关于else的有关问题。

2012-11-06

关于else的问题。。。。。。#include stdio.h#include math.hmain(){int a,b,cfloat x1,x2if(b*b - 4*a*c

关于else的问题。。。。。。
#include <stdio.h>
#include <math.h>

main()
{
int a,b,c;
float x1,x2;
if
(b*b - 4*a*c >= 0);
{
printf("input a,b,c:");
scanf("&lf,&lf,&lf,a,b,c");
x1 = (sqrt(b*b-4*a*c)-b)/2*a;
x2 = (-sqrt(b*b-4*a*c)-b)/2*a;
printf("x1 = f%,x2 = f%\n",x1,x2);
}
else
{
printf("no solution");
}

}
运行错误，怎么改。。。

[解决办法]

C/C++ code

#include <stdio.h>#include <math.h>main() {  int a=3,b=4,c=5;  float x1,x2;  if(b*b - 4*a*c >= 0)  {  printf("input a,b,c:");  scanf("&lf,&lf,&lf,a,b,c");  x1 = (sqrt(b*b-4*a*c)-b)/2*a;  x2 = (-sqrt(b*b-4*a*c)-b)/2*a;  printf("x1 = f%,x2 = f%\n",x1,x2);  }  else  {  printf("no solution");  } }
[解决办法]
C/C++ code#include <stdio.h>#include <math.h>main() {  int a,b,c;  float x1,x2;  if(b*b - 4*a*c >= 0)  {      printf("input a,b,c:");      scanf("lf,lf,lf,&a,&b,&c");      x1 = (sqrt(b*b-4*a*c)-b)/2*a;      x2 = (-sqrt(b*b-4*a*c)-b)/2*a;      printf("x1 = f%,x2 = f%\n",x1,x2);  }  else  {      printf("no solution");  }       }
[解决办法]
C/C++ code#include <stdio.h>#include <math.h>main() {  int a,b,c;  float x1,x2;    printf("input a,b,c:");  scanf("%d,%d,%d",&a,&b,&c);  if(b*b - 4*a*c >= 0)  {  x1 = (sqrt(b*b-4*a*c)-b)/2*a;  x2 = (-sqrt(b*b-4*a*c)-b)/2*a;  printf("x1 = %.2f,x2 = %.2f\n",x1,x2);  }  else  {  printf("no solution");  } }
[解决办法]
scanf函数目测用错了，此外，scanf应在if前面

热点排行

C语言

关于else的有关问题。