C ++不需要5/9并且似乎类型转换 [英] C++ does not take 5/9 and seems to typecast it

问题描述

为什么

  float bla =（5/9）*（100  -  32）result in 0 * 100  - > ; 0？ 
  

和

  float bla = 5 *（100-32）/ 9; 
  

运行？
他似乎看到5/9作为一个整数，并使其为0 immidiatly，但结果变量是一个浮点数，所以编译器应该知道他可以使用5/9或？

  
 > float bla = 5.0f *（100.0f-32.0f）/ 9.0f; 
  

请注意以下简要列表

  5 // int 
 5l // long 
 5.0 // double 
 5.0f // float 
 5ul // unsigned long 
  

更全面的整数文字和浮点文字 a>

在您的示例中，整个右侧在 int 操作中执行，隐式转换为 float 。到时为时已晚。

请记住，使用正确的文字很重要！

Why does

 float bla = ( 5/9 )* (100 - 32) result in 0 * 100 -> 0 ?

and

float bla= 5*(100- 32) / 9;

work out? He seems to see 5/9 as a integer and make it to 0 immidiatly but the result variable is a float so the compiler should know he can work with 5/9 or?

解决方案

Use the correct literals so you get the correct mathematical behavior

float bla = 5.0f * (100.0f - 32.0f) / 9.0f;

Note the following brief list

5      // int
5l     // long
5.0    // double
5.0f   // float
5ul    // unsigned long

A more comprehensive list of integer literals and floating point literals

In your example, the entire right hand side is performed in int operations, then the final result is implicitly converted to float. By then it is too late.

Remember, using the correct literal is important!

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