为什么是一个字符串的字节大小比长度更久？ [英] Why is a string's byte size longer than the length?

问题描述

为什么说的sizeof（条例）是 5 ，但的sizeof（字符）是 1 ？

不应该使的sizeof（条例）是 4 因为字符串的长度为4字符（ 4×1 ）？

Shouldn't that make sizeof("Bill") be 4 since the length of the string is 4 chars (4 x 1)?

我认为它可能有一些做与条例是一个字符数组，但为什么不增加字节大小？

I believe it may have something to do with "Bill" being an array of characters, but why does that increase the byte size?

推荐答案

C字符串是空终止。有在该字符串的末尾零的字节。假设ASCII，条例看起来像这样在内存中：

C strings are null terminated. There is a zero byte at the end of that string. Assuming ASCII, "Bill" looks like this in memory:

'B'  'i'  'l'  'l'  '\0'
0x42 0x69 0x6c 0x6c 0x00

从C标准，第 6.4.5字符串字面，第7段：

在翻译阶段7，零值字节或code追加到从字符串字面的结果或文字每个多字节字符序列。

In translation phase 7, a byte or code of value zero is appended to each multibyte character sequence that results from a string literal or literals.

如果你想获得 4 的答案长度，你应该使用的strlen（条例），而不是的sizeof 。

If you want to get an answer of 4 for the length, you should use strlen("Bill"), rather than sizeof.

如果你真的不想空终止，这也是有可能的，虽然可能是不明智的。这样的定义：

If you really don't want the null-terminator, that's possible too, though probably ill-advised. This definition:

char bill[4] = "Bill";

将产生一个4字节数组含法案只是字符'B'，我，L和L，没有空终止子。

will yield a 4-byte array bill containing just the characters 'B', 'i', 'l', and 'l', with no null-terminator.

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