这个程序错在哪儿啊

#include<stdio.h>
#include<math.h>
void main()
{
float a,b,c,disc,x1,x2,realpart,imagpart;
scanf("%f,%f,%f",&a,&b,&c);
printf("The equation");
if(fabs(a)<=1e-6)
printf("is not a quadratic\n");
else
{
disc=b*b-4*a*c;
if(fabs(disc)<=1e-6)
printf("has two equal roots:%8.4f\n",-b/(2*a));
else
if(disc>1e-6)
{
x1=(-b+sqrt(disc))/(2*a);
x2=(-b-sqrt(disc))/(2*a);
printf("has distinct real roots:%8.4fand%8.4f\n",x1,x2);
}
else
{
realpart=-b/(2*a);
imagpart=sqrt(-disc)/(2*a);
printf("has complex roots: \n");
printf("%8.4f+%8.4fi\n",realpart,imagpart);
printf("%8.4f-%8.4fi\n",realpart,imagpart);
}
}
}

最简单的方法 就是把float 换成double不就行了

这个warnning就是说double类型的结果转换成float类型 可能会丢失精度

没错，只不过sqrt这个函数操作的是double，返回的也是double，从double到float赋值可能会造成精度丢失，所以编译器给了你个警告，换用sqrff就没有了： #include<stdio.h> #include<math.h> void main() { float a,b,c,disc,x1,x2,realpart,imagpart; scanf("%f,%f,%f",&a,&b,&c); printf("The equation"); if(fabs(a)<=1e-6) printf("is not a quadratic\n"); else { disc=b*b-4*a*c; if(fabs(disc)<=1e-6) printf("has two equal roots:%8.4f\n",-b/(2*a)); else if(disc>1e-6) { x1=(-b+sqrtf(disc))/(2*a); x2=(-b-sqrtf(disc))/(2*a); printf("has distinct real roots:%8.4fand%8.4f\n",x1,x2); } else { realpart=-b/(2*a); imagpart=sqrtf(-disc)/(2*a); printf("has complex roots: \n"); printf("%8.4f+%8.4fi\n",realpart,imagpart); printf("%8.4f-%8.4fi\n",realpart,imagpart); } } }