将文字定义为另一个的倍数会导致错误的结果

问题描述 投票:0回答:1

[当在1.0/FOUR_PI中使用分母使用4 * PI时,为什么下面的代码给出错误的结果?

#include <stdio.h>
#define PI 3.14159265358979323846
#define FOUR_PI 4.0*PI

int main() {
  printf("4.0*PI   = %.20lf\n", 4.0*PI);
  printf("FOUR_PI  = %.20lf\n", FOUR_PI);
  printf("4.0*PI - FOUR_PI  = %.60lf\n", 4.0*PI - FOUR_PI);
  printf("1.0/(4.0*PI) = %.20lf\n", 1.0/(4.0*PI));
  printf("1.0/FOUR_PI  = %.20lf\n", 1.0/FOUR_PI);
  return 0;
}
c literals
1个回答
0
投票

结果是,在解析文字时,其表达式将替换为代码,而不是其值。即,1.0/FOUR_PI1.0/4.0*PI代替。请参见以下示例进行说明:

#include <stdio.h>
#define PI 3.14159265358979323846
#define FOUR_PI (4.0*PI)
#define FOUR_PI2 4.0*PI

int main() {
  printf("4.0*PI   = %.20lf\n", 4.0*PI);
  printf("FOUR_PI  = %.20lf\n", FOUR_PI);
  printf("FOUR_PI2 = %.20lf\n", FOUR_PI2);
  printf("4.0*PI - FOUR_PI  = %.20lf\n", 4.0*PI - FOUR_PI);
  printf("4.0*PI - FOUR_PI2 = %.20lf\n", 4.0*PI - FOUR_PI2);
  printf("1.0/(4.0*PI)  = %.20lf\n", 1.0/(4.0*PI));
  printf("1.0/FOUR_PI   = %.20lf\n", 1.0/FOUR_PI);
  printf("1.0/(FOUR_PI) = %.20lf\n", 1.0/(FOUR_PI));
  printf("1.0/FOUR_PI2  = %.20lf\n", 1.0/FOUR_PI2);
  return 0;
}

输出为:

4.0*PI   = 12.56637061435917246399
FOUR_PI  = 12.56637061435917246399
FOUR_PI2 = 12.56637061435917246399
4.0*PI - FOUR_PI  = 0.00000000000000000000
4.0*PI - FOUR_PI2 = 0.00000000000000000000
1.0/(4.0*PI)  = 0.07957747154594767280
1.0/FOUR_PI   = 0.07957747154594767280
1.0/(FOUR_PI) = 0.07957747154594767280
1.0/FOUR_PI2  = 0.78539816339744827900
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