这个表达有什么问题？ [英] What is wrong in this expression ?

问题描述

嗨我有这个代码，我不知道它有什么问题，你能帮我解决一下吗？谢谢

Hi i have this code, i don't know what is wrong in it, Can you help me to fix ? Thanks

#include<stdio.h>
#include<conio.h>
#define PI 3.1415

float aria(float);
float perimetru(float);


void main(void)
{
float L, a, l, P;

printf("\nIntroduceti valoarea L: ");
scanf("%f", &l);
printf("\nIntroduceti valoarea l: ");
scanf("%f", &L);


printf("\n=======================================");

printf("\nPatratul are:");
printf("\nl = %f ", l);
printf("\nL = %f ", L);
printf("\nperimetru = %f ", perimetru(l, L));
printf("\naria = %f", aria(l && L));

getch();
}
float aria(float)
{ float a, l, L;
a = L * l;
return a;
}
float perimetru(float)
{ float P, l , L;
P = 2*(l+L);
return P;
}

推荐答案

你想要一个清单吗？

因为编译器会给你一个免费，你不必等待......



让我们先看看你的功能来计算结果：

Would you like a list?
Because the compiler will give you one for free, and you don't have to wait at all...

Lets start by looking at your functions for working out the results:

float aria(float)
{ float a, l, L;
a = L * l;
return a;
}
float perimetru(float)
{ float P, l , L;
P = 2*(l+L);
return P;
}

您希望这些值返回什么值？

因为您在本地声明所有变量，所以 aria 相当于：

What values do you expect these to return?
Because you are declaring all your variables locally, so aria is the equivalent of:

float aria()
   {
   float a;
   a = 0.0 * 0.0;
   return a;
   }

您无法从 main 函数中的声明中访问L，a，L或P，因为它们只是在范围内，并在主函数本身内可用。

相反，试试这个：

You can't access L, a, L, or P from the declaration in your main function, because they are only in scope and available within the main function itself.
Instead, try this:

float aria(float l1, float l2)
   {
   return l1 * l2;
   }

你需要更改原型才能匹配。

然后这样称呼它：

You will need to change the prototype to match.
Then call it like this instead:

printf("\naria = %f", aria(l, L));

然后对perimetru做同样的事。



并帮自己一个忙：不要使用单个字符变量名，永远，永远，使用两个仅因情况而异的变量！将l和L放在相同的范围内是一个非常愚蠢的想法，很容易输入错误的...

Then do the same fro perimetru.

And do yourself a favour: don't use single character variable names, and never, ever, ever use two variables that only differ by case! It is a very silly idea to have "l" and "L" in the same scope as it is far, far too easy to type the wrong one...

确切的答案是。

Exact answer is.

#include<stdio.h>
#include<conio.h>
#define PI 3.1415

float aria(float,float);
float perimetru(float,float);


void main(void)
{
float L, a, l, P;

printf("\nIntroduceti valoarea L: ");
scanf("%f", &l);
printf("\nIntroduceti valoarea l: ");
scanf("%f", &L);


printf("\n=======================================");

printf("\nPatratul are:");
printf("\nl = %f ", l);
printf("\nL = %f ", L);
printf("\nperimetru = %f ",perimetru(l, L));
printf("\naria = %f", aria(l,L));

getch();
}
float aria(float l1,float L1)
{float a;
a = L1*l1;
return a;
}
float perimetru(float l2,float L2)
{ float P;
P = 2*(l2+L2);
return P;
}

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