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/questio ... -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
https://bbs.emath.ac.cn/static/image/smiley/1/victory.gif
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？ 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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