当使用int64_t或uint64_t操作数的乘法运算在C中溢出时,是否有任何高效便捷的方式？
例如,为了添加uint64_t,我可以做：
if (UINT64_MAX - a < b) overflow_detected(); else sum = a + b;
但是我不能得到一个类似的简单的乘法表达式.
所有这一切发生在我身上,将操作数分解成高低的uint32_t部分,并执行这些部分的乘法,同时检查溢出,真的很丑陋,也可能是低效的.
基准计划：
#include <stdio.h> #include <stdint.h> #include <stdlib.h> #define N 100000000 int d = 2; #define POW_2_64 ((double)(1 << 31) * (double)(1 << 31) * 4) #define calc_b (a + c) // #define calc_b (a + d) int main(int argc,char *argv[]) { uint64_t a; uint64_t c = 0; int o = 0; int opt; if (argc != 2) exit(1); opt = atoi(argv[1]); switch (opt) { case 1: /* faked check,just for timing */ for (a = 0; a < N; a++) { uint64_t b = a + c; if (c > a) o++; c += b * a; } break; case 2: /* using division */ for (a = 0; a < N; a++) { uint64_t b = a + c; if (b && (a > UINT64_MAX / b)) o++; c += b * a; } break; case 3: /* using floating point,unreliable */ for (a = 0; a < N; a++) { uint64_t b = a + c; if ((double)UINT64_MAX < (double)a * (double)b) o++; c += b * a; } break; case 4: /* using floating point and division for difficult cases */ for (a = 0; a < N; a++) { uint64_t b = a + c; double m = (double)a * (double)b; if ( ((double)(~(uint64_t)(0xffffffff)) < m ) && ( (POW_2_64 < m) || ( b && (a > UINT64_MAX / b) ) ) ) o++; c += b * a; } break; case 5: /* Jens Gustedt method */ for (a = 0; a < N; a++) { uint64_t b = a + c; uint64_t a1,b1; if (a > b) { a1 = a; b1 = b; } else { a1 = b; b1 = a; } if (b1 > 0xffffffff) o++; else { uint64_t a1l = (a1 & 0xffffffff) * b1; uint64_t a1h = (a1 >> 32) * b1 + (a1l >> 32); if (a1h >> 32) o++; } c += b1 * a1; } break; default: exit(2); } printf("c: %lu,o: %u\n",c,o); }
到目前为止,使用浮点过滤大多数情况的情况4是最快的,当假定溢出是非常不寻常的,至少在我的电脑上,它比无人操作的情况下慢两倍.
情况5比4慢了30％,但它总是执行相同的,没有任何特殊的案例编号需要较慢的处理,就像4发生的那样.
如果你想避免在Ambroz的答案中划分：
原文链接：https://www.f2er.com/windows/371631.html首先你必须看到两个数字中的较小的一个,比如a,小于232,否则结果会溢出.令b被分解为b = c 232 d的两个32位字.
那么计算不是那么难,我发现：
uint64_t mult_with_overflow_check(uint64_t a,uint64_t b) {
if (a > b) return mult_with_overflow_check(b,a);
if (a > UINT32_MAX) overflow();
uint32_t c = b >> 32;
uint32_t d = UINT32_MAX & b;
uint64_t r = a * c;
uint64_t s = a * d;
if (r > UINT32_MAX) overflow();
r <<= 32;
return addition_with_overflow_check(s,r);
}
所以这是两次乘法,两班,一些补充和条件检查.这可能比划分更有效率,因为例如两个乘法可以并行流水线.您必须进行基准测试,才能看到更适合您的功能.
