C ++单精度和双精度混合计算的规则是什么? [英] What are the rules governing C++ single and double precision mixed calculations?
问题描述
result(double)
a(double)
b(float) )
c(float)
d(double)
一个简单的计算: p>
result = a *(b + c)* d
转换的类型和时间,以及如何计算每个计算执行的精度?
一个使用不同类型的程序,编译器会自动升级 ONE 参数,使它们都是相同的。
在这种情况下,升级为双打:
pre code $ result $ a b(b + c)d
$ b $ float tmp1 = b + c; / /加上操作完成浮动。
//所以结果是一个浮点数
double tmp2 = a *(double)tmp1; //在double上完成乘法(因为`a`是双精度)
//所以tmp1将被上变换为double。
double tmp3 = tmp2 * d; //在双打上进行乘法运算
//所以结果是双
result = tmp3; //由于tmp3与结果类型相同,因此不进行转换。
For example, these variables:
result (double)
a (double)
b (float)
c (float)
d (double)
A simple calculation:
result = a * (b + c) * d
How and when are the types converted and how do I figure out what precision each calculation is performed at?
All operations are done on objects of the same type (assuming normal arithmetic operations).
If you write a program that uses different types then the compiler will auto upgrade ONE parameter so that they are both the same.
In this situations floats will be upgraded to doubles:
result = a * (b + c) * d
float tmp1 = b + c; // Plus operation done on floats.
// So the result is a float
double tmp2 = a * (double)tmp1; // Multiplication done on double (as `a` is double)
// so tmp1 will be up converted to a double.
double tmp3 = tmp2 * d; // Multiplication done on doubles.
// So result is a double
result = tmp3; // No conversion as tmp3 is same type as result.
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