将浮点数/双精简舍入到最接近的1/10 [英] rounding a float/double to nearest 1/10th
问题描述
任何人都有一个更好/更简单的方法来将浮点数舍入到最近的
1/10？这是我正在使用的，并且必须有更好的方式，或者可能是我不知道的罐装方法。
double z = atof（arg [1]）;
z = z * 100.0;
int zi =（int）floor（（double）z）;
int ri = zi％10;
zi - = ri;
zi + =（ri< 5）？ 0:10;
z =（double）zi /（double）100;
注意：最初双打是浮点数，但1.5 * 100 = 149.9999时
使用花车时，使用双打时给出的正确答案。
任何建议都会非常感谢。
~
Any one have a better/simpler method for rounding a float to the nearest
1/10th? This is currently what I am using, and there must be a better
way, or perhaps a canned method that I am not aware of.
double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;
NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.
Any suggestions would be great thank.
~S
推荐答案
Shea Martin写道：
Shea Martin wrote:
任何人都有一个更好/更简单的方法来舍入浮点数到最近的
1/10？这是我正在使用的，并且必须有更好的方法，或者可能是我不知道的罐装方法。
double z = atof（arg [1]） ;
z = z * 100.0;
int zi =（int）floor（（double）z）;
int ri = zi％10;
zi - = ri;
zi + =（ri <5）？ 0：10;
z =（double）zi /（double）100;
注意：最初双打是浮点数，但是当使用浮点数时1.5 * 100 = 149.9999，使用双打时给出的正确答案。
任何建议都会非常感谢。
~S
Any one have a better/simpler method for rounding a float to the nearest
1/10th? This is currently what I am using, and there must be a better
way, or perhaps a canned method that I am not aware of.
double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;
NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.
Any suggestions would be great thank.
~S
float z2 = atof（arg [1]）;
z2 =（浮动）楼层（z2 * 10 + 0.5）/ 10;
我认为这是一个好多了。
~S
float z2 = atof(arg[1]);
z2 = (float)floor(z2*10+0.5)/10;
I thinks this is a lot better.
~S
> float z2 = atof（arg [1]）;
> float z2 = atof(arg[1]);
z2 =（float）floor（z2 * 10 + 0.5）/ 10;
z2 = (float)floor(z2*10+0.5)/10;
更好
z2 =（浮动）（楼层（z2 * 10 + 0.5）/ 10）
所以价值被砍成了一个分裂后浮动（如果有的话）。
better
z2 = (float)(floor( z2 * 10 + 0.5 ) / 10)
So the value gets chopped to a float (if at all) after the division.
Shea Martin写道：
Shea Martin wrote:
任何人都有一个更好/更简单的方法来将浮点数舍入到最近的1/10？这是我正在使用的，并且必须有更好的方法，或者可能是我不知道的罐装方法。
double z = atof（arg [1]） ;
z = z * 100.0;
int zi =（int）floor（（double）z）;
int ri = zi％10;
zi - = ri;
zi + =（ri <5）？ 0：10;
z =（double）zi /（double）100;
注意：最初双打是浮点数，但是当使用浮点数时1.5 * 100 = 149.9999，使用双打时给出的正确答案。
Any one have a better/simpler method for rounding a float to the nearest
1/10th? This is currently what I am using, and there must be a better
way, or perhaps a canned method that I am not aware of.
double z = atof(arg[1]);
z = z*100.0;
int zi = (int)floor((double)z);
int ri = zi%10;
zi -= ri;
zi += ( ri < 5 ) ? 0 : 10;
z = (double)zi/(double)100;
NOTE: originally the doubles were floats, but 1.5*100 = 149.9999 when
using floats, correct answer given when doubles used.
z = 0.1 * round（z * 10.0）;
z = 0.1 * round( z * 10.0 );
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