求助 C++ 的 mode(TI)

wayne

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;

gxqcn

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:

1. typedef unsigned int dword_t __attribute__((mode(DI)));

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On 32 bit systems and

1. typedef unsigned int dword_t __attribute__((mode(TI)));

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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:

1. QI: An integer that is as wide as the smallest addressable unit, usually 8 bits.
2. HI: An integer, twice as wide as a QI mode integer, usually 16 bits.
3. SI: An integer, four times as wide as a QI mode integer, usually 32 bits.
4. DI: An integer, eight times as wide as a QI mode integer, usually 64 bits.
5. SF: A floating point value, as wide as a SI mode integer, usually 32 bits.
6. DF: A floating point value, as wide as a DI mode integer, usually 64 bits.

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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).

wayne

2# gxqcn 我看过了,不知道老大平时用过这个语句没

wayne

感觉把C++用到这种程度的一定是火星人

gxqcn

3# wayne 没用过。我倒是非常需要这样的数据结构。哪里有例程？

wayne

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:

1. #include <cstdio>
2. #include <set>
3. #include <cmath>
4. #include <vector>
5. #include <cassert>
6. //1805315691966540?
8. int const cutoff = 200000;//1638400000;
9. int64_t const len = 1000000/*cutoff+2*/; //(int64_t)cutoff*cutoff+2;
11. int primeFac[len];
12. int primes[len/10];
13. int numPrimes;
15. std::vector<int64_t> factors(int64_t v, int64_t q)
16. {
17. if(v == 1)
18. return std::vector<int64_t>(1, 1);
20. //int p = primeFac[v];
21. int64_t p = v;
22. for(int index = 0; primes[index] < q && index < numPrimes; ++index)
23. if(v%primes[index] == 0)
24. {
25. p = primes[index];
26. break;
27. }
28. int count = 0;
29. //std::printf("* %ld %ld %ld\n", v, p, q);
30. //assert(v < len);
31. //while(p == primeFac[v])
32. while(v%p == 0)
33. {
34. ++count;
35. v /= p;
36. //std::printf(" %d %d\n", v, p);
37. }
38. std::vector<int64_t> ret = factors(v, std::min(2+(int64_t)sqrt(v), q));
39. int siz = ret.size();
40. for(int c = 0; c < count; ++c)
41. for(int s = 0; s < siz; ++s)
42. ret.push_back(ret[s+c*siz]*p);
43. //std::printf("/ %ld %ld %ld\n", v, p, q);
44. return ret;
45. }
47. int main()
48. {
49. for(int i = 0; i < len; ++i)
50. primeFac[i] = i;
51. for(int i = 2; i < len; ++i)
52. if(primeFac[i] == i)
53. for(int j = i; j < len; j += i)
54. primeFac[j] = i;
56. numPrimes = 0;
57. for(int i = 2; i < len; ++i)
58. if(primeFac[i] == i)
59. primes[numPrimes++] = i;
61. std::vector<int64_t> test = factors(120, 120);
62. for(std::vector<int64_t>::iterator i = test.begin(); i != test.end(); ++i)
63. std::printf("%ld ", *i);
64. std::printf("\n\n");
66. std::set<uint64_t> A;
68. // s = p+q
69. for(int64_t q = 1; q <= cutoff; ++q)
70. {
71. int64_t zeroModS = q*q+1;
72. std::vector<int64_t> S = factors(zeroModS, q);//std::min(q, 2000000000000000/(q*q)));
73. if(q == 100000000)
74. {
75. std::printf("q = %ld\nS.size() = %zu\n", q, S.size());
76. for(std::vector<int64_t>::iterator i = S.begin(); i != S.end(); ++i)
77. std::printf(" %lu\n", *i);
78. }
79. //for(int sign = -1; sign <= 1; sign += 2)
80. int sign = -1;
81. for(std::vector<int64_t>::iterator i = S.begin(); i != S.end(); ++i)
82. {
83. int64_t s = sign * *i;
84. int64_t p = s-q;
85. int64_t r = zeroModS/s-q; //-(p*q-1)/(p+q);
86. int64_t __attribute__((mode(TI))) a = p*q; a *= r;
87. if(a > 1 && a < 2000000000000000 && p < 2000000000000000/*(1LL<<62)*/)
88. A.insert(a);
89. }
90. }
92. unsigned n = 0;
93. for(std::set<uint64_t>::iterator i = A.begin(); i != A.end(); ++i)
94. {
95. ++n;
96. std::printf("%lu\n", *i);
97. if(n == 150000)
98. {
99. std::printf("Done\n");
100. break;
101. }
102. }
103. std::printf("set size = %zu\n", A.size());
104. }
106. /*
107. int main()
108. {
109. int cutoff = 1000000000;
110. std::set<uint64_t> a;
111. for(int p = 1; p < cutoff; ++p)
112. {
113. int range = std::min(cutoff/p, p);
114. for(int q = -range; q <= range; ++q)
115. if(q != 0 && p+q != 0)
116. {
117. if(((int64_t)p*q-1) % (p+q) == 0)
118. {
119. int r = -(p*q-1)/(p+q);
120. if(r < cutoff)
121. a.insert(std::max((int64_t)p*q*r, -(int64_t)p*q*r));
122. }
123. }
124. }
125. a.erase(0);
126. unsigned n = 0;
127. for(std::set<uint64_t>::iterator i = a.begin(); i != a.end(); ++i)
128. {
129. ++n;
130. std::printf("%lu\n", *i);
131. if(n == 150000)
132. {
133. std::printf("Done\n");
134. break;
135. }
136. }
137. std::printf("set size = %zu\n", a.size());
138. }
139. */

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gxqcn

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？

wayne

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。