错误：表达式语法，警告：R被分配一个从未使用过的值。，plz help [英] Error: expression syntax, warning: R is assigned a value that is never used., plz help

问题描述

  #include   <   stdio.h  >  
   #include   <   conio.h  >  
   #include   <   math.h  >  
   void  main（）
 {
  int  p =  20 ，q =  15 ，r，s; 
 clrscr（）; 
 r = ++ p  -  q; 
 s = q +++ p; 
 printf（  p =％dq =％ds =％d \ n， p，q，s）; 
 printf（  p * = q =％d \ n，p * = q）; 
 printf（ ％d \ n，（r> s）？！：  0 ）; 
} 

我的尝试：



plz帮我解决这个问题，我是新手，我还没有尝试过任何东西，尽快回复

解决方案

错误和警告信息包含行号，通常还包含指示错误发生位置的列号。看看指示的线并检查它们。



此行可能出现错误：

 s = q +++ p; 

C / C ++没有 +++ 运算符。编译器不知道你的意思是

 s = q + ++ p; 

或

 s = q ++ + p; 

警告只是为了通知您代码中可能存在逻辑错误，因为您已将值赋给 r 但之后从未使用过它。代码将编译并运行，但您可能无法获得预期的结果。





你可以看到它是一个很好的想提一下行号（或者更好地在发布的代码中表明它）。所以我错过了真正的错误

 printf（％d \ n，（r> s）？！：0）; 

参见还有其他解决方案。

警告是由于该错误造成的：由于错误，编译器未评估（r> s）处理 r 因此未使用。



我还必须更正上述关于的陈述+++ 。编译器应将其处理为

 s = q ++ + p; 

因为 ++ 运算符的优先级高于add运算符。

[/ EDIT]

引用：

s = q +++ p;

在这里你必须做出决定，

 s = q + ++ p; 

或

 s = q ++ + p; 

Quote：

printf（％d \ n，（r> s）?! ：0）;

这是另一种语法错误。你的意思是

 printf（％d \ n，（r> s）？1：0）; 

？

看第12行：

 printf（ ％d \ n，（r> s）？！： 0 ）; 

！它不是一个价值，而是一个运营商。我认为你的意思是：

 printf（ ％d \ n，（r> s）？1： 0 ）; 

#include<stdio.h>
#include<conio.h>
#include<math.h>
void main( )
{
int p = 20, q = 15, r,s;
clrscr( );
r = ++p – q;
s = q+++p;
printf("p = %d q = %d s = %d \n", p, q, s);
printf("p*=q=%d\n",p*=q);
printf("%d \n", (r>s)?!:0);
}

What I have tried:

plz help me solve this issue, i am novice here, i havent tried anything yet, reply as soon as possible

解决方案

Error and warning messages contain a line number and often also a column number indicating where the error occurred. Have a look at the indicated lines and inspect them.

The error occured probably in this line:

s = q+++p;

C/C++ does not have a +++ operator. The compiler did not know if you mean

s = q + ++p;

or

s = q++ + p;

The warning is just to inform you that you might have a logical error in your code because you have assigned a value to r but never used it afterwards. The code will compile and run but you may no get the expected result.

[EDIT]
As you can see it is a good idea to mention the line number (or better indicate it in the posted code). So I missed the real error at

printf("%d \n", (r>s)?!:0);

See also the other solutions.
The warning is a result of that error: The compiler did not evaluate (r>s) due to the error and treated r therefore as unused.

I must also correct my above statement about the +++. The compiler should process it as

s = q++ + p;

because the ++ operator has a higher precedence than the add operator.
[/EDIT]

Quote:

s = q+++p;

Here you have to take a decision, either

s = q + ++p;

or

s = q++ + p;

Quote:

printf("%d \n", (r>s)?!:0);

That's another syntax error. Did you mean

printf("%d \n", ( r > s ) ? 1 : 0);

?

Look at line 12:

printf("%d \n", (r>s)?!:0);

"!" is not a value, it's an operator. I Think you meant:

printf("%d \n", (r>s)?1:0);

这篇关于错误：表达式语法，警告：R被分配一个从未使用过的值。，plz help的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！