C++ char* 和 char[] 之间的区别

Difference between char* and char[]

提问by Nemo

I know this is a very basic question. I am confused as to why and how are the following different.

我知道这是一个非常基本的问题。我很困惑为什么以及以下内容有何不同。

char str[] = "Test";
char *str = "Test";

采纳答案by K-ballo

char str[] = "Test";

Is an array of chars, initialized with the contents from "Test", while

是一个数组chars，用“Test”中的内容初始化，而

char *str = "Test";

is a pointer to the literal (const) string "Test".

是指向文字（常量）字符串“Test”的指针。

The main difference between them is that the first is an array and the other one is a pointer. The array owns its contents, which happen to be a copy of "Test", while the pointer simply refers to the contents of the string (which in this case is immutable).

它们之间的主要区别在于第一个是数组，另一个是指针。数组拥有它的内容，它恰好是 的副本"Test"，而指针只是指向字符串的内容（在这种情况下是不可变的）。

回答by Jon Reid

A pointer can be re-pointed to something else:

一个指针可以重新指向别的东西：

char foo[] = "foo";
char bar[] = "bar";

char *str = foo;  // str points to 'f'
str = bar;        // Now str points to 'b'
++str;            // Now str points to 'a'

The last example of incrementing the pointer shows that you can easily iterate over the contents of a string, one element at a time.

增加指针的最后一个示例表明您可以轻松地迭代字符串的内容，一次一个元素。

回答by Luka Rahne

One is pointer and one is array. They are different type of data.

一种是指针，一种是数组。它们是不同类型的数据。

int main ()
{
   char str1[] = "Test";
   char *str2 = "Test";
   cout << "sizeof array " << sizeof(str1) << endl;
   cout << "sizeof pointer " << sizeof(str2) << endl;
}

output

输出

sizeof array 5
sizeof pointer 4

回答by rnrneverdies

The diference is the STACK memory used.

不同之处在于使用的堆栈内存。

For example when programming for microcontrollers where very little memory for the stack is allocated, makes a big difference.

例如，当为微控制器编程时，为堆栈分配的内存很少，这会产生很大的不同。

char a[] = "string"; // the compiler puts {'s','t','r','i','n','g', 0} onto STACK 

char *a = "string"; // the compiler puts just the pointer onto STACK 
                    // and {'s','t','r','i','n','g',0} in static memory area.

回答by Peter Olson

The first

首先

char str[] = "Test";

is an array of five characters, initialized with the value "Test"plus the null terminator '\0'.

是一个包含五个字符的数组，用值"Test"加上空终止符初始化'\0'。

The second

第二

char *str = "Test";

is a pointer to the memory location of the literal string "Test".

是指向文字字符串内存位置的指针"Test"。

回答by Matthieu Brucher

Starting from C++11, the second expression is now invalid and must be written:

从 C++11 开始，第二个表达式现在无效，必须写成：

const char *str = "Test";

The relevant section of the standard is Appendix C section 1.1:

该标准的相关章节是附录C第1.1节：

Change: String literals made const

The type of a string literal is changed from “array of char” to “array of const char.” The type of a char16_t string literal is changed from “array of some-integer-type” to “array of const char16_t.” The type of a char32_t string literal is changed from “array of some-integer-type” to “array of const char32_t.” The type of a wide string literal is changed from “array of wchar_t” to “array of const wchar_t.”

Rationale: This avoids calling an inappropriate overloaded function, which might expect to be able to modify its argument.

Effect on original feature: Change to semantics of well-defined feature.

更改：字符串文字常量

字符串文字的类型从“字符数组”更改为“常量字符数组”。char16_t 字符串文字的类型从“某种整数类型的数组”更改为“const char16_t 的数组”。char32_t 字符串文字的类型从“某种整数类型的数组”更改为“const char32_t 的数组”。宽字符串文字的类型从“wchar_t 数组”更改为“const wchar_t 数组”。

基本原理：这避免了调用不合适的重载函数，该函数可能期望能够修改其参数。

对原始特征的影响：更改为明确定义的特征的语义。

回答by Tim

"Test"is an array of five characters (4 letters, plus the null terminator.

"Test"是一个包含五个字符的数组（4 个字母，加上空终止符。

char str1[] = "Test";creates that array of 5 characters, and names it str1. You can modify the contents of that array as much as you like, e.g. str1[0] = 'B';

char str1[] = "Test";创建 5 个字符的数组，并将其命名为str1. 您可以根据需要修改该数组的内容，例如str1[0] = 'B';

char *str2 = "Test";creates that array of 5 characters, doesn't name it, and also creates a pointer named str2. It sets str2to point at that array of 5 characters. You can follow the pointer to modify the array as much as you like, e.g. str2[0] = 'B';or *str2 = 'B';. You can even reassign that pointer to point someplace else, e.g. str2 = "other";.

char *str2 = "Test";创建 5 个字符的数组，不命名它，并创建一个名为str2. 它设置str2为指向该 5 个字符的数组。您可以按照指针随意修改数组，例如str2[0] = 'B';或*str2 = 'B';。您甚至可以重新分配该指针以指向其他地方，例如str2 = "other";.

An array isthe text in quotes. The pointer merely points at it. You can do a lot of similar things with each, but they are different:

数组是引号中的文本。指针只是指向它。你可以用它们做很多类似的事情，但它们是不同的：

char str_arr[] = "Test";
char *strp = "Test";

// modify
str_arr[0] = 'B'; // ok, str_arr is now "Best"
strp[0] = 'W';    // ok, strp now points at "West"
*strp = 'L';      // ok, strp now points at "Lest"

// point to another string
char another[] = "another string";
str_arr = another;  // compilation error.  you cannot reassign an array
strp = another;     // ok, strp now points at "another string"

// size
std::cout << sizeof(str_arr) << '\n';  // prints 5, because str_arr is five bytes
std::cout << sizeof(strp) << '\n';     // prints 4, because strp is a pointer

for that last part, note that sizeof(strp) is going to vary based on architecture. On a 32-bit machine, it will be 4 bytes, on a 64-bit machine it will be 8 bytes.

对于最后一部分，请注意 sizeof(strp) 将因架构而异。在 32 位机器上，它将是 4 个字节，在 64 位机器上它将是 8 个字节。