ios ios中的Mod操作符
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Mod operator in ios
提问by nuteron
have been searching for a mod operator in ios, just like the %in c, but no luck in finding it. Tried the answer in this linkbut it gives the same error.
I have a float variable 'rotationAngle' whose angle keeps incrementing or decrementing based on the users finger movement.
Some thing like this:
一直在 ios 中寻找 mod 运算符，就像%在 c 中一样，但没有找到它。尝试了此链接中的答案，但它给出了相同的错误。我有一个浮动变量“rotationAngle”，它的角度会根据用户手指的移动不断增加或减少。像这样的事情：
if (startPoint.x < pt.x) {
if (pt.y<936/2)
rotationAngle += pt.x - startPoint.x;
else
rotationAngle += startPoint.x - pt.x;
}
rotationAngle = (rotationAngle % 360);
}
I just need to make sure that the rotationAngle doesnot cross the +/- 360 limit. Any help any body. Thanks
我只需要确保rotationAngle 不超过+/- 360 的限制。任何帮助任何身体。谢谢
回答by Hailei
You can use fmod(for double) and fmodf(for float) of math.h:
您可以使用math.h 的fmod(for double) 和fmodf(for float)：
#import <math.h>
rotationAngle = fmodf(rotationAngle, 360.0f);
回答by kuba
Use the fmodfunction, which does a floation-point modulo, for definition see here: http://www.cplusplus.com/reference/clibrary/cmath/fmod/. Examples of how it works (with the return values):
使用执行fmod浮点模数的函数，定义见这里：http: //www.cplusplus.com/reference/clibrary/cmath/fmod/。它是如何工作的示例（带有返回值）：
fmodf(100, 360); // 100
fmodf(300, 360); // 300
fmodf(500, 360); // 140
fmodf(1600, 360); // 160
fmodf(-100, 360); // -100
fmodf(-300, 360); // -300
fmodf(-500, 360); // -140
fmodftakes "float" as arguments, fmodtakes "double" and fmodltakes "double long", but they all do the same thing.
fmodf将“float”作为参数，fmod采用“double”和fmodl“double long”，但它们都做同样的事情。
回答by Jason McTaggart
I cast it to an int first
我首先将它转换为 int
rotationAngle = (((int)rotationAngle) % 360);
if you want more accuracy use
如果你想要更准确的使用
float t = rotationAngle-((int)rotationAngle);
rotationAngle = (((int)rotationAngle) % 360);
rotationAngle+=t;
