你能帮我解决这个问题吗？ [英] Could you help me solve this problem?

问题描述

 {
  int  x，y; 
  float  z; 
 printf（ 写2个整数：）; 
 scanf（ ％d％d，& x，& y）; 
 
  switch （x& y）
 {
 案例 x + y> 0&& x + y< =  4 ：
 z = exp（x + y）; 
 printf（ ％f是输出。\ n，z）; 
  break ; 
 
 
  case  x + y ==  8  || x + y ==  10 ：
z =（pow（x， 2 ）+ y） /（XY）; 
 printf（ ％f是输出。\ n，z）; 
  break ; 
 
 默认：
 z =  0 ; 
 printf（ ％f是输出。\ n，z）; 
} 
  return   0 ; 
} 

我的尝试：



我试图添加If语句，但我不能

解决方案

您的案例陈述无效。案例使用常量值，而不是表达式。您的代码应如下所示：

  int  x，y; 
  float  z; 
 printf（ 写2个整数：）; 
 scanf（ ％d％d，& x，& y）; 
 
  if （x + y> 0&& x + y< =  4 ）
 {
z = exp（x + y）; 
} 
 其他 如果（x + y ==  8  || x + y ==  10 ）
 {
z =（pow（x， 2 ）+ Y）/（XY）; 
} 
  else  
 z =  0 ; 
 
  //  你最后只需要一个printf来打印z的值。 
 printf（ ％f是输出。\ n，z ）; 
 

解决方案1 ​​为您提供切换的可能更正，但由于你没有告诉我们它应该做什么，我们无法确定。

显然，你不知道<$的用法c $ c>开关。

你应该尽快正确地学习这门语言。

建议：

- 阅读语言文档，按照教程。

- 调试器是你的朋友。它会告诉你你的代码到底在做什么。

一步一步地执行，检查变量，你会发现它有一个停止做你期望的点。

掌握Visual Studio 2010中的调试 - 初学者指南 [ ^ ]

http://docs.oracle .com / javase / 7 / docs / technotes / tools / windows / jdb.html [ ^ ]

https://www.jetbrains.com/idea/help/debugging-your-first-java-application.html [ ^ ]

< pre lang =c ++> #include < < span class =code-leadattribute> stdio.h >
main（）
{
int x，y，w; // ad另一个整合
float z;
printf（ 写2个整数：）;
scanf（ ％d％d，& x，& y）;
if （（x + y == 8 || x + y == 10 ））w = 1 ; // < span class =code-comment> state condition 0
if （x + y> 0&& x + y< = 4 ）w = 0 ; // 州条件1

开关（w） // 在此替换为w
{
case 0 ： // 替换为w
z = exp（x + y）;
printf（ ％f是输出。\ n，z）;
break ;


case 1 ： // 在此替换为w
z =（pow（x， 2 ）+ Y）/（XY）;
printf（ ％f是输出。\ n，z）;
break ;

默认：
z = 0 ;
printf（ ％f是输出。\ n，z）;
}
return 0 ;
}

{
    int x,y;
    float z;
    printf("Write 2 integers: ");
    scanf("%d%d",&x,&y);

    switch (x && y)
    {
    case x+y>0 && x+y<=4:
        z = exp(x+y);
        printf("%f is an output.\n",z);
        break;
    
    
    case x+y==8 || x+y==10:
    z = (pow(x,2)+y)/(x-y);
    printf("%f is an output.\n",z);
    break;
    
    default:
        z = 0;
        printf("%f is an output.\n",z);
    }
    return 0;
}

What I have tried:

I tried to add If statement but i couldnt

解决方案

Your case statements are not valid. Cases use constant values, not expressions. Your code should look like:

int x,y;
float z;
printf("Write 2 integers: ");
scanf("%d%d",&x,&y);

if (x+y>0 && x+y<=4)
{
    z = exp(x+y);
}
else if (x+y==8 || x+y==10)
{
    z = (pow(x,2)+y)/(x-y);
}
else
    z = 0;

// you only need one printf at the end to print the value of z.
printf("%f is an output.\n",z);

Solution 1 gives you a probable correction of switch, but since you didn't tell us what it is supposed to do, we can't be sure.
Obviously, you don't know the usage of switch.
You should properly learn the language as soon as possible.
Advice:
- read language documentation, follow tutorials.
- The debugger is your friend. It will show you what your code is really doing.
Follow the execution step by step, inspect variables and you will see that there is a point where it stop doing what you expect.
Mastering Debugging in Visual Studio 2010 - A Beginner's Guide[^]
http://docs.oracle.com/javase/7/docs/technotes/tools/windows/jdb.html[^]
https://www.jetbrains.com/idea/help/debugging-your-first-java-application.html[^]

#include<stdio.h>
main()
{
    int x,y,w; //ad another integar
    float z;
    printf("Write 2 integers: ");
    scanf("%d%d",&x,&y);
    if((x+y==8 || x+y==10)) w=1;//state condition 0
    if(x+y>0 && x+y<=4) w=0;//state condition 1

    switch (w) // replace here by w
    {
    case 0:      // replace here by w      
        z = exp(x+y);
        printf("%f is an output.\n",z);
        break;


    case 1 :   // replace here by w
    z = (pow(x,2)+y)/(x-y);
    printf("%f is an output.\n",z);
    break;

    default:
        z = 0;
        printf("%f is an output.\n",z);
    }
    return 0;
}

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