计算sqrt6的近似值，运行结果sqrt6=2.5，不符合精度要求，哪里出问题了？

Question

#include
#include
int main()
{
int n,z=2;
n=6;
double x[123];
x[1]=n/2.0;
do
{
x[z]=(x[z-1]+n/x[z-1])/2;
printf("sqrt6=%lf",x[z]);
}
while(fabs(x[z]-x[z-1])<1e-5);
return 0;
}

Answer 1

1.z从头到尾没有变，应该再循环里让z自增来增加精度，
2.while条件错误，当精度满足要求是，跳出，不满足要求是继续循环，你大于小于弄反了

int main()
{
    int n, z = 1;
    n = 6;
    double x[123];
    x[1] = n / 2.0;
    do
    {
        ++z;
        x[z] = (x[z - 1] + n / x[z - 1]) / 2;
        printf("sqrt6=%lf\n", x[z]);
    } while (fabs(x[z] - x[z - 1])>1e-5);
    return 0;
}