Question

代码：

    int a1, b1, c1, a2, b2, c2;

    printf("This is the program to help you solve \n First Degree In Two Variable Equation \n");

    printf("The standard of \n First Degree In Two Variable Equation is follow: "
        " \n a1 * x + b1 * y = c1\n a2 * x + b2 * y = c2  ");

    printf("Please enter a1");
    scanf("%d", &a1);

    printf("Please enter b1");
    scanf("%d", &b1);

    printf("Please enter c1");
    scanf("%d", &c1);

    printf("Please enter a2");
    scanf("%d", &a2);

    printf("Please enter b2");
    scanf("%d", &b2);

    printf("Please enter c2");
    scanf("%d", &c2);
    if (a1 * b2 == a2 * b1)
        printf("Coefficient's VALUE ARE NOT ALLOW please enter again");
    else
    {
        double x = (double)(b2 * c1 - b1 * c2) / (a1 * b2 - a2 * b1);
        double y = (double)(a1 * c2 - a2 * c1) / (a1 * b2 - a2 * b1);
        printf(" x = " + x + "\n y = " + y);
    }
    return 0;
}

尝试将此代码放在Visual Studio中，您将在代码printf的末尾看到（＆＃34; x =＆＃34; + * x * +＆＃34; \ ny =＆＃34; + y ）;你可以看到x我用*来阻止显示错误信息Expression必须有整数或无范围的枚举类型！我该如何解决？

Answer 1

尝试以下代码：

int a1, b1, c1, a2, b2, c2;

printf("This is the program to help you solve \n First Degree In Two Variable Equation \n");

printf("The standard of \n First Degree In Two Variable Equation is followed: "
    " \n a1 * x + b1 * y = c1\n a2 * x + b2 * y = c2  ");

printf("Please enter a1");
scanf("%d", &a1);

printf("Please enter b1");
scanf("%d", &b1);

printf("Please enter c1");
scanf("%d", &c1);

printf("Please enter a2");
scanf("%d", &a2);

printf("Please enter b2");
scanf("%d", &b2);

printf("Please enter c2");
scanf("%d", &c2);
if (a1 * b2 == a2 * b1)
    printf("Coefficient's VALUE ARE NOT ALLOW please enter again");
else
{
    double x = (double)(b2 * c1 - b1 * c2) / (a1 * b2 - a2 * b1);
    double y = (double)(a1 * c2 - a2 * c1) / (a1 * b2 - a2 * b1);
    printf(" x = %f\ny = %f\n", x, y);
}
return 0;
}

错误消息＆＃34;表达式必须具有整数或未整合的枚举类型＆＃34;

1 个答案: