C语言 浮点类型的限制？
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Limits for floating point types?
提问by ferz
#include <stdio.h>
#include <limits.h>
int main(void){
printf("Type Size Min Max\n----------------------------------------------------------------------\n");
printf("%-20s%-10d%-20ld%-20ld\n", "long", sizeof(long), LONG_MIN, LONG_MAX);
printf("%-20s%-10d%-20lu%-20lu\n", "unsigned long", sizeof(long), 0, ULONG_MAX);
return 0;
}
where double? i.e. variable LONG_MIN be in file limits.h. in which type double?
哪里双？即变量 LONG_MIN 位于文件limits.h 中。在哪种类型双？
int i, min, max;
for (i = 1.0; i > 0; ++i)
{
max = i;
};
min = i;
printf ("int: min: %d max: %d \n", min, max);
how do for float and double? how min calculated this variable? sorry bad english
float 和 double 怎么做？min 是如何计算这个变量的？抱歉英语不好
回答by Clifford
回答by vmpstr
on linux, I have float.h which has FLT_MAX and DBL_MAX defined for maximum float and double values respectively. I'm not sure how "standard" that is, though...
在 linux 上，我有 float.h，它分别为最大浮点数和双精度值定义了 FLT_MAX 和 DBL_MAX。不过，我不确定那是多么“标准”...
回答by Nick Banks
I think this is what you want:
我认为这就是你想要的：
float: %f
浮点数：%f
long float (double): %lf
长浮点数（双精度）：%lf
You may also want to see it in exponential notation: %E
您可能还想以指数表示法查看它：%E
For min/max for float and double here is what you want
对于 float 和 double 的最小值/最大值，这是您想要的
Here's a snippet from float.h:
这是来自 float.h 的片段：
#define DBL_MAX 1.7976931348623158e+308 /* max value */
#define DBL_MIN 2.2250738585072014e-308 /* min positive value */
#define FLT_MAX 3.402823466e+38F /* max value */
#define FLT_MIN 1.175494351e-38F /* min positive value */
回答by BlackBear
To compute the limits for a given data-type, you have to compute simply (2^(sizeof(type) * 8)) - 1, which is (2^number_of_bits) - 1.
要计算给定数据类型的限制，您必须简单地计算(2^(sizeof(type) * 8)) - 1，即(2^number_of_bits) - 1。
Then, if you consider this type to be signed min and max values are -2^(number_of_bits - 1)and (2^(number_of_bits - 1)) - 1or if they're unsigned MIN will be 0 and MAX (2^number_of_bits) - 1.
然后，如果您认为这种类型是有符号的 min 和 max 值是-2^(number_of_bits - 1)和(2^(number_of_bits - 1)) - 1或者如果它们是无符号的 MIN 将是 0 和 MAX (2^number_of_bits) - 1。
This applies only to integer types, so not for floats and doubles, and then only for Two's Complement integer representations.
这仅适用于整数类型，因此不适用于浮点数和双精度数，然后仅适用于二进制补码整数表示。
