Why doesn't g++ optimize local constexpr array access?
1
I have following c++ code:
inline int choose(int i){
static constexpr int arr={1,3,3,2,4,1,4};
return arr[i];
}
void f(int);
int main(){
for(int i=0;i<5;i++){
f(choose(i));
}
}
When I compile this with g++ 8.2 with option -O3, it produces well-optimized asm code. However, when I change the second line from static constexpr to constexpr, which should be semantically the same, it no longer optimizes array access, and produces somewhat inefficient asm code.
Does anyone have any idea why this happens?
c++ g++ compiler-optimization
asked Nov 26 '18 at 13:45
eivoureivour
726616
add a comment |
1
I have following c++ code:
inline int choose(int i){
static constexpr int arr={1,3,3,2,4,1,4};
return arr[i];
}
void f(int);
int main(){
for(int i=0;i<5;i++){
f(choose(i));
}
}
When I compile this with g++ 8.2 with option -O3, it produces well-optimized asm code. However, when I change the second line from static constexpr to constexpr, which should be semantically the same, it no longer optimizes array access, and produces somewhat inefficient asm code.
Does anyone have any idea why this happens?
c++ g++ compiler-optimization
asked Nov 26 '18 at 13:45
eivoureivour
726616
My guess, when the array is static the compiler knows where it will be whenever the function is called so it can unroll the loop and optimize away the function call and array access. When it is not static to location of the array can change so it can't/wont provide the optimization.
– NathanOliver
Nov 26 '18 at 13:48
@NathanOliver But when we writeconstexprvariable we don't expect it to be "real variable" (hence have memory address in stack or anywhere), so the compiler should be able to put it wherever it likes...
– eivour
Nov 26 '18 at 13:57
constexpris there mostly to express intent, so that the compiler can stop you when you try something that can't be done compile-time. It's more for the programmer than the compiler, which can (should) already see what can and cannot be optimized. As to the question - would "it's an improvement opportunity in GCC" be a sufficient answer?
– rustyx
Nov 26 '18 at 14:07
add a comment |
1
1
1
I have following c++ code:
inline int choose(int i){
static constexpr int arr={1,3,3,2,4,1,4};
return arr[i];
}
void f(int);
int main(){
for(int i=0;i<5;i++){
f(choose(i));
}
}
When I compile this with g++ 8.2 with option -O3, it produces well-optimized asm code. However, when I change the second line from static constexpr to constexpr, which should be semantically the same, it no longer optimizes array access, and produces somewhat inefficient asm code.
Does anyone have any idea why this happens?
c++ g++ compiler-optimization
asked Nov 26 '18 at 13:45
eivoureivour
726616
I have following c++ code:
inline int choose(int i){
static constexpr int arr={1,3,3,2,4,1,4};
return arr[i];
}
void f(int);
int main(){
for(int i=0;i<5;i++){
f(choose(i));
}
}
When I compile this with g++ 8.2 with option -O3, it produces well-optimized asm code. However, when I change the second line from static constexpr to constexpr, which should be semantically the same, it no longer optimizes array access, and produces somewhat inefficient asm code.
Does anyone have any idea why this happens?
c++ g++ compiler-optimization
c++ g++ compiler-optimization
asked Nov 26 '18 at 13:45
eivoureivour
726616
asked Nov 26 '18 at 13:45
eivoureivour
726616
asked Nov 26 '18 at 13:45
eivoureivour
726616
asked Nov 26 '18 at 13:45
eivoureivour
726616
asked Nov 26 '18 at 13:45
eivoureivour
726616
726616
My guess, when the array is static the compiler knows where it will be whenever the function is called so it can unroll the loop and optimize away the function call and array access. When it is not static to location of the array can change so it can't/wont provide the optimization.
– NathanOliver
Nov 26 '18 at 13:48
@NathanOliver But when we writeconstexprvariable we don't expect it to be "real variable" (hence have memory address in stack or anywhere), so the compiler should be able to put it wherever it likes...
– eivour
Nov 26 '18 at 13:57
constexpris there mostly to express intent, so that the compiler can stop you when you try something that can't be done compile-time. It's more for the programmer than the compiler, which can (should) already see what can and cannot be optimized. As to the question - would "it's an improvement opportunity in GCC" be a sufficient answer?
– rustyx
Nov 26 '18 at 14:07
add a comment |
My guess, when the array is static the compiler knows where it will be whenever the function is called so it can unroll the loop and optimize away the function call and array access. When it is not static to location of the array can change so it can't/wont provide the optimization.
– NathanOliver
Nov 26 '18 at 13:48
@NathanOliver But when we writeconstexprvariable we don't expect it to be "real variable" (hence have memory address in stack or anywhere), so the compiler should be able to put it wherever it likes...
– eivour
Nov 26 '18 at 13:57
constexpris there mostly to express intent, so that the compiler can stop you when you try something that can't be done compile-time. It's more for the programmer than the compiler, which can (should) already see what can and cannot be optimized. As to the question - would "it's an improvement opportunity in GCC" be a sufficient answer?
– rustyx
Nov 26 '18 at 14:07
My guess, when the array is static the compiler knows where it will be whenever the function is called so it can unroll the loop and optimize away the function call and array access. When it is not static to location of the array can change so it can't/wont provide the optimization.
– NathanOliver
Nov 26 '18 at 13:48
My guess, when the array is static the compiler knows where it will be whenever the function is called so it can unroll the loop and optimize away the function call and array access. When it is not static to location of the array can change so it can't/wont provide the optimization.
– NathanOliver
Nov 26 '18 at 13:48
@NathanOliver But when we write
constexpr variable we don't expect it to be "real variable" (hence have memory address in stack or anywhere), so the compiler should be able to put it wherever it likes...– eivour
Nov 26 '18 at 13:57
@NathanOliver But when we write
constexpr variable we don't expect it to be "real variable" (hence have memory address in stack or anywhere), so the compiler should be able to put it wherever it likes...– eivour
Nov 26 '18 at 13:57
constexpr is there mostly to express intent, so that the compiler can stop you when you try something that can't be done compile-time. It's more for the programmer than the compiler, which can (should) already see what can and cannot be optimized. As to the question - would "it's an improvement opportunity in GCC" be a sufficient answer?– rustyx
Nov 26 '18 at 14:07
constexpr is there mostly to express intent, so that the compiler can stop you when you try something that can't be done compile-time. It's more for the programmer than the compiler, which can (should) already see what can and cannot be optimized. As to the question - would "it's an improvement opportunity in GCC" be a sufficient answer?– rustyx
Nov 26 '18 at 14:07
add a comment |
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My guess, when the array is static the compiler knows where it will be whenever the function is called so it can unroll the loop and optimize away the function call and array access. When it is not static to location of the array can change so it can't/wont provide the optimization.
– NathanOliver
Nov 26 '18 at 13:48
@NathanOliver But when we write
constexprvariable we don't expect it to be "real variable" (hence have memory address in stack or anywhere), so the compiler should be able to put it wherever it likes...– eivour
Nov 26 '18 at 13:57
constexpris there mostly to express intent, so that the compiler can stop you when you try something that can't be done compile-time. It's more for the programmer than the compiler, which can (should) already see what can and cannot be optimized. As to the question - would "it's an improvement opportunity in GCC" be a sufficient answer?– rustyx
Nov 26 '18 at 14:07