rand () function C++

本文关键字：C++ function rand 更新时间：2023-10-16

我需要生成4个随机数，每个都在[-45+45]度之间。如果rand%2=0，那么我想要结果(生成的随机数等于-angle)。一旦生成了4个随机数，就需要扫描这些角度并找到一个锁(角度相交的点)。还有-3，-2，-1+如果语句中的循环中的3指示锁定发生在6度波束宽度内。代码有效。但是它可以简化吗？目标是通过扫描两个点的仰角和方位角，在两个点之间建立锁定。

#include <iostream>
#include <conio.h>
#include <time.h>
using namespace std;
class Cscan
{
public:
int gran, lockaz, lockel;
};
int main()
{
srand (time(NULL));
int az1, az2, el1, el2, j, k;

BS1.lockaz = rand() % 46;
BS1.lockel = rand() % 46;
BS2.lockaz = rand() % 46;
BS2.lockel = rand() % 46;
k = rand() % 2;
if(k == 0)
k = -1;
BS1.lockaz = k*BS1.lockaz;
k = rand() % 2;
if(k == 0)
k = -1;
BS1.lockel = k*BS1.lockel;
k = rand() % 2;
if(k == 0)
k = -1;
BS2.lockaz = k*BS2.lockaz;
k = rand() % 2;
if(k == 0)
k = -1;
BS2.lockel = k*BS2.lockel;
for(az1=-45; az1<=45; az1=az1+4)
{
for(el1=-45; el1<=45; el1=el1+4)
{
for(az2=-45; az2<=45; az2=az2+4)
{
for(el2=-45; el2<=45; el2=el2+4)
{
if((az1==BS1.lockaz-3||az1==BS1.lockaz-2||az1==BS1.lockaz-1||az1==BS1.lockaz||az1==BS1.lockaz+1||az1==BS1.lockaz+2||az1==BS1.lockaz+3)&&
(az2==BS2.lockaz-3||az2==BS2.lockaz-2||az2==BS2.lockaz-1||az2==BS2.lockaz||az2==BS2.lockaz+1||az2==BS2.lockaz+2||az2==BS2.lockaz+3)&&
(el1==BS1.lockel-3||el1==BS1.lockel-2||el1==BS1.lockel-1||el1==BS1.lockel||el1==BS1.lockel+1||el1==BS1.lockel+2||el1==BS1.lockel+3)&&
(el2==BS2.lockel-3||el2==BS2.lockel-2||el2==BS2.lockel-1||el2==BS2.lockel||el2==BS2.lockel+1||el2==BS2.lockel+2||el2==BS2.lockel+3))
{      
cout << "locked n" << BS1.lockaz << " " << BS1.lockel << " " << BS2.lockaz << " " << BS2.lockel <<endl
< az1 << " " << el1 << " " << az2 << " " << el2 << endl;
k = 1;
break;
}
if(k==1)
break;
}
if(k==1)
break;
}
if(k==1)
break;
}
if(k==1)
break;
}
_getch();
}

BS1.lockaz = rand() % 91 - 45;
BS1.lockel = rand() % 91 - 45;
BS2.lockaz = rand() % 91 - 45;
BS2.lockel = rand() % 91 - 45;

以度为单位的整数角度？非常值得怀疑。像角度这样的"物理"通常最好用浮点数表示，所以我首先要更改

typedef double angle;
struct Cscan {  // why class? This is clearly POD
int gran; //I don't know what gran is. Perhaps this should also be floating-point.
angle lockaz, lockel;
};

乍一看，这似乎让它变得更加困难，因为%的随机范围选择既不起作用，也没有太多用处来比较浮点值是否相等然而，这是一件好事，因为所有这些实际上都是非常糟糕的做法。

如果您想继续使用rand()作为随机数生成器(尽管我建议使用std::uniform_real_distribution)，请编写一个函数来实现这一点：

const double pi = 3.141592653589793;  // Let's use radians internally, not degrees.
const angle rightangle = pi/2.;      //  It's much handier for real calculations.
inline angle deg2rad(angle dg) {return dg * rightangle / 90.;}
angle random_in_sym_rightangle() {
return rightangle * ( ((double) rand()) / ((double) RAND_MAX) - .5 );
}

现在你只需要做

BS1.lockaz = random_in_sym_rightangle();
BS1.lockel = random_in_sym_rightangle();
BS2.lockaz = random_in_sym_rightangle();
BS2.lockel = random_in_sym_rightangle();

然后你需要做这个范围检查。这又是一个需要放入专用功能的东西

bool equal_in_margin(angle theta, angle phi, angle margin) {
return (theta > phi-margin && theta < phi+margin);
}

然后您对锁进行彻底的搜索。这肯定可以更有效地完成，但这是一个算法问题，与语言无关。坚持使用for循环，通过避免这种显式的中断检查，仍然可以使它们看起来更好。一种方法是好的旧goto，我建议你把它放在一个额外的函数中，完成后返回

#define TRAVERSE_SYM_RIGHTANGLE(phi) 
for ( angle phi = -pi/4.; phi < pi/4.; phi += deg2rad(4) )
int lock_k  // better give this a more descriptive name
( const Cscan& BS1, const Cscan& BS2, int k ) {
TRAVERSE_SYM_RIGHTANGLE(az1) {
TRAVERSE_SYM_RIGHTANGLE(el1) {
TRAVERSE_SYM_RIGHTANGLE(az2) {
TRAVERSE_SYM_RIGHTANGLE(el2) {
if( equal_in_margin( az1, BS1.lockaz, deg2rad(6.) )
&& equal_in_margin( el1, BS1.lockel, deg2rad(6.) )
&& equal_in_margin( az2, BS1.lockaz, deg2rad(6.) )
&& equal_in_margin( el2, BS2.lockel, deg2rad(6.) ) ) {
std::cout << "locked n" << BS1.lockaz << " " << BS1.lockel << " " << BS2.lockaz << " " << BS2.lockel << 'n'
<< az1 << " " << el1 << " " << az2 << " " << el2 << std::endl;
return 1;
}
}
}
}
}
return k;
}