C语言 C将浮点数转换为int
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C convert floating point to int
提问by SergSoftwares
I'm using C(not C++).
我正在使用C（不是 C++）。
I need to convert a float number into an int. I do not want to round to the the nearest number, I simply want to eliminate what is after the integer part. Something like
我需要将浮点数转换为int. 我不想四舍五入到最接近的数字，我只想消除整数部分之后的内容。就像是
4.9 -> 4.9-> 4
回答by Zach P
my_var = (int)my_var;
As simple as that. Basically you don't need it if the variable is int.
就如此容易。如果变量是int，基本上你不需要它。
回答by user3161739
Use in C
在 C 中使用
int C = var_in_float;
They will convert implicit
他们将转换隐式
Thank you
谢谢
回答by AdriZ
If you want to round it to lower, just cast it.
如果你想把它四舍五入到更低，就投吧。
float my_float = 42.8f;
int my_int;
my_int = (int)my_float; // => my_int=42
For other purpose, if you want to round it to nearest, you can make a little function or a define like this:
出于其他目的，如果你想把它四舍五入到最近，你可以做一个小函数或这样的定义：
#define FLOAT_TO_INT(x) ((x)>=0?(int)((x)+0.5):(int)((x)-0.5))
float my_float = 42.8f;
int my_int;
my_int = FLOAT_TO_INT(my_float); // => my_int=43
Be careful, ideally you should verify float is between INT_MIN and INT_MAX before casting it.
小心，理想情况下，您应该在转换之前验证浮点数介于 INT_MIN 和 INT_MAX 之间。
回答by Legu
double a = 100.3;
printf("%f %d\n", a, (int)(a* 10.0));
Output Cygwin 100.3 1003
Output MinGW: 100.3 1002
Using (int) to convert double to int seems not to be fail-safe
使用 (int) 将 double 转换为 int 似乎不是故障安全的
You can find more about that here: Convert double to int?
您可以在此处找到更多相关信息：Convert double to int?
回答by Tomi Ollila
Good guestion! -- where I have not yet found a satisfying answer for my case, the answer I provide here works for me, but may not be future proof...
好客！- 在我还没有为我的案例找到令人满意的答案的地方，我在这里提供的答案对我有用，但可能不是未来的证明......
If one uses gcc (clang?) and have -Werrorand -Wbad-function-castdefined,
如果一个人使用 gcc (clang?) 并且有-Werror和 -Wbad-function-cast定义，
int val = (int)pow(10,9);
will result:
将导致：
error: cast from function call of type 'double' to non-matching type 'int' [-Werror=bad-function-cast]
error: cast from function call of type 'double' to non-matching type 'int' [-Werror=bad-function-cast]
(for a good reason, overflow and where values are rounded needs to be thought out)
（有充分的理由，需要考虑溢出和四舍五入的位置）
After properly thought out (that parameters to pow() are good), int val = pow(10,9);seems to work with gcc 9.2 x86-64 ...
经过适当的考虑（pow() 的参数很好），int val = pow(10,9);似乎适用于 gcc 9.2 x86-64 ...
but note:
但请注意：
printf("%d\n", pow(10,4));
may output e.g.
可以输出例如
-1121380856
(did for me) where
（为我做的）在哪里
int i = pow(10,4); printf("%d\n", i);
printed
打印
10000
in one particular case I tried.
在我尝试过的一种特殊情况下。
