想要关注以下c程序的建议。 [英] Want a suggestion for following c program.

问题描述

大家好，这是我的第二个问题而且我不理睬我，但我没有得到它所以请帮助我。

在下面的程序中我知道输出但是我很困惑已完成。

计划1

Hello guys this is my second question and it is quit foolish of me, but I am not getting it so please help me.
In the following program I know the output but I am confused how it is done.
Program 1

#include<stdio.h>
#include<conio.h>
main()
{
      int x=2;
      if(x=3)
      {
      printf("%d",x++);
      }
      else
      {
      printf("%d",--x);
      }
getch();
}

以下程序的输出 3 但它是怎么来的，我不知道得不到。



计划2

The output of the following program is 3 but how it comes, I don''t get it.

Program 2

#include<stdio.h>
#include<conio.h>
int main()
{
      int x=01234;
      printf("The result is  %d",x);
      return 0;
      getch();
}

以下程序的输出 668 并且没有得到它也好。

所以请给我一些提示。



谢谢大家......：）

The Output of the following program is 668 and don''t get it too.
So please give me some hint.

Thanks guys...:)

推荐答案

在程序1中，x = 3赋值隐藏在if语句中。程序员常常使用赋值（=）运算符代替等号（==）运算符。



在程序2中，当你指定一个变量x的值你使用八进制形式（当数字文字以0开头时它是一个八进制数字）

In the program 1 the x=3 assignment is hidden inside the if statement. It is a common mistake by the programmers to use assignment (=) operator in place of equal (==) operator.

In the program 2, when you assign a value to variable x you use an octal form (when a numeric literal begins with 0 it is an octal number)

这实际上是一个糟糕的问题。出于什么目的编写代码？您是否在Code项目中测试成员？听program1，如果条件没有比较。即，看你的代码

This is realy a bad question. For what purpose the code is write? Are you test members in Code project? Listen program1, there is no comparison in if condition. ie, watch your code

if( x = 3 )

。 Actualy这里将值3分配给x。没有比较。

不要在代码项目中发布这类问题。您可以调试代码并确保发生的事情。

. Actualy here value 3 is assigned to x. no comparison is done.
Don''t post this type of question in code project. You can debug the code and ensure what happend.

在程序1中，x = 3赋值隐藏在if语句中。程序员常常使用赋值（=）运算符代替等号（==）运算符。

In the program 1 the x=3 assignment is hidden inside the if statement. It is a common mistake by the programmers to use assignment (=) operator in place of equal (==) operator.

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