C++：使用 sizeof 的字符数组的大小
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C++: size of a char array using sizeof
提问by Saad
Look at the following piece of code in C++:
看下面一段 C++ 代码：
char a1[] = {'a','b','c'};
char a2[] = "abc";
cout << sizeof(a1) << endl << sizeof(a2) << endl;
Though sizeof(char)is 1 byte, why does the output show sizeof(a2)as 4 and not 3 (as in case of a1)?
虽然sizeof(char)是 1 个字节，但为什么输出显示sizeof(a2)为 4 而不是 3（如a1）？
回答by Pubby
C-strings contain a null terminator, thus adding a character.
C 字符串包含一个空终止符，因此添加了一个字符。
Essentially this:
本质上是这样的：
char a2[] = {'a','b','c','##代码##'};
回答by Component 10
That's because there's an extra null '\0'character added to the end of the C-string, whereas the first variable, a1is an array of three seperate characters.
这是因为'\0'在 C 字符串的末尾添加了一个额外的空字符，而第一个变量a1是一个由三个单独字符组成的数组。
sizeofwill tell you the byte size of a variable, but prefer strlenif you want the length of a C-string at runtime.
sizeof会告诉您变量的字节大小，但strlen如果您想要运行时 C 字符串的长度，则更喜欢。
回答by miks
For a2, this is a string so it also contains the '\n'
对于 a2，这是一个字符串，因此它还包含 '\n'
Correction, after Ethan & Adam comment, this is not '\n' of course but null terminator which is '\0'
更正，在 Ethan & Adam 评论之后，这当然不是 '\n' 而是空终止符，即 '\0'
