[英]C++; Floating Point exception; Without ./ operator
The example below, is from OpenCv documentation[1]. 下面的示例来自OpenCv文档[1]。
Mat H(100, 100, CV_64F);
for(int i = 0; i < H.rows; i++)
for(int j = 0; j < H.cols; j++)
H.at<double>(i,j)=1./(i+j);
This works perfectly fine. 这工作得很好。 But in the last line what is ./ operator?
但是最后一行是./运算符? And if I replace it with / it gives me floating point exception.
如果我用/替换它,则会给我浮点异常。
So, in both cases we have infinity when i and j are 0; 因此,在两种情况下,当i和j为0时,我们都是无穷大。 then why do we get floating point exception for the second case?
那为什么第二种情况我们会得到浮点异常呢?
[1] http://docs.opencv.org/trunk/d3/d63/classcv_1_1Mat.html [1] http://docs.opencv.org/trunk/d3/d63/classcv_1_1Mat.html
./
is not an operator. ./
不是运算符。 The dot binds with the 1
, making it a double constant. 点与
1
绑定,使其成为双常量。 It's equivalent to this: 等效于:
1.0 / (i+j+1)
Only a bit shorter. 只是短一点。
When you omit the dot, the expression is evaluated using integer arithmetic, giving all zero for all entries but 0, 0
. 省略点时,将使用整数算术对表达式求值,除
0, 0
所有条目的全零。
The .
的
.
character is part of the 1.
double literal. 字符是
1.
double文字的一部分。 /
is an arithmetic operator so the right hand side expression becomes: /
是算术运算符,因此右侧表达式变为:
1. / (i+j+1);
and the result is a value of type double. 结果是double类型的值。 Omitting the
.
省略
.
character makes it an integer literal of 1
and the expression becomes: 字符使得文字的整数
1
和表达式变为:
1 / (i+j+1);
where both operands are integer values and the result is an integer value. 其中两个操作数都是整数值,结果是整数值。 Spaces in C++ code make no difference to compiler.
C ++代码中的空格对编译器没有影响。 For readability reasons the statement should include spaces where appropriate:
出于可读性原因,该语句应在适当的地方包含空格:
H.at<double>(i,j) = 1. / (i+j+1);
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.