为什么这句话int a = 012945;在C ++中给我一个错误，如“无效的八进制数字” ？ [英] Why this statement int a= 012945; in C++ gives me an error like "invalid octal digit" ?
问题描述
我正在使用2017年的视觉工作室，所以是因为它给我一个错误无效的八进制数字？
我有什么尝试过：
如果我给出一个以0开头的整数变量值，那么只有这样的错误才会发生这样的事情：e = 0129612;
但是如果我在第一个地方不使用0就不会给出任何错误。
I'm using visual studio 2017,so is it because of that it gives me an error "invalid octal digit"?
What I have tried:
if i give a integer variable value that starts with 0 then only this error comes something like this int e = 0129612;
but it doesn't gives any error if i just don't use 0 on the first place.
推荐答案
八进制数字的范围可以是0到7这是因为八进制是基数8.以0开头的文字整数值被认为是八进制。如果以0x开头，则为十六进制值。
Octal digits can range from 0 to 7. This is because octal is base 8. A literal integer value starting with 0 is considered to be octal. If it starts with 0x it is a hexadecimal value.
C / C ++知道三种类型的整数文字：八进制，十进制和十六进制。
自C + +14有第四种类型：二进制。
一个以零开头，后跟另一个数字的文字被视为八进制值（基数为8），这是不允许的包含数字8和9.
另请参见 integer literal - cppreference.com [ ^ ]。
C/C++ knows three types of integer literals: octal, decimal, and hex.
Since C++14 there is a fourth type: binary.
A literal beginning with a zero and followed by another digit is treated as octal value (base 8) which is not allowed to contain the digits 8 and 9.
See also integer literal - cppreference.com[^].
如果我给出一个以0开头的整数变量值，那么只有这个错误来自这样的int e = 0129612;
但如果我不在第一个地方使用0，它不会给出任何错误。
if i give a integer variable value that starts with 0 then only this error comes something like this int e = 0129612;
but it doesn't gives any error if i just don't use 0 on the first place.
因为它是原则：从0开始的整数是八进制（基数为8）
Because it is the principle: integers starting with 0 are octal (base 8)
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