constexpr 问题,为什么这两个不同的程序在 g++ 中运行的时间如此不同?
我正在使用 gcc 4.6.1,并且遇到了一些有趣的行为,涉及调用 constexpr
函数.该程序运行良好,并立即打印出 12200160415121876738
.
I'm using gcc 4.6.1 and am getting some interesting behavior involving calling a constexpr
function. This program runs just fine and straight away prints out 12200160415121876738
.
#include <iostream>
extern const unsigned long joe;
constexpr unsigned long fib(unsigned long int x)
{
return (x <= 1) ? 1 : (fib(x - 1) + fib(x - 2));
}
const unsigned long joe = fib(92);
int main()
{
::std::cout << "Here I am!
";
::std::cout << joe << '
';
return 0;
}
这个程序需要很长时间才能运行,我从来没有耐心等待它打印出一个值:
This program takes forever to run and I've never had the patience to wait for it to print out a value:
#include <iostream>
constexpr unsigned long fib(unsigned long int x)
{
return (x <= 1) ? 1 : (fib(x - 1) + fib(x - 2));
}
int main()
{
::std::cout << "Here I am!
";
::std::cout << fib(92) << '
';
return 0;
}
为什么会有这么大的差异?我在第二个程序中做错了吗?
Why is there such a huge difference? Am I doing something wrong in the second program?
我在 64 位平台上使用 g++ -std=c++0x -O3
进行编译.
I'm compiling this with g++ -std=c++0x -O3
on a 64-bit platform.
推荐答案
joe
是一个整型常量表达式;它必须在数组范围内可用.因此,合理的编译器会在编译时对其进行评估.
joe
is an Integral Constant Expression; it must be usable in array bounds. For that reason, a reasonable compiler will evaluate it at compile time.
在您的第二个程序中,即使编译器可以在编译时计算它,也没有理由必须这样做.
In your second program, even though the compiler could calculate it at compile time, there's no reason why it must.
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