求助 C++ 的 mode(TI)
看到某人的C++ 代码,里面有一行看不懂, 求解释~~int64_t __attribute__((mode(TI))) a = p*q; a *= r; 引用自：http://stackoverflow.com/questions/4559025/what-does-gcc-attribute-modexx-actually-do
Q:Hi All,
This arose from a question earlier today on the subject of bignum libraries and gcc specific hacks to the c language. Specifically, these two declarations were used:typedef unsigned int dword_t __attribute__((mode(DI)));On 32 bit systems andtypedef unsigned int dword_t __attribute__((mode(TI)));On 64-bit systemms.
I assume given this is an extension to the C language that there exists no way to achieve whatever it achieves in current (C99) standards. So my questions are simple: is that assumption correct? And what do these statements do to the underlying memory? I think the result is I have 2xsizeof(uint32_t) for a dword in 32-bit systems and 2*sizeof(uint64_t) for 64-bit systems, am I correct?
Thanks for all your help,
A:These allow you to explicitly specify a size for a type without depending on compiler or machine semantics, such as the size of 'long' or 'int'.
They are described fairly well on this page.
I quote from that page:QI: An integer that is as wide as the smallest addressable unit, usually 8 bits.
HI: An integer, twice as wide as a QI mode integer, usually 16 bits.
SI: An integer, four times as wide as a QI mode integer, usually 32 bits.
DI: An integer, eight times as wide as a QI mode integer, usually 64 bits.
SF: A floating point value, as wide as a SI mode integer, usually 32 bits.
DF: A floating point value, as wide as a DI mode integer, usually 64 bits.So DI is essentially sizeof(char) * 8.
Further explanation, including TI mode, can be found here (possibly better than the first link, but both provided for reference).
So TI is essentially sizeof(char) * 16 (128 bits). 2# gxqcn
:victory:
我看过了,不知道老大平时用过这个语句没 感觉把C++用到这种程度的一定是火星人 3# wayne
没用过。我倒是非常需要这样的数据结构。哪里有例程？ 本帖最后由 wayne 于 2011-4-11 08:30 编辑
5# gxqcn
好像是gcc编译器带的,这有文档:
http://gcc.gnu.org/onlinedocs/gcc/X86-Built_002din-Functions.html
问题源自 projctEuler 221题的 讨论专栏里.
http://projecteuler.net/index.php?section=forum&id=221
答案是1884161251122450 ,你输进去即可查看所有人的讨论,
在我的楼下,有一位C++火星人提供的代码如下:#include <cstdio>
#include <set>
#include <cmath>
#include <vector>
#include <cassert>
//1805315691966540?
int const cutoff = 200000;//1638400000;
int64_t const len = 1000000/*cutoff+2*/; //(int64_t)cutoff*cutoff+2;
int primeFac;
int primes;
int numPrimes;
std::vector<int64_t> factors(int64_t v, int64_t q)
{
if(v == 1)
return std::vector<int64_t>(1, 1);
//int p = primeFac;
int64_t p = v;
for(int index = 0; primes < q && index < numPrimes; ++index)
if(v%primes == 0)
{
p = primes;
break;
}
int count = 0;
//std::printf("* %ld %ld%ld\n", v, p, q);
//assert(v < len);
//while(p == primeFac)
while(v%p == 0)
{
++count;
v /= p;
//std::printf("%d %d\n", v, p);
}
std::vector<int64_t> ret = factors(v, std::min(2+(int64_t)sqrt(v), q));
int siz = ret.size();
for(int c = 0; c < count; ++c)
for(int s = 0; s < siz; ++s)
ret.push_back(ret*p);
//std::printf("/ %ld %ld%ld\n", v, p, q);
return ret;
}
int main()
{
for(int i = 0; i < len; ++i)
primeFac = i;
for(int i = 2; i < len; ++i)
if(primeFac == i)
for(int j = i; j < len; j += i)
primeFac = i;
numPrimes = 0;
for(int i = 2; i < len; ++i)
if(primeFac == i)
primes = i;
std::vector<int64_t> test = factors(120, 120);
for(std::vector<int64_t>::iterator i = test.begin(); i != test.end(); ++i)
std::printf("%ld ", *i);
std::printf("\n\n");
std::set<uint64_t> A;
// s = p+q
for(int64_t q = 1; q <= cutoff; ++q)
{
int64_t zeroModS = q*q+1;
std::vector<int64_t> S = factors(zeroModS, q);//std::min(q, 2000000000000000/(q*q)));
if(q == 100000000)
{
std::printf("q = %ld\nS.size() = %zu\n", q, S.size());
for(std::vector<int64_t>::iterator i = S.begin(); i != S.end(); ++i)
std::printf(" %lu\n", *i);
}
//for(int sign = -1; sign <= 1; sign += 2)
int sign = -1;
for(std::vector<int64_t>::iterator i = S.begin(); i != S.end(); ++i)
{
int64_t s = sign * *i;
int64_t p = s-q;
int64_t r = zeroModS/s-q; //-(p*q-1)/(p+q);
int64_t __attribute__((mode(TI))) a = p*q; a *= r;
if(a > 1 && a < 2000000000000000 && p < 2000000000000000/*(1LL<<62)*/)
A.insert(a);
}
}
unsigned n = 0;
for(std::set<uint64_t>::iterator i = A.begin(); i != A.end(); ++i)
{
++n;
std::printf("%lu\n", *i);
if(n == 150000)
{
std::printf("Done\n");
break;
}
}
std::printf("set size = %zu\n", A.size());
}
/*
int main()
{
int cutoff = 1000000000;
std::set<uint64_t> a;
for(int p = 1; p < cutoff; ++p)
{
int range = std::min(cutoff/p, p);
for(int q = -range; q <= range; ++q)
if(q != 0 && p+q != 0)
{
if(((int64_t)p*q-1) % (p+q) == 0)
{
int r = -(p*q-1)/(p+q);
if(r < cutoff)
a.insert(std::max((int64_t)p*q*r, -(int64_t)p*q*r));
}
}
}
a.erase(0);
unsigned n = 0;
for(std::set<uint64_t>::iterator i = a.begin(); i != a.end(); ++i)
{
++n;
std::printf("%lu\n", *i);
if(n == 150000)
{
std::printf("Done\n");
break;
}
}
std::printf("set size = %zu\n", a.size());
}
*/ 这个限定是不是通知编译器可以向量优化？比如用MMX、SIMD、3DNow! 等汇编指令。
还有，标准C语言原生态整数类型是否最高仅支持到64位？128位的四则运算、位运算等还需要程序员自定义？ 7# gxqcn
stackoverflow.com上的回答好像是在说，直接显式的指定一个类型的大小，而与编译器和机器类型无关。
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