如何获得char *(char数组)的真实长度和总长度?

How to get the real and total length of char * (char array)?(如何获得char *(char数组)的真实长度和总长度?)

问题描述

对于 char []，我可以通过以下方式轻松获取其长度:

For a char [], I can easily get its length by:

char a[] = "aaaaa";
int length = sizeof(a)/sizeof(char); // length=6

但是，我不能这样做来获取 char * 的长度:

However, I cannot do like this to get the length of a char * by:

char *a = new char[10];
int length = sizeof(a)/sizeof(char);

因为，我知道，这里的 a 是一个指针，因此这里的 length 将始终是 4 (或其他不同的系统).

because, I know, a here is a pointer, such that length here will be always be 4 (or something other in different systems).

我的问题是我怎样才能获得 char * 之后的长度?我知道有人可能会质疑我你已经知道它的 10 因为你刚刚创建了它.我想知道这一点，因为获取它的长度的这个步骤可能距离它的创建还有很长的路要走，我不想很长一段时间回来检查这个数字.另外，我也想知道它的真实长度.

My question is that how can I get the length of a char * afterwards? I know someone may challenge me that you already know its 10 because you just created it. I want to know this because this step of getting its length may come long long way from its creation and I don't want to come long long way back to check this number. Moreover, I also want to know its real length.

更具体

• 我怎样才能得到它真正的length=5?
• 我怎样才能得到它的总length=10?

以下示例:

char *a = new char[10]; 
strcpy(a, "hello");

推荐答案

你不能.无论如何，不​​是100％准确.指针没有长度/大小，只有它自己的.它所做的只是指向内存中保存一个字符的特定位置.如果该字符是字符串的一部分，那么您可以使用 strlen 来确定当前指向的字符后面的字符，但这并不意味着您的 array箱子那么大.
基本上:

You can't. Not with 100% accuracy, anyway. The pointer has no length/size but its own. All it does is point to a particular place in memory that holds a char. If that char is part of a string, then you can use strlen to determine what chars follow the one currently being pointed to, but that doesn't mean the array in your case is that big.
Basically:

指针不是数组，所以它不需要 知道数组的大小.指针可以指向单个值，因此指针可以在没有数组的情况下存在.它甚至不关心它指向的内存位于何处(只读，堆或堆栈......无关紧要).指针除了自身之外没有长度.指针只是...
考虑一下:

A pointer is not an array, so it doesn't need to know what the size of the array is. A pointer can point to a single value, so a pointer can exist without there even being an array. It doesn't even care where the memory it points to is situated (Read only, heap or stack... doesn't matter). A pointer doesn't have a length other than itself. A pointer just is...
Consider this:

char beep = 'a';
void alert_user(const char *msg, char *signal); //for some reason
alert_user("Hear my super-awsome noise!", &beep); //passing pointer to single char!

void alert_user(const char *msg, char *signal)
{
    printf("%s%c
", msg, *signal);
}

指针可以是单个字符，也可以是数组的开头、结尾或中间...
将字符视为结构.您有时会在堆上分配一个结构.这也创建了一个没有数组的指针.

A pointer can be a single char, as well as the beginning, end or middle of an array...
Think of chars as structs. You sometimes allocate a single struct on the heap. That, too, creates a pointer without an array.

仅使用指针来确定它指向的数组有多大是不可能的.最接近它的是使用 calloc 并计算您可以通过指针找到的连续 字符的数量.当然，一旦您将内容分配/重新分配给该数组的键，这将不起作用，并且如果数组的 外部 内存恰好保存 也是.所以使用这种方法是不可靠的、危险的，而且通常是愚蠢的.别.做.它.

char a_str[] = "hello";//{h,e,l,l,o,}
char *arr_ptr = &a_str[0];
printf("Get length of string -> %d
", strlen(arr_ptr));

int length = sizeof(a)/sizeof(char);//sizeof char is guaranteed 1, so sizeof(a) is enough

size_t length = sizeof(a)/sizeof(*a);//best use ptr's type -> good habit