如何在 C++ 中声明一个字符串数组？

How to declare an array of strings in C++?

提问by naumcho

I am trying to iterate over all the elements of a static array of strings in the best possible way. I want to be able to declare it on one line and easily add/remove elements from it without having to keep track of the number. Sounds really simple, doesn't it?

我试图以最好的方式遍历静态字符串数组的所有元素。我希望能够在一行中声明它并轻松地从中添加/删除元素，而不必跟踪数字。听起来很简单，不是吗？

Possible non-solutions:

可能的非解决方案：

vector<string> v;
v.push_back("abc");
b.push_back("xyz");

for(int i = 0; i < v.size(); i++)
    cout << v[i] << endl;

Problems - no way to create the vector on one line with a list of strings

问题 - 无法使用字符串列表在一行上创建向量

Possible non-solution 2:

可能的非解决方案 2：

string list[] = {"abc", "xyz"};

Problems - no way to get the number of strings automatically (that I know of).

问题 - 无法自动获取字符串的数量（我知道）。

There must be an easy way of doing this.

必须有一个简单的方法来做到这一点。

回答by Anthony Cramp

C++ 11 added initialization lists to allow the following syntax:

C++ 11 添加了初始化列表以允许以下语法：

std::vector<std::string> v = {"Hello", "World"};

Support for this C++ 11 feature was added in at least GCC 4.4and only in Visual Studio 2013.

至少在GCC 4.4 中添加了对此 C++ 11 功能的支持，并且仅在Visual Studio 2013 中添加。

回答by Tyler

You can concisely initialize a vector<string>from a statically-created char*array:

您可以vector<string>从静态创建的char*数组简洁地初始化 a ：

char* strarray[] = {"hey", "sup", "dogg"};
vector<string> strvector(strarray, strarray + 3);

This copies all the strings, by the way, so you use twice the memory. You can use Will Dean's suggestion to replace the magic number 3 here with arraysize(str_array) -- although I remember there being some special case in which that particular version of arraysize might do Something Bad (sorry I can't remember the details immediately). But it very often works correctly.

顺便说一下，这会复制所有字符串，因此您使用了两倍的内存。您可以使用 Will Dean 的建议，用 arraysize(str_array) 替换这里的幻数 3——尽管我记得有一些特殊情况，其中特定版本的 arraysize 可能会做一些坏事（抱歉，我无法立即记住细节） . 但它经常正常工作。

Also, if you're really gung-ho about the one line thingy, you can define a variadic macro so that a single line such as DEFINE_STR_VEC(strvector, "hi", "there", "everyone");works.

此外，如果您真的很喜欢单行代码，您可以定义一个可变参数宏，以便像这样的单行DEFINE_STR_VEC(strvector, "hi", "there", "everyone");工作。

回答by Will Dean

Problems - no way to get the number of strings automatically (that i know of).

问题 - 无法自动获取字符串的数量（我知道）。

There is a bog-standard way of doing this, which lots of people (including MS) define macros like arraysizefor:

有一种沼泽标准的方法可以做到这一点，很多人（包括 MS）定义了如下宏arraysize：

#define arraysize(ar)  (sizeof(ar) / sizeof(ar[0]))

回答by Will Dean

Declare an array of strings in C++ like this : char array_of_strings[][]

在 C++ 中声明一个字符串数组，如下所示： char array_of_strings[][]

For example : char array_of_strings[200][8192];

例如 ： char array_of_strings[200][8192];

will hold 200 strings, each string having the size 8kb or 8192 bytes.

将保存 200 个字符串，每个字符串的大小为 8kb 或 8192 字节。

use strcpy(line[i],tempBuffer);to put data in the array of strings.

用于strcpy(line[i],tempBuffer);将数据放入字符串数组中。

回答by Eclipse

One possiblity is to use a NULL pointer as a flag value:

一种可能性是使用 NULL 指针作为标志值：

const char *list[] = {"dog", "cat", NULL};
for (char **iList = list; *iList != NULL; ++iList)
{
    cout << *iList;
}

回答by Ross Smith

You can use the beginand endfunctions from the Boost range library to easily find the ends of a primitive array, and unlike the macro solution, this will give a compile error instead of broken behaviour if you accidentally apply it to a pointer.

您可以使用Boost 范围库中的begin和end函数轻松找到原始数组的末尾，并且与宏解决方案不同，如果您不小心将其应用于指针，这将产生编译错误而不是破坏行为。

const char* array[] = { "cat", "dog", "horse" };
vector<string> vec(begin(array), end(array));

回答by Shadow2531

Here's an example:

下面是一个例子：

#include <iostream>
#include <string>
#include <vector>
#include <iterator>

int main() {
    const char* const list[] = {"zip", "zam", "bam"};
    const size_t len = sizeof(list) / sizeof(list[0]);

    for (size_t i = 0; i < len; ++i)
        std::cout << list[i] << "\n";

    const std::vector<string> v(list, list + len);
    std::copy(v.begin(), v.end(), std::ostream_iterator<string>(std::cout, "\n"));
}

回答by Matthew Crumley

You can use Will Dean's suggestion [#define arraysize(ar) (sizeof(ar) / sizeof(ar[0]))] to replace the magic number 3 here with arraysize(str_array) -- although I remember there being some special case in which that particular version of arraysize might do Something Bad (sorry I can't remember the details immediately). But it very often works correctly.

您可以使用 Will Dean 的建议 [ #define arraysize(ar) (sizeof(ar) / sizeof(ar[0]))] 用 arraysize(str_array) 替换这里的幻数 3——尽管我记得有一些特殊情况，在这种情况下，特定版本的 arraysize 可能会做一些坏事（对不起，我不记得细节了立即地）。但它经常正常工作。

The case where it doesn't work is when the "array" is really just a pointer, not an actual array. Also, because of the way arrays are passed to functions (converted to a pointer to the first element), it doesn't work across function calls even if the signature looks like an array — some_function(string parameter[])is really some_function(string *parameter).

它不起作用的情况是“数组”实际上只是一个指针，而不是实际的数组。此外，由于数组传递给函数的方式（转换为指向第一个元素的指针），即使签名看起来像一个数组，它也不能跨函数调用工作——some_function(string parameter[])实际上是some_function(string *parameter).

回答by DrPizza

Instead of that macro, might I suggest this one:

而不是那个宏，我可以建议这个：

template<typename T, int N>
inline size_t array_size(T(&)[N])
{
    return N;
}

#define ARRAY_SIZE(X)   (sizeof(array_size(X)) ? (sizeof(X) / sizeof((X)[0])) : -1)

1) We want to use a macro to make it a compile-time constant; the function call's result is not a compile-time constant.

1) 我们想用一个宏使它成为一个编译时常量；函数调用的结果不是编译时常量。

2) However, we don't want to use a macro because the macro could be accidentally used on a pointer. The function can only be used on compile-time arrays.

2) 但是，我们不想使用宏，因为宏可能会被意外地用在指针上。该函数只能用于编译时数组。

So, we use the defined-ness of the function to make the macro "safe"; if the function exists (i.e. it has non-zero size) then we use the macro as above. If the function does not exist we return a bad value.

因此，我们使用函数的定义性来使宏“安全”；如果函数存在（即它的大小非零），那么我们使用上面的宏。如果函数不存在，我们将返回一个错误的值。

回答by DrPizza

#include <boost/foreach.hpp>

const char* list[] = {"abc", "xyz"};
BOOST_FOREACH(const char* str, list)
{
    cout << str << endl;
}