问题是代码在beetleSimulation
下的while循环中,它会永远存在,而不是在x/yCount
超出边界时退出。 x
和y
比20更进一步,任何人都可以帮我解释原因吗?
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define PI 3.14159265
void beetleSimulation(int, int);
int main ( int argc, char *argv[] )
{
if ( argc != 3 ) // argc should be 2 for correct execution
{
// If the number of arguments is not 2
printf("Program only has %d arguments ", argc);
return 0;
}
else
{
//run the beetle simulation
beetleSimulation(atoi(argv[1]), atoi(argv[2]) );
return 0;
}
}
void beetleSimulation(int size, int iterations){
int i;
double xCount = 0;
double yCount = 0;
int timeCount = 0;
int overallCount = 0;
int degree;
double radian;
for(i=0; i < 10; i++){
while(xCount < 20 || xCount > -20 || yCount <20 || yCount >-20){
timeCount += 1;
degree = rand() % 360;
radian = degree / (180 * PI);
xCount += sin(radian);
yCount += cos(radian);
printf("X and Y are %f and %f\n", xCount, yCount);
}
//when beetle has died, add time it took to overall count, then go through for loop again
overallCount += timeCount;
}
//calculate average time
double averageTime = overallCount/iterations;
printf("Average Time is %f",averageTime);
}
答案 0 :(得分:2)
目前您的循环条件将始终为真,请记住,如果任何条件为真,则使用or运算符,整个表达式将计算为true。
您希望在while循环条件中使用ands。有了它们,只有在所有条件都成立的情况下,循环才会继续。
while(xCount > -20 && xCount < 20 && yCount > -20 && yCount < 20)