Question

#include <stdio.h>

#define GA_OF_PA_NEED 267.0

int getSquareFootage(int squareFootage);
double calcestpaint(int squareFootage);
double printEstPaint(double gallonsOfPaint);

int main(void)

{
  //Declaration
  int squareFootage = 0;
  double gallonsOfPaint = 0;


  //Statements
  getSquareFootage(squareFootage);
  gallonsOfPaint = calcestpaint(squareFootage);
  gallonsOfPaint = printEstPaint(gallonsOfPaint);
  system("PAUSE");
  return 0;
}

int getSquareFootage(int squareFootage)
{
  printf("Enter the square footage of the surface: ");
  scanf("%d", &squareFootage);
  return squareFootage;
}

double calcestpaint( int squareFootage)
{
  return (double) (squareFootage * GA_OF_PA_NEED);    
}
double printEstPaint(double gallonsOfPaint)
{

  printf("The estimate paint is: %lf\n",gallonsOfPaint);
  return gallonsOfPaint;
}

为什么我的输出显示gallonsOfPaint为0.0，没有错误，一切似乎在逻辑上正确。看来calc函数中的calculate语句有问题。

Answer 1

您需要指定getSquareFootage(squareFootage);的结果：

squareFootage = getSquareFootage(squareFootage);

由于squareFootage是按值传递的，而不是通过引用传递的，换句话说，在函数中更改它并不重要，它在函数外部没有任何影响。或者，您可以通过引用传递它：

void getSquareFootage(int * squareFootage)
{
     printf("Enter the square footage of the surface: ");
     scanf("%d", squareFootage);
}

将被这样调用：

getSquareFootage(&squareFootage);

Answer 2

更正squareFootage=getSquareFootage();

无需传递参数。

Answer 3

您没有更新变量square Footage。当你调用calcestpaint（square Footage）时，你传递值0作为参数。

输出返回错误的结果

3 个答案: