#定义输出不正确 [英] #define output is not correct

问题描述

我输入了以下代码，但无法理解输出内容

i have typed the following code but not able understand the output

#include<iostream.h>
#include<conio.h>
#define AREA(a) a*a;
int area(int a)
   {
   return a*a;
   }
void main()
   {
   cout<<AREA(5+2);
   cout<<"\t"<<area(5+2);
   }

</conio.h></iostream.h>

输出为17和49
我无法理解为什么宏将ans设为17而不是49
请通过您的宝贵意见解决这些问题

[edit]已添加代码块，格式为-OriginalGriff [/edit]
@griff
先生

为什么将这样的代码替换为5 + 2 * 5 + 2
能否请您解释一下

and the output is 17 and 49
im not able to understand why macro giving ans as 17 instead of 49
please solve this ques with your valueable comments

[edit]Code blocks added, formatting - OriginalGriff[/edit]
@griff
sir

why this code is replaced like this 5+2*5+2
can you please explain

推荐答案

这很简单-运算符优先级和预处理程序:
使用area(5+2)时，编译器进行数学运算，并将其更改为对area(7)的调用-返回7*7或49.
当您使用AREA(5+2)时，预处理程序会进行文本替换，并用5+2*5+2
替换它. 然后，运算符在"+"上的"*"优先级将其编译为5 + (2*5) + 2或17.

要解决此问题，请用
替换您的定义

It''s pretty simple - operator precidence and preprocessors:
When you use area(5+2) the compiler does the math, and changes it to a call to area(7) - which returns 7*7 or 49.
When you use AREA(5+2) the preprocessor does a text subsitution, and replaces it with 5+2*5+2
Operator precidence of the "*" over the "+" then gets it compiled as 5 + (2*5) + 2 or 17.

To fix it, replace your definition by

#define AREA(a) ((a)*(a))

杰夫·威廉姆斯(Geoff Williams)正确地指出，为了真正安全，您需要在整个宏定义周围加上方括号，并加以感谢.我发现您绝对不需要分号，最后用腮红将其删除...-OriginalGriff [/edit]

[edit]Geoff Williams correctly pointed out that you need brackets round the whole macro definition to be truely safe, added with thanks. I spotted that you definitely don''t need the semicolon on the end, removed with blushes... - OriginalGriff[/edit]

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