// t: current time, b: begInnIng value, c: change In value, d: duration
def: 'easeOutQuad',
swing: function (x, t, b, c, d) {
//alert(jQuery.easing.default);
return jQuery.easing[jQuery.easing.def](x, t, b, c, d);
},
easeInQuad: function (x, t, b, c, d) {
return c*(t/=d)*t + b;
},
easeOutQuad: function (x, t, b, c, d) {
return -c *(t/=d)*(t-2) + b;
},
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我试图将Robert Penner的Easing Functions转换为python并卡住!任何帮助或其他任何人之前做过这个?
https://github.com/danro/jquery-easing/blob/master/jquery.easing.js
我需要为一个代表三角矩阵的非常大的数组分配内存.我写了以下代码:
const int max_number_of_particles=20000;
float **dis_vec;
dis_vec = new float **[max_number_of_particles];
for (i = 0; i<max_number_of_particles; i++)
dis_vec[i] = new float *[i];
for (i = 0; i<max_number_of_particles; i++)
for (j = 0; j<i; j++)
dis_vec[i][j] = new float[2];
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问题是这样做所需的时间(分配内存)随着矩阵大小的增加而迅速增加.有谁知道这个问题更好的解决方案?
谢谢.
所以,我有一个bash脚本,其中我想要一个条件,这取决于perl脚本返回的内容.我的代码背后的想法如下:
for i in $(ls); do
if $(perl -e "if (\$i =~ /^.*(bleh|blah|bluh)/) {print 'true';}"); then
echo $i;
fi;
done
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目前,这总是返回true,当我在if语句周围尝试[[]]时,我遇到了错误.任何人的想法?
Ps我知道我可以用grep做到这一点,但这只是一个例子.我想知道如何让Bash一般使用Perl输出
Pps我知道我可以用两行来做,将perl输出设置为变量,然后测试该变量值,但我宁愿避免使用那个额外的变量.似乎很浪费.
我对如何做到这一点的最佳实践感到有些困惑.假设我有一个类,例如分配一些内存.我希望它像汽车一样自我毁灭,但也因为某种原因未知而将它放在矢量中.
#include <iostream>
#include <vector>
class Test {
public:
Test();
Test(int a);
virtual ~Test();
int counter;
Test * otherTest;
};
volatile int count = 0;
Test::Test(int a) {
count++;
counter = count;
std::cout << counter << "Got constructed!\n";
otherTest = new Test();
otherTest->counter = 999;
}
Test::Test() {
count++;
counter = count;
std::cout << counter << "Alloced got constructed!\n";
otherTest = NULL;
}
Test::~Test() {
if(otherTest != 0){
std::cout << otherTest->counter << " 1Got destructed" << counter << "\n";
otherTest->counter …Run Code Online (Sandbox Code Playgroud) 我有以下代码:
#include<stdio.h>
#include<conio.h>
#include<math.h>
#include<graphics.h>
void main()
{
int gd=DETECT,gm;
int dx,dy,p,end;
float x1,x2,y1,y2,x,y;
initgraph(&gd,&gm,"");
printf("\nEnter the value of x1: ");
scanf("%f",&x1);
printf("\nEnter the value of y1: ");
scanf("%f",&y1);
printf("\nEnter the value of x2: ");
scanf("%f",&x2);
printf("\nEnter the value of y2: ");
scanf("%f",&y2);
dx=abs(x1-x2);
dy=abs(y2-y1);
p=2*dy-dx;
if(x1>x2)
{
x=x2;
y=y2;
end=x1;
}
else
{
x=x1;
y=y1;
end=x2;
}
putpixel(x,y,10);
while(x<end)
{
x=x+1;
if(p<0)
{
p=p+2*dy;
}
else
{
y=y+1;
p=p+2*(dy-dx);
}
putpixel(x,y,10);
}
getch();
closegraph();
}
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代码主要用于创建一条线.但是当我运行这个程序时,我在控制台中收到错误消息(我使用的是Ubuntu 10.04版本):
test.c:2: fatal …