st(0)
fsqrt
fucomi %st(0), %st
jp .L7
fstp %st(1)
.cfi_startproc
subq $40, %rsp
.cfi_def_cfa_offset 48
movslq (%rdi), %rax
fildl vars(,%rax,4)
fld %st(0)
fsqrt
fucomi %st(0), %st
jp .L12
fstp %st(1)
来源：https://stackoverflow.com/questions/7958242/do-mainstream-compilers-convert-passed-by-reference-basic-types-into-pass-by-cop