求解两个单精度浮点数的乘积
求解两个单精度浮点数的乘积
float val=2.22f;
float val2=val*3.33333f;//第一种
float val3=2.22*3.33333;//第二种
哪一个可以保证精度不丢失？
第二：你们一般用什么法子求解两个浮点数的乘积
第一种？第二种？
[解决办法]
貌似第二种精度高一点。
第二种，乘法运算的是double相乘。
C89
3.2.1.5 Usual arithmetic conversions
Many binary operators that expect operands of arithmetic type cause conversions and yield result types in a similar way. The purpose is to yield a common type, which is also the type of the result. This pattern is called the usual arithmetic conversions: First, if either operand has type long double, the other operand is converted to long double. Otherwise, if either operand has type double, the other operand is converted to double. Otherwise, if either operand has type float, the other operand is converted to float. Otherwise, the integral promotions are performed on both operands. Then the following rules are applied: If either operand has type unsigned long int, the other operand is converted to unsigned long int. Otherwise, if one operand has type long int and the other has type unsigned int, if a long int can represent all values of an unsigned int, the operand of type unsigned int is converted to long int ; if a long int cannot represent all the values of an unsigned int, both operands are converted to unsigned long int. Otherwise, if either operand has type long int, the other operand is converted to long int. Otherwise, if either operand has type unsigned int, the other operand is converted to unsigned int. Otherwise, both operands have type int.
The values of operands and of the results of expressions may be represented in greater precision and range than that required by the type; the types are not changed thereby.
[解决办法]
