C 代码编译期间的错误

Question

我在使用此代码时遇到了一些问题。我已在代码末尾包含错误。

#include <stdio.h>

int main()

{
void addition(double number1, double number2); /* create the functions */
void subtraction(double number1, double number2);
void division(double number1, double number2);
void multiplication(double number1, double number2);
int inputfunc=1;
double inputnum1=0;
double inputnum2=0;
int number1;
int number2;
int answer;

while (inputfunc >= 1 && inputfunc <= 4) /* If function to be performed are those   below then continue performing loop */

{
printf("Press 1 to add two numbers.\n");
printf("Press 2 to subtract two numbers.\n");
printf("Press 3 to multiply two numbers.\n");
printf("Press 4 to divide two numbers.\n");
printf("Press 5 to exit.\n");
printf("Enter your choice\n");
scanf_s("%d", &inputfunc);

if( inputfunc == 5) /* Exit program if requested via 5 function */
return(0);

printf("Enter both numbers with a space in between.");
scanf_s("%lf %lf", &inputnum1, &inputnum2);

void(*func[4])(double, double)={&addition, &subtraction, &division, &multiplication};
  (*func[inputfunc-1])(inputnum1, inputnum2); 
  return(0);
  }

}

void addition(double number1, double number2) 
{
  double answer; 
  answer=number1+number2;
  printf("Addition of the two numbers = %lf + %lf = %lf\n", number1, number2, answer); 
  return;
} 

void subtraction(double number1, double number2)
{ 
  double answer; 
  answer=number1-number2;
  printf("By subtracting the two numbers results are %lf - %lf = %lf\n", number1,
    number2, answer); 
  return;
} 

void multiplication(double number1, double number2) 
{ 
  double answer; 
  answer=number1*number2;
  printf("By multiplying the two numbers results are %lf * %lf = %lf\n", number1, 
      number2, answer); 
  return;
} 

void division(double number1, double number2) 
{ 
  double answer; 
  answer=number1/number2;
  printf("By dividing the two numbers results are %lf / %lf = %lf\n", number1, 
        number2,  answer); 
  return ;
}

错误 C2143：语法错误：缺少 ';'在“类型”之前 错误 C2065：“func”：未声明的标识符 错误C2109：下标需要数组或指针类型

Answer 1

在我从您的代码下面发布的那一行中，您将用第二个大括号结束 main 方法。 这不是一个好主意，因为你在那个大括号下面有代码。

void(*func[4])(double, double)={&addition, &subtraction, &division, &multiplication};
(*func[inputfunc-1])(inputnum1, inputnum2); 
return(0);
} //end while

}  //end main method

Answer 2

代码编译并工作 - 见it in action here

几点：

• 这是 C 代码，不是 C#
• return(0); 被放置在while 循环中 - 这将防止它向用户询问不止一次运行。将此移动到 while 循环的末尾下方。
• 包含算术运算符的各种函数指针的数组的声明不应放置在 while 循环中 - 它不会在每次迭代之间更改 - 即将 void(*func[4])(double, double) = { &addition, &subtraction, &division, &multiplication }; 移动到 while 上方
• 更好的缩进将使代码更具可读性
• int number1; int number2; int answer; 的最顶层声明是多余的，应删除（尤其是在 4 个算术函数中用作具有不同类型的局部变量名称时）。

我对 sn-p 进行了上述更改（scanf_s 被简单的 scanf 替换，因为 IdeOne 不使用 MS 编译器）。

Answer 3

// the scanf_s function is the secure function for string input, using scanf

// this version compiles without any warnings, etc under linux gcc

// this version checks for input errors 
// (except the actual value of the variable inputFunc)
// I think the use of a switch() statement would be much more robust
// rather than the use of the function ptr table, although not quite as flexable

// Notice the function prototypes are outside of any function
// so the compiler will create the proper code

#include <stdio.h>
#include <stdlib.h> // contains exit() and EXIT_FAILURE

// function prototypes
void addition(double number1, double number2);
void subtraction(double number1, double number2);
void division(double number1, double number2);
void multiplication(double number1, double number2);


void(*func[4])(double, double)={&addition, &subtraction, &division, &multiplication};

int main()
{

    int inputFunc=1;
    double inputnum1=0;
    double inputnum2=0;

    /* If function to be performed are those   
       below then continue performing loop */

    // note:
    // if the inputFunc is (for instance) 6 then this while loop is exited
    // however, there was no return statement
    // for that execution path

    while (inputFunc >= 1 && inputFunc <= 4) 
    {
        printf("Press 1 to add two numbers.\n");
        printf("Press 2 to subtract two numbers.\n");
        printf("Press 3 to multiply two numbers.\n");
        printf("Press 4 to divide two numbers.\n");
        printf("Press 5 to exit.\n");
        printf("Enter your choice\n");

        if( 1 != scanf(" %d", &inputFunc) )
        {
            perror( "scanf_s" );
            exit( EXIT_FAILURE ) ;
        }

        // implied else, scanf for which command was successful

        // note: there should be some checking here 
        //       to assure that the input was in the valid range '1...5'

        if( inputFunc == 5)
        { /* then, Exit while loop if requested via 5 function */
          // note: good program practice is to put the return
          //       at the bottom of the function
             break;
        }

        printf("Enter both numbers with a space in between.");
        if( 2 != scanf(" %lf %lf", &inputnum1, &inputnum2) )
        {
            perror( "scanf for 2 input numbers" );
            exit(EXIT_FAILURE);
        }

        // implied else, scanf for two input numbers successful

        // exec the desired function
        (*func[inputFunc-1])(inputnum1, inputnum2);
    }
    return(0); // to avoid compiler warning
}

void addition(double number1, double number2)
{
  double answer;
  answer=number1+number2;
  printf("Addition of the two numbers = %lf + %lf = %lf\n", number1, number2, answer);
  return;
}

void subtraction(double number1, double number2)
{
    double answer;
    answer=number1-number2;
    printf("By subtracting the two numbers results are %lf - %lf = %lf\n",
        number1,
        number2,
        answer);
}

void multiplication(double number1, double number2)
{
    double answer;
    answer=number1*number2;
    printf("By multiplying the two numbers results are %lf * %lf = %lf\n",
        number1,
        number2,
        answer);
}

void division(double number1, double number2)
{
    // note:  this should check that number2 is NOT 0, to avoid divide by zero error
    double answer;
    answer=number1/number2;
    printf("By dividing the two numbers results are %lf / %lf = %lf\n",
        number1,
        number2,
        answer);
}