谁能解释为什么这段代码输出“真实”？ [英] can anyone please explain why this code is outputting "true"?

问题描述

  #include   <   stdio.h  >  
   int  main（）{
  float  a =  0 。  7 ; 
  if （a< 0。 7 ）{
 printf（  true）; 
} 
  else  {
 printf（ 假）; 
} 
  return   0 ; 
} 

解决方案

你在代码中写的任何东西都必须以二进制形式存储在内存中（0,1形式）

当你写浮点数a = 0.7然后必须得到0.7的二进制值。

二进制0.7是

 b0。 1011001100110011001100110011001100110011001100110011001100110  ... 

但是，0.7是双精度字面值，其值为0.7舍入到最接近的可表示的双精度值，即：

 b0。 10110011001100110011001100110011001100110011001100110  

十进制，这正是：

  0 。  6999999999999999555910790149937383830547332763671875  

当写一个浮点数a = 0.7时，该双精度值再次舍入为单精度，并且a获取二进制值：

 b0。  101100110011001100110011  

这正是

 < span class =code-digit> 0 。 699999988079071044921875  

十进制。

进行比较时（a< 0.7），您将这个单精度值（转换为double，不会舍入，因为所有单精度值都可以双精度表示）与原始双精度值进行比较。

< pre lang =c＃> 0 。 699999988079071044921875 < 0 。 6999999999999999555910790149937383830547332763671875

比较正确返回true。

请访问此链接进行浮点运算airth>

浮点Airthmatic [ ^ ]

你好，

在C编程语言中，浮点常数默认是双重类型，这就是0.7 i的原因s double类型，除非使用'f'或'F'后缀来表示浮点类型。因此浮点变量a小于双常数0.7。

 #include< stdio。 h取代; 
 int main（）
 {
浮动a = 0.7; 
 printf（％。10f％.10f \ n，0.7，a）; 
返回0; 
} 
< /stdio.h> 

O / P将是

 0.7000000000 0.6999999881 

因此满足if条件并打印true

#include<stdio.h>
int main(){
float a=0.7;
if(a<0.7){
printf("true");
}
else{
printf("false");
}
return 0;
}

解决方案

Any thing you write in code must be stored in memory in binary form (0,1 form)
When you write float a=0.7 then a must get the binary value of 0.7.
In binary 0.7 is

b0.1011001100110011001100110011001100110011001100110011001100110...

However, 0.7 is a double-precision literal, whose value is 0.7 rounded to the closest representable double-precision value, which is:

b0.10110011001100110011001100110011001100110011001100110

In decimal, that's exactly:

0.6999999999999999555910790149937383830547332763671875

When you write float a = 0.7, that double value is rounded again to single-precision, and a gets the binary value:

b0.101100110011001100110011

which is exactly

0.699999988079071044921875

in decimal.
When you do the comparison (a < 0.7), you are comparing this single-precision value (converted to double, which does not round, because all single-precision values are representable in double precision) to the original double-precision value.

0.699999988079071044921875 < 0.6999999999999999555910790149937383830547332763671875

the comparison correctly returns true.
Please visit this link for Floating point airthmatic operation
Floating Point Airthmatic[^]

Hello ,
In C programming language, the floating point constant is double type by default, that's why 0.7 is double type, unless use 'f' or 'F' suffix to indicate float type.Hence the float variable a is less than double constant 0.7.

#include<stdio.h>
int main()
{
    float a=0.7;
    printf("%.10f %.10f\n",0.7, a);
    return 0;
}
</stdio.h>

O/P will be

0.7000000000 0.6999999881

Hence the if condition is satisfied and it prints "true"

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