Adi*_*una 4 c++ casting rounding
假设我有一个双数据类型,变量名为'x'.是否有必要将double转换为long long数据类型给出与转换为long long floor(x)相同的结果.
不,这不一样.
转换截断(向零舍入),floor函数向下舍入.
#include <iostream>
#include <math.h>
int main()
{
double x = -1.4;
std::cout << "(long long)x = " << ((long long)x) << "\n";
std::cout << "(long long)floor(x) = " << ((long long)floor(x)) << "\n";
return 0;
}
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