C++过早地长满

发布于 2024-08-22 21:51:26 字数 609 浏览 4 评论 0原文

我在使用 C++ 时遇到了一个奇怪的问题，其中长数据类型早在应该溢出的时候就溢出了。我正在做的（到目前为止已经成功）是让整数表现得像浮点数，以便范围 [-32767,32767] 映射到 [-1.0,1.0]。它的失败之处在于代表浮点数大于 1.0 的较大参数：

inline long times(long a, long b) {
  printf("a=%ld b=%ld ",a,b);
  a *= b;
  printf("a*b=%ld ",a);
  a /= 32767l;
  printf("a*b/32767=%ld\n",a);
  return a;
}

int main(void) {
  printf("%ld\n",times(98301l,32767l));
}

我得到的输出是：

a=98301 b=32767 a*b=-1073938429 a*b/32767=-32775
-32775

所以 times(98301,32767) 类似于 3.0*1.0。当 times 参数小于 32767 (1.0) 时，此代码可以完美运行，但使用上述参数的任何中间步骤都不应溢出 64 位 long。

有什么想法吗？

原文

I'm having a bizarre problem with C++ where the long data type is overflowing long before it should. What I'm doing (with success so far) is to have integers behave like floats, so that the range [-32767,32767] is mapped to [-1.0,1.0]. Where it stumbles is with larger arguments representing floats greater than 1.0:

inline long times(long a, long b) {
  printf("a=%ld b=%ld ",a,b);
  a *= b;
  printf("a*b=%ld ",a);
  a /= 32767l;
  printf("a*b/32767=%ld\n",a);
  return a;
}

int main(void) {
  printf("%ld\n",times(98301l,32767l));
}

What I get as output is:

a=98301 b=32767 a*b=-1073938429 a*b/32767=-32775
-32775

So times(98301,32767) is analogous to 3.0*1.0. This code works perfectly when the arguments to times are less than 32767 (1.0), but none of the intermediate steps with the arguments above should overflow the 64 bits of long.

Any ideas?

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