在C ++中，float值被从double截断 [英] In C++ float value being truncated from double

问题描述

我以前使用float变量进行编码，从来没有遇到过这个问题.

I've coded using float variables before and never had this problem.

float  a, b, subtotal, stx;
a=15.95;
b=24.95;
subtotal=a+b;
stx=subtotal*.07;

cout << "Item 1: $" << a << endl;
cout << "Item 2: $" << b << endl;
cout << "\nSubtotal: $" <<subtotal<< endl;
cout << "Sales Tax: $" << stx << endl; 
cout << "Total: $"  << subtotal+stx << endl;

相对海峡前向代码

warning C4305: '=' : truncation from 'double' to 'float'

我了解数据被截断的想法(并且我也知道您可以在变量的末尾写入f.但是，如果将变量声明为float，为什么编译器将字面值解释为双精度，如果它被声明为浮点数.

I understand the idea of data being truncated (and I also know that you can write the f at the end of the variable. But if variables are declared as float why is the compiler interpreting the literal values as as doubles if it was declared as floats.

我查询了其他一些票证，但它们与众不同，因此我的查询似乎找不到解决方案，即如果将数据声明为浮点数，为什么将其读取为double值.

I looked up a few other tickets and they were different then my inquiry I can't seem to find a solution as to why the data is being read as a double if its declared as a float.

推荐答案

为什么编译器将文字值解释为双精度值

why is the compiler interpreting the literal values as as doubles

因为这就是文字的解释方式，除非您添加修饰符以指定其他类型.

Because that's how literals are interpreted, unless you add modifiers to specify a different type.

a=15.95f;
       ^ gives the literal "float" type

但是如果变量被声明为float ...

But if variables are declared as float...

表达式的类型从不依赖于表达式的使用方式.因此15.95的类型均为double，无论您做什么.如果需要，可以将类型转换为在较大的表达式中使用，这就是在这种情况下发出警告的原因.

The type of an expression never depends on how the expression is used; so 15.95 has type double whatever you do with it. The type is converted for use in a larger expression, if necessary, and that's what gives the warning in this case.

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