便携式int？ [英] Portable int?

问题描述

我只是好奇，为什么我们只是使用uint8_t或uint16_t例如

来声明一个整数类型的存储，当我们知道int总是

平台依赖？除了M $ Windows和Linux之外，我没有其他

平台的编程经验。


谢谢。

I''m just curious, why don''t we just use uint8_t or uint16_t for example
to declare an integer type storage when we know int is always
platform-dependent? I do not have programming experience on other
platform except M$ Windows and Linux.

Thank you.

推荐答案

Windows和Linux。


谢谢。

Windows and Linux.

Thank you.

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我只是好奇，为什么我们不要只使用uint8_t或uint16_t来声明一个整数类型的存储，当我们知道int总是
platform-依赖？我没有其他平台的编程经验，除了M

I''m just curious, why don''t we just use uint8_t or uint16_t for example
to declare an integer type storage when we know int is always
platform-dependent? I do not have programming experience on other
platform except M

Windows和Linux。

Windows and Linux.

Cuz''愚蠢的。原因是int并不总是一个固定的大小[它确实有b $ b b b最小btw]是它意味着一个理想的平台大小。例如。对于

8086它是16位，而对于68000它是32位。


这个想法是我想要一些简单的东西喜欢


int x;

for（x = 0; x< strlen（buf）; x ++）{

buf [x] = tolower（buf [x]）;

}


[或其他] ......变量x应该易于管理[例如不是

在8086上是32位而在68k上不是16位]。


Along总是意味着稍微不方便但范围更大。

但情况并非总是这样。在68k / 386编译器上长对于int来说通常是

的同义词[除非存在数组表达式，因此我们都

知道它们必须是int]。


Tom

Cuz that''s stupid. The reason "int" isn''t always a fixed size [it does have
minimums btw] is that it''s meant to be an ideal platform size. E.g. for the
8086 it''s 16-bits while for the 68000 it''s 32-bits.

The idea is if I want something simple like

int x;
for (x = 0; x < strlen(buf); x++) {
buf[x] = tolower(buf[x]);
}

[or whatever].... the variable "x" should be easily manageable [e.g. not
32-bits on an 8086 and not 16-bits on a 68k].

A "long" was always meant to be "slightly less convenient but larger range".
That''s not always the case though. on 68k/386 compilers "long" is typically
synonymous for "int" [except when array expressions are present cuz we all
know they must be int].

Tom

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