public class Test {
public static void main(String[] args) {
methode1();
}
static void methode1() {
int ubnd = Integer.MAX_VALUE;
int lbnd = ubnd;
while ((float)lbnd == (float)ubnd) {
--lbnd;
}
System.out.println((++lbnd)
我正在移植一些比较浮点数的代码,以处理64位双倍,而不是32位浮点数,但我对代码中使用的一些神奇数字感到有点困惑。
来源:
下面是代码:
bool AlmostEqual2sComplement(float A, float B, int maxUlps)
{
// Make sure maxUlps is non-negative and small enough that the
// default NAN won't compare as equal to anything.
assert(maxUlps > 0 && maxUlp
在C中,复数是浮点数或双复数,与规范类型有相同的问题:
#include <stdio.h>
#include <complex.h>
int main(void)
{
double complex a = 0 + I * 0;
double complex b = 1 + I * 1;
for (int i = 0; i < 10; i++) {
a += .1 + I * .1;
}
if (a == b) {
puts("Ok");
}
else {
假设我想要一个接受两个浮点数(x和y)的函数,并且我希望不使用它们的float表示,而是使用它们作为32位unsigned int的逐位表示来比较它们。也就是说,像-495.5这样的数字具有位表示0b11000011111001011100000000000000或0xC3E5C000作为float,而我有一个具有相同位表示的unsigned int (对应于十进制值3286614016,我不关心它)。有没有什么简单的方法可以让我只使用它们各自的unsigned int副本中包含的信息来对这些浮点数执行像<=这样的操作?
int main()
{
float a = 0.7;
float b = 0.5;
if (a < 0.7)
{
if (b < 0.5) printf("2 are right");
else printf("1 is right");
}
else printf("0 are right");
}
我原以为这段代码的输出是0 are right。但令我沮丧的是,输出结果是1 is right,为什么?
我目前正在收到以下错误:
错误
OpenCV Error: Assertion failed (0 <= i && i < (int)vv.size()) in getMat_, file
/tmp/binarydeb/ros-kinetic-opencv3-3.1.0/modules/core/src/matrix.cpp, line 1200
terminate called after throwing an instance of 'cv::Exception'
what(): /tmp/binarydeb/ros-kinetic-
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