英文:
Java vs C++ speed on simple for loop
问题
#include <iostream>
#include <ctime>
int main()
{
long long x = 0;
clock_t begin = clock();
for (long long i = 0; i < 2147483647; i++)
x += i;
clock_t end = clock();
double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
std::cout << "Time elapsed: " << elapsed_secs << std::endl;
std::cout << "x = " << x << std::endl;
return 0;
}
英文:
Why in simple for loop the same code in Java works 4 times faster than in C++? i.e. in Java this code completes in 700-800 ms and in C++ 4-5 SECONDS. Although C++ usually considered much faster than Java, especially with CPU-bound workloads. Have i lost sight of some important moment ???
Java:
import java.time.Duration;
import java.time.Instant;
public class Main {
public static void main(String[] args) {
long x = 0;
Instant start = Instant.now();
for (long i = 0; i < 2147483647; i++)
x += i;
Instant end = Instant.now();
Duration result = Duration.between(start, end);
System.out.println("TIME: " + result.toMillis());
System.out.println("X = " + x);
}
}
Output:
TIME: 799
X = 2305843005992468481
C++:
#include <iostream>
#include <ctime>
int main()
{
long long x = 0;
clock_t begin = clock();
for (long long i = 0; i < 2147483647; i++)
x += i;
clock_t end = clock();
double elapsed_secs = double(end - begin) / CLOCKS_PER_SEC;
std::cout << "Time elapsed: " << elapsed_secs << std::endl;
std::cout << "x = " << x << std::endl;
return 0;
}
Output:
Time elapsed: 4.59629
x = 2305843005992468481
答案1
得分: 5
C++在开启优化时可以运行得很快。默认情况下，编译时不会进行优化，因为编译C++需要时间。你需要使用`-O`标志，`-O2`应该是可以的。
你测量的循环是一种常见的模式，并且有一个直接的解决方案：
int sum = 0;
for (int i = 0; i < n; ++i)
sum += i;
// 这将得到相同的结果：
sum = ((n+1) * n) / 2;
而且编译器知道这个技巧（它们可能不使用那个公式，因为对于`(n+1)*n`，可能会溢出，而最终结果不会溢出）。对于gcc，下面的代码：
#include <iostream>
int main(){
long long x = 0;
for (long long i = 0; i < 2147483647; ++i)
x += i;
std::cout << x;
}
[翻译为](https://godbolt.org/z/EWE4rx)：
main:
sub rsp, 8
mov edi, OFFSET FLAT:_ZSt4cout
movabs rsi, 2305843005992468481
call std::basic_ostream<char, std::char_traits<char> > & std::basic_ostream<char, std::char_traits<char> >::_M_insert<long long>(long long)
xor eax, eax
add rsp, 8
ret
_GLOBAL__sub_I_main:
sub rsp, 8
mov edi, OFFSET FLAT:_ZStL8__ioinit
call std::ios_base::Init::Init() [complete object constructor]
mov edx, OFFSET FLAT:__dso_handle
mov esi, OFFSET FLAT:_ZStL8__ioinit
mov edi, OFFSET FLAT:_ZNSt8ios_base4InitD1Ev
add rsp, 8
jmp __cxa_atexit
注意，这里没有循环！不像这样的情况需要几秒钟才能执行。
在这里查看当你关闭优化（默认情况）时会发生什么：https://godbolt.org/z/8KncEj
英文:
C++ can be fast when you turn on optimizations. The default is compiling without optimizations because compiling C++ takes time. You need to use the -O flag, -O2 should be ok.
The loop you measure is a common pattern and has a direct solution:
int sum = 0;
for (int i=0;i < n; ++i) sum += i
// this will give the same result:
sum = ((n+1)*n)/2;
And compilers know about this trick (they probably don't use that formula, because it can overflow for (n+1)*n while the final result is no overflow). With gcc, this:
#include <iostream>
int main(){
long long x = 0;
for (long long i = 0; i < 2147483647; ++i)
x += i;
std::cout << x;
}
main:
sub rsp, 8
mov edi, OFFSET FLAT:_ZSt4cout
movabs rsi, 2305843005992468481
call std::basic_ostream<char, std::char_traits<char> >& std::basic_ostream<char, std::char_traits<char> >::_M_insert<long long>(long long)
xor eax, eax
add rsp, 8
ret
_GLOBAL__sub_I_main:
sub rsp, 8
mov edi, OFFSET FLAT:_ZStL8__ioinit
call std::ios_base::Init::Init() [complete object constructor]
mov edx, OFFSET FLAT:__dso_handle
mov esi, OFFSET FLAT:_ZStL8__ioinit
mov edi, OFFSET FLAT:_ZNSt8ios_base4InitD1Ev
add rsp, 8
jmp __cxa_atexit
Note that there is no loop! Unlikely this takes seconds to execute.
See here what happens when you turn off optimiziations (the default): https://godbolt.org/z/8KncEj
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