如何在C中打印尽可能大的float和double? [英] How would I print the largest possible float and double in C?

问题描述

对于以下代码，

 #include <stdio.h>
#include <limits.h>
#include <float.h>

int main(void) {
    printf("double max = %??\n", DBL_MAX);
    printf("double min = %??\n", DBL_MIN);
    printf("double epsilon  = %??\n", DBL_EPSILON);
    printf("float epsilon  = %??\n", FLT_EPSILON);
    printf("float max = %??\n", FLT_MAX);
    printf("float min = %??\n\n", FLT_MIN);
    return 0;
}
 

为了使printf将各种数量显示为适当大小的十进制数字，我必须使用哪些说明符代替???

解决方案

使用与这些类型的其他任何值相同的格式:

 #include <float.h>
#include <stdio.h>
int main(void) {
    printf("FLT_MAX = %g\n", FLT_MAX);
    printf("DBL_MAX = %g\n", DBL_MAX);
    printf("LDBL_MAX = %Lg\n", LDBL_MAX);
}
 

对于printf等可变函数，类型float的参数被提升为double，这就是为什么您对两者使用相同格式的原因.

%f使用不带指数的十进制表示法来打印浮点值，这将为您提供一个很长的字符串(对于大多数值来说是无关紧要的).

%e强制使用指数.

%g根据要打印的数字的大小使用%f或%e.

在我的系统上，上面显示了以下内容:

 FLT_MAX = 3.40282e+38
DBL_MAX = 1.79769e+308
LDBL_MAX = 1.18973e+4932
 

正如Eric Postpischil在评论中指出的那样，以上内容仅显示这些值的近似值.您可以通过指定精度来打印更多数字(您需要的位数取决于类型的精度)；例如，您可以将%g替换为%.20g.

或者，如果您的实现支持它，则C99可以根据需要以十六进制的精度打印浮点值的功能:

 printf("FLT_MAX = %a\n", FLT_MAX);
printf("DBL_MAX = %a\n", DBL_MAX);
printf("LDBL_MAX = %La\n", LDBL_MAX);
 

但是结果并不像通常的十进制格式那样容易被人类理解:

FLT_MAX = 0x1.fffffep+127
DBL_MAX = 0x1.fffffffffffffp+1023
LDBL_MAX = 0xf.fffffffffffffffp+16380

(注意:main()是一个过时的定义；请使用int main(void).)

For the following code,

#include <stdio.h>
#include <limits.h>
#include <float.h>

int main(void) {
    printf("double max = %??\n", DBL_MAX);
    printf("double min = %??\n", DBL_MIN);
    printf("double epsilon  = %??\n", DBL_EPSILON);
    printf("float epsilon  = %??\n", FLT_EPSILON);
    printf("float max = %??\n", FLT_MAX);
    printf("float min = %??\n\n", FLT_MIN);
    return 0;
}

what specifiers would I have to use in place of the ??'s in order for printf to display the various quantities as appropriately-sized decimal numbers?

解决方案

Use the same format you'd use for any other values of those types:

#include <float.h>
#include <stdio.h>
int main(void) {
    printf("FLT_MAX = %g\n", FLT_MAX);
    printf("DBL_MAX = %g\n", DBL_MAX);
    printf("LDBL_MAX = %Lg\n", LDBL_MAX);
}

Arguments of type float are promoted to double for variadic functions like printf, which is why you use the same format for both.

%f prints a floating-point value using decimal notation with no exponent, which will give you a very long string of (mostly insignificant) digits for very large values.

%e forces the use of an exponent.

%g uses either %f or %e, depending on the magnitude of the number being printed.

On my system, the above prints the following:

FLT_MAX = 3.40282e+38
DBL_MAX = 1.79769e+308
LDBL_MAX = 1.18973e+4932

As Eric Postpischil points out in a comment, the above prints only approximations of the values. You can print more digits by specifying a precision (the number of digits you'll need depends on the precision of the types); for example, you can replace %g by %.20g.

Or, if your implementation supports it, C99 added the ability to print floating-point values in hexadecimal with as much precision as necessary:

printf("FLT_MAX = %a\n", FLT_MAX);
printf("DBL_MAX = %a\n", DBL_MAX);
printf("LDBL_MAX = %La\n", LDBL_MAX);

But the result is not as easily human-readable as the usual decimal format:

FLT_MAX = 0x1.fffffep+127
DBL_MAX = 0x1.fffffffffffffp+1023
LDBL_MAX = 0xf.fffffffffffffffp+16380

(Note: main() is an obsolescent definition; use int main(void) instead.)

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