問題描述

我有一個非常奇怪的問題，如果我像這樣聲明一個 int

I have this very strange problem where if I declare an int like so

int time = 0110;

然后顯示到控制臺返回的值為72.但是，當我刪除前面的 0 以便 int time = 110; 控制臺然后像預期的那樣顯示 110 .

and then display it to the console the value returned is 72. However when I remove the 0 at the front so that int time = 110; the console then displays 110 like expected.

我想知道兩件事，首先為什么它在 int 的開頭使用前面的 0 來執行此操作，并且有沒有辦法阻止它，以便 0110 至少等于110?
其次，有沒有什么辦法可以讓0110返回0110?
如果您對變量名稱進行猜測，我會嘗試在 24 小時內進行操作，但此時 1000 之前的任何時間都會因此導致問題.

Two things I'd like to know, first of all why it does this with a preceding 0 at the start of the int and is there a way to stop it so that 0110 at least equals 110?
Secondly is there any way to keep it so that 0110 returns 0110?
If you take a crack guess at the variable name I'm trying to do operations with 24hr time, but at this point any time before 1000 is causing problems because of this.

提前致謝！

推薦答案

從 0 開始的整數字面量定義了八進制整數字面量.現在在 C++ 中有四類整數文字

An integer literal that starts from 0 defines an octal integer literal. Now in C++ there are four categories of integer literals

integer-literal:
    decimal-literal integer-suffixopt
    octal-literal integer-suffixopt
    hexadecimal-literal integer-suffixopt
    binary-literal integer-suffixopt

八進制整數字面量定義如下

And octal-integer literal is defined the following way

octal-literal:
    0 octal-literal
    opt octal-digit

就是從0開始.

因此這個八進制整數文字

Thus this octal integer literal

0110

對應下面的十進制數

8^2 + 8^1 

即等于 72.

您可以通過運行以下簡單程序來確定八進制表示中的 72 等于 110

You can be sure that 72 in octal representation is equivalent to 110 by running the following simple program

#include <iostream>
#include <iomanip>

int main() 
{
    std::cout << std::oct << 72 << std::endl;

    return 0;
}

輸出是

110

這篇關于前面為 0 的 C++ int 更改整個值的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網！