
c语言中float、double的区别和用法? - 知乎
C语言中,float和double都属于 浮点数。区别在于:double所表示的范围,整数部分范围大于float,小数部分,精度也高于float。 举个例子: 圆周率 3.1415926535 这个数字,如果用float …
What is the difference between float and double? - Stack Overflow
2021年12月31日 · I've read about the difference between double precision and single precision. However, in most cases, float and double seem to be interchangeable, i.e. using one or the …
C语言中,double**和double (*) [5]有什么区别?怎么理解? - 知乎
2019年11月24日 · double** 是指向 double* 类型的指针类型。 区别不是很显然的吗? double [5] 类型能隐式转换成 double* 类型,但它们不是同一类型。 可以考虑一下 short 和 long 这两种 …
java - Cast Double to Integer - Stack Overflow
494 A Double is not an Integer, so the cast won't work. Note the difference between the Double class and the double primitive. Also note that a Double is a Number, so it has the method …
c++ - Double precision - decimal places - Stack Overflow
2017年10月20日 · From what I have read, a value of data type double has an approximate precision of 15 decimal places. However, when I use a number whose decimal representation …
c++ - double and accuracy - Stack Overflow
2012年6月19日 · A double which is usually implemented with IEEE 754 will be accurate to between 15 and 17 decimal digits. Anything past that can't be trusted, even if you can make …
Correct format specifier for double in printf - Stack Overflow
Format %lf in printf was not supported in old (pre-C99) versions of C language, which created superficial "inconsistency" between format specifiers for double in printf and scanf.
double 和 long double 有哪些区别? - 知乎
The long double function prototypes are identical to the prototypes for their double counterparts, except that the longdouble data type replaces the double data type. The long double versions …
What's the difference between a single precision and double …
The term double precision is something of a misnomer because the precision is not really double. The word double derives from the fact that a double-precision number uses twice as many bits …
Integer division: How do you produce a double? - Stack Overflow
double d = ((double) num) / denom; But is there another way to get the correct double result? I don't like casting primitives, who knows what may happen.