printf打印浮点数和cout输入结果不一样
- C/C++ code
float val=2.0f; float val2=( sqrt(val)*sqrt(val) )-val; printf("%f\n",val2); cout<<val2;
[解决办法]
去深入了解这两个吧
一个是函数 一个是对象
http://www.cnblogs.com/zdl110110/archive/2010/12/14/1905801.html
http://www.cnblogs.com/meimei888/archive/2010/10/20/1856832.html
[解决办法]
f double类型;十进制小数[-]m.dddddd,其中d的个数由精度指定 (默认值为6)
e double类型;[-]m.dddddd e +/- xx,其中d的个数由精度指定 (默认值为6)
你的printf 按照转化规格(<<0.0000001) 已经为0了
改成
printf(“%.20f\n”,val2);
结果如下:
0.00000000000000027344
2.73436e-16
[解决办法]
C:\Program Files\Microsoft Visual Studio 10.0\VC\include\float.h
- C/C++ code
...#define DBL_DIG 15 /* # of decimal digits of precision */#define DBL_EPSILON 2.2204460492503131e-016 /* smallest such that 1.0+DBL_EPSILON != 1.0 */#define DBL_MANT_DIG 53 /* # of bits in mantissa */#define DBL_MAX 1.7976931348623158e+308 /* max value */#define DBL_MAX_10_EXP 308 /* max decimal exponent */#define DBL_MAX_EXP 1024 /* max binary exponent */#define DBL_MIN 2.2250738585072014e-308 /* min positive value */#define DBL_MIN_10_EXP (-307) /* min decimal exponent */#define DBL_MIN_EXP (-1021) /* min binary exponent */#define _DBL_RADIX 2 /* exponent radix */#define _DBL_ROUNDS 1 /* addition rounding: near */#define FLT_DIG 6 /* # of decimal digits of precision */#define FLT_EPSILON 1.192092896e-07F /* smallest such that 1.0+FLT_EPSILON != 1.0 */#define FLT_GUARD 0#define FLT_MANT_DIG 24 /* # of bits in mantissa */#define FLT_MAX 3.402823466e+38F /* max value */#define FLT_MAX_10_EXP 38 /* max decimal exponent */#define FLT_MAX_EXP 128 /* max binary exponent */#define FLT_MIN 1.175494351e-38F /* min positive value */#define FLT_MIN_10_EXP (-37) /* min decimal exponent */#define FLT_MIN_EXP (-125) /* min binary exponent */#define FLT_NORMALIZE 0#define FLT_RADIX 2 /* exponent radix */#define FLT_ROUNDS 1 /* addition rounding: near */...
[解决办法]
提什么精度,这问题跟精度P关系啊
明显是%f参数默认只有8位(包括小数点不包括符号位)