在C ++中是默认`int long int`的`int`？ [英] Is `int` by default `signed long int` in C++?

问题描述

在C ++中 int 默认为 signed long int ？

它是平台和/或编译器依赖吗？如果是，如何？

以下是否确保重复？ p>

  signed short int 
 signed int 
 signed long int 
 signed long long int 
 unsigned short int 
 unsigned int 
 unsigned long int 
 unsigned long long int 
  

的所有整数都是不同的，也就是说，你可以安全地重载所有的函数，并且不会产生任何冲突。然而，有时使用相同数量的位用于它们的表示。即使它们使用相同数量的位，有符号和无符号类型始终具有不同的范围。除了 char ，使用任何不带签名的整数类型相当于使用 signed ，即 signed int 和 int 是等价的。 是与 signed char 和不同的类型但 char 具有与 signed char 或 unsigned char 。您可以使用 std :: numeric_limits< char> :: is_signed 来确定它使用的是哪个。

更有趣的方面。以下条件都是真的：

• 7< = std :: numeric_limits< signed char&


• sizeof（char）== 1


• code> sizeof（char）== sizeof（signed char）


• ）


• 15< = std :: numeric_limits< short> :: digits


• sizeof（char）< = sizeof（short）


• short）< = sizeof（int）


• 31< = std :: numeric_limits< long& c $ c>


• sizeof（int）< = sizeof（long）


• 63< = std :: numeric_limits< long long> :: digits


• sizeof ; = sizeof（long long）


• sizeof（X）== sizeof（signed X）


• sizeof（signed X）== sizeof（unsigned X）

（其中X是 char ， short ， int ， long 和 long long ）。
$ b

这意味着所有整数类型的大小可以相同，只要此类型至少包含64位（显然 Cray X-MP 是这样的野兽）。在现代机器上通常 sizeof（int）== sizeof（long）但有一些机器 sizeof（int） code>。 long 是32位还是64位取决于实际架构，两种类型目前都在。

Is int by default signed long int in C++?

Is it platform and/or compiler dependent? If so, how?

[EDIT]

Are any of the following guaranteed to be duplicate?

signed short int
signed int
signed long int
signed long long int
unsigned short int
unsigned int
unsigned long int
unsigned long long int

解决方案

All of the integer types are different, i.e. you can safely overload functions for all of them and you won't get any conflict. However, some times use the same number of bits for their representation. Even if they use the same number of bits signed and unsigned types always have a different range. Except for char, using any integer type without signed is equivalent to using it with signed, i.e. signed int and int are equivalent. char is a different type as signed char and unsigned char but char has the same representation and range of either signed char or unsigned char. You can use std::numeric_limits<char>::is_signed to find out which it uses.

On to the more interesting aspects. The following conditions are all true:

• 7 <= std::numeric_limits<signed char>::digits
• sizeof(char) == 1
• sizeof(char) == sizeof(signed char)
• sizeof(char) == sizeof(unsigned char)
• 15 <= std::numeric_limits<short>::digits
• sizeof(char) <= sizeof(short)
• sizeof(short) <= sizeof(int)
• 31 <= std::numeric_limits<long>::digits
• sizeof(int) <= sizeof(long)
• 63 <= std::numeric_limits<long long>::digits
• sizeof(long) <= sizeof(long long)
• sizeof(X) == sizeof(signed X)
• sizeof(signed X) == sizeof(unsigned X)

(where "X" is one of char, short, int, long, and long long).

This means that the size of all integer types can be the same as long as this types hold at least 64 bits (and apparently the Cray X-MP was such a beast). On contemporary machines typically sizeof(int) == sizeof(long) but there are machines where sizeof(int) == sizeof(short). Whether long is 32 or 64 bits depends on the actual architecture and both kinds are currently around.

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