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一元减号运算符如何与 C++ 中的 integer 文字一起使用？

[英]How the unary minus operator works with integer literals in C++?

原文 2019-11-04 05:57:32 3 1 c++/ literals

I read that decimal literals are signed by default.我读到十进制文字是默认签名的。
For simplicity, assume that the values that int can hold are the integers [-128,127], and long can hold the integer 128. Now, what happens if I code the literal -128?为简单起见，假设int可以容纳的值是整数 [-128,127]，而long可以容纳 integer 128。现在，如果我对文字 -128 进行编码会发生什么？ What I know is that the literal here is simply '128', which can't fit into an int but rather would go into a long ?我所知道的是这里的文字只是'128'，它不能适合int而是 go 适合long吗？ or does the unary minus operator do something else?还是一元减号运算符做其他事情？
So, how does the unary minus sign work with integer literals?那么，一元减号如何与 integer 文字一起使用？

1 个解决方案

From cppreference.com :从cppreference.com ：

The type of the integer literal is the first type in which the value can fit, from the list of types which depends on which numeric base and which integer-suffix was used. integer 文字的类型是该值可以适合的第一种类型，来自类型列表，这取决于使用的数字基数和整数后缀。

When using a decimal base and no suffix, as in your example, the possible types are int , long int , and long long int .如您的示例所示，使用十进制基数且无后缀时，可能的类型为int 、 long int和long long int 。 If the value (ignoring the minus sign) fits in a long but not in an int , then the type of the value is long .如果值（忽略减号）适合long但不适合int ，则值的类型是long 。

After the type is determined, the unary minus operator is applied as normal.确定类型后，将照常应用一元减号运算符。 Applying unary minus to a long results in a long (even if the result could fit in an int ).将一元减号应用于long会导致long （即使结果可能适合int ）。