如何获得char *的真实长度? strlen和sizeof(target)/sizeof(target *)之间有区别吗? [英] How can I get the real length of char*? Different between strlen and sizeof(target)/sizeof(target*)?

问题描述

这是我的测试代码.

string str=string("def");
const char* c_str=str.c_str();
char* charString=(char*)"abc";
char* target;

cout << "str : "<< str<< endl;
cout << "c_str : "<< c_str << endl;
cout << "charString : " << charString << endl << endl;


cout << "string length : " << str.length() << endl;
cout << "c_str length : " << strlen(c_str) << endl;
cout << "c_str array length : " << (sizeof(c_str)/sizeof(*c_str)) << endl;
cout << "charString length : " << strlen(charString) << endl;
cout << "charString array length " << (sizeof(charString)/sizeof(*charString)) << endl<< endl;

target = (char*) malloc(sizeof(char)*2);
target = (char*)"ab";
cout << "target : " << target << endl;
cout << "target strlen : " << strlen(target) << endl;
cout << "target arr length : " << (sizeof(target)/sizeof(char)) << endl;

这是上述代码的结果.

str : def
c_str : def
charString : abc

string length : 3
c_str length : 3
c_str array length : 8
charString length : 3
charString array length 8

target : ab
target strlen : 2
target arr length : 8

在str，c_str，charString的情况下，strlen为3，数组长度为8. 8包括编码和最后的'\ 0'字符.对吧?

In str, c_str, charString case, strlen is 3 and array length is 8. 8 includes encoding and '\0' character at last. Right?

所以，我真的不了解目标案例.

So, I really don't understand target case.

我分配了2个字节的内存. 并分配2个字符. strlen是正确的.

I allocate memory 2 byte. And allocate 2 characters. strlen is right.

• 但是为什么目标数组长度是8，就像c_str和charString的数组长度?为什么目标的数组长度不小于c_str和charString?

• But why target array length is 8 like array length of c_str and charString? Why target's array length isn't less than c_str and charString ?

最后一个问题，我在消息头中发送消息char *，包括消息长度.如果我想从string.c_str()计算消息长度，是否需要像(sizeof(target)/sizeof(* target))那样进行计算?

Last question, I send message char* including message length in message header. If I want to calculate the message length from string.c_str(), do I have to calculate like (sizeof(target)/sizeof(*target)) ?

推荐答案

sizeof(x)/(sizeof(*x) 基本上是一个错误.这种基本构造唯一有意义的方法是x是数组-但是在这种情况下，*x仅通过将数组的名称转换为指向其第一个元素的指针才有意义(是的，确实会发生隐式)，然后除以该项目的大小.

sizeof(x)/(sizeof(*x) is basically a mistake. The only way this basic construct makes sense is if x is an array--but in this case, *x only makes sense by converting the name of the array to a pointer to its first element (which, yes, does happen implicitly), then dividing by the size of that item.

至少在理论上，sizeof(x)/sizeof(x[0])更好一些.实际使用中没有可观察到的差异，但是至少对于观察它的人来说，使用[0]可能会提供一些指示，表明它确实打算(仅)应用于数组，而不是指针.

At least theoretically, sizeof(x)/sizeof(x[0]) is a little better. There's no observable difference in actual use, but at least to somebody looking at it, the use of [0] might provide some indication that this is really intended to be applied (only) to an array, not a pointer.

假设将它们应用于数组，它们仍然具有与strlen根本不同的效果. strlen假定其参数为NUL终止的字节序列.这意味着它将从您传递的地址开始计数字节，直到遇到包含值'\0'的字节为止.它可能比阵列更短，或者(如果阵列不包含NTBS)可能更长.简而言之，strlen试图告诉您字符串的当前长度，而sizeof(x)/sizeof(x[0])试图告诉您特定数组可能具有的最大字符串大小.当应用于初始化数组，例如char foo[] = "Something";.

Assuming they are applied to arrays, they still have an effect that's fundamentally different from strlen. strlen assumes its argument is a NUL-terminated byte sequence. That means it counts bytes starting from the address you pass, until it encounters a byte containing the value '\0'. That might be shorter than the array, or (if the array doesn't contain an NTBS) potentially much longer. In short, strlen tries to tell you the current length of a string, whereas the sizeof(x)/sizeof(x[0]) attempts to tell you the maximum string size a particular array could potentially hold. These will coincide primarily when applied to an initialized array, like: char foo[] = "Something";.

还有一个小的区别:strlen仅包含终止NUL之前 的字符数，其中sizeof(x)/sizeof(x[0])包含数组的所有存储，大于最大长度您可以将NUL终止的字符串存储在此处.

One other minor difference as well: strlen only includes the number of characters before the terminating NUL, where sizeof(x)/sizeof(x[0]) includes all the storage of the array, one larger than the maximum length of NUL-terminated string you can store there.

如果将其中任何一个应用于指针而不是数组，则可以计划通常会得到不好的结果(通常"的狡猾的措辞是为了掩盖极端情况:如果碰巧数组的大小完全相同作为指针，它可以正常工作，就像破碎的时钟每天提供两次正确的时间一样.

If you apply either of these to a pointer instead of an array, you can plan on normally getting bad results (the "normally" weasel wording being to cover the corner case: if you happen to have an array exactly the same size as a pointer, it'll work, much like a broken clock giving the right time twice a day).

在C ++中，通常使用以下方法会更好:

In C++, you're usually better off with something like this:

template <class T, size_t N>
constexpr size_t elements(T (&array)[N]) {
    return N;
}

这要求您通过引用将其传递给数组.如果尝试传递指针而不是数组，则会得到编译器错误，而不是错误的答案.

This requires that you pass it an array by reference. If you attempt to pass a pointer instead of an array, you'll get a compiler error instead of an incorrect answer.

当然，在C ++中，您通常也希望避免使用strlen-您通常可以使用std::string(或std::string_view)并使用其length()或size()成员.

Of course, in C++ you generally want to avoid using strlen as well--you can normally use std::string (or std::string_view) and use its length() or size() member.

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