C语言 如何以字节为单位获取字符串大小？

How to get the string size in bytes?

提问by artaxerxe

As the title implies, my question is how to get the size of a string in C. Is it good to use sizeofif I've declared it (the string) in a function without mallocin it? Or, if I've declared it as a pointer? What if I initialized it with malloc? I would like to have an exhaustive response.

正如标题所暗示的，我的问题是如何在C. 它的好处是使用sizeof不，如果我已经声明它的功能（串）malloc的呢？或者，如果我已将其声明为指针？如果我用 初始化它malloc怎么办？我想得到详尽的答复。

回答by nmikhailov

You can use strlen. Size is determined by the terminating null-character, so passed string should be valid.

您可以使用strlen。大小由终止空字符决定，因此传递的字符串应该是有效的。

If you want to get size of memory buffer, that contains your string, and you have pointer to it:

如果您想获取包含您的字符串的内存缓冲区的大小，并且您有指向它的指针：

• If it is dynamic array(created with malloc), it is impossible to get it size, since compiler doesn't know what pointer is pointing at. (check this)
• If it is static array, you can use sizeofto get its size.

• 如果它是动态数组（用 malloc 创建），就不可能得到它的大小，因为编译器不知道指针指向什么。（检查这个）
• 如果是静态数组，则可以使用sizeof获取其大小。

If you are confused about difference between dynamic and static arrays, check this.

如果您对动态数组和静态数组之间的区别感到困惑，请查看此。

回答by Bernhard Barker

Use strlento get the length of a null-terminated string.

使用strlen获得的长度空终止字符串。

sizeofreturns the length of the arraynot the string. If it's a pointer (char *s), not an array (char s[]), it won't work, since it will return the size of the pointer (usually 4 bytes on 32-bit systems). I believe an array will be passed or returned as a pointer, so you'd lose the ability to use sizeofto check the size of the array.

sizeof返回数组的长度而不是字符串。如果它是一个指针 ( char *s)，而不是一个数组 ( char s[])，它将不起作用，因为它将返回指针的大小（在 32 位系统上通常为 4 个字节）。我相信数组将作为指针传递或返回，因此您将失去sizeof用于检查数组大小的能力。

So, only if the string spans the entire array(e.g. char s[] = "stuff"), would using sizeoffor a statically definedarray return what you want (and be faster as it wouldn't need to loop through to find the null-terminator) (if the last character is a null-terminator, you will need to subtract 1). If it doesn't span the entire array, it won't return what you want.

因此，只有当整个阵列的字符串跨度（例如char s[] = "stuff"），将使用sizeof一个静态定义你想要什么回报阵列（且更快，因为它不会需要遍历找到空终止符）（如果最后一个字符是空终止符，您需要减去 1)。如果它不跨越整个数组，它不会返回你想要的。

An alternative to all this is actually storing the size of the string.

所有这些的替代方法实际上是存储字符串的大小。

回答by hyde

While sizeofworks for this specific type of string:

虽然sizeof适用于这种特定类型的字符串：

char str[] = "content";
int charcount = sizeof str - 1; // -1 to exclude terminating '
(strlen(string) + 1 ) * sizeof(char)
'

It does not work if stris pointer (sizeofreturns size of pointer, usually 4 or 8) or array with specified length (sizeofwill return the byte count matching specified length, which for char type are same).

如果str是指针（sizeof返回指针的大小，通常为 4 或 8）或具有指定长度的数组（sizeof将返回匹配指定长度的字节数，对于 char 类型相同），则不起作用。

Just use strlen().

只需使用strlen().

回答by Aniket Bhanja

If you use sizeof()then a char *strand char str[]will return different answers. char str[]will return the length of the string(including the string terminator) while char *strwill return the size of the pointer(differs as per compiler).

如果您使用sizeof()then a char *strandchar str[]将返回不同的答案。char str[]将返回字符串的长度（包括字符串终止符），同时char *str将返回指针的大小（因编译器而异）。

回答by A.Clymer

I like to use:

我喜欢使用：

const char* message = "%s, World!";
char* string = (char*)malloc((strlen(message)+1))*sizeof(char));
snprintf(string, (strlen(message)+1))*sizeof(char), message, "Hello");

This will give you the buffer size in bytes. You can use this with snprintf() may help:

这将为您提供以字节为单位的缓冲区大小。您可以将其与 snprintf() 一起使用可能会有所帮助：

##代码##

Cheers! Function: size_t strlen (const char *s)

干杯! 功能：size_t strlen (const char *s)