[英]Rounding floating point program
I'm trying to write a C program to find the smallest positive integer x so that (1/x) * x is not equal to 1, using single precision. 我正在尝试编写一个C程序以使用单精度查找最小的正整数x,以使(1 / x)* x不等于1。 And I do it again with double precision. 然后我再次以双精度进行。 I know that the x for single precision is 41, however when I test it by writing C code, I still get 1.00000 我知道单精度的x是41,但是当我通过编写C代码进行测试时,我仍然得到1.00000
This is my test code 这是我的测试代码
int main()
{
float x = 41;
float div = 1/x;
float test = div * x;
printf("%f\n", test);
}
I get 1.0000 instead of a 0.999999 我得到1.0000而不是0.999999
You are not seeing the problem using 您没有看到使用的问题
printf("%f\n", test);
since the default precision used by printf
for floating point numbers is 6
. 因为printf
用于浮点数的默认精度是6
。 If you increase the precision to 10
, you will see that the number is not 1.0
. 如果将精度提高到10
,则会看到该数字不是1.0
。
printf("%.10f\n", test);
prints 0.9999999404
for me. 为我打印0.9999999404
。
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