blocks|key|2461495|text|如果您只知道运行时的大小，则无法将成员声明为数组，因为可变长度数组不是C%2B%2B特性。只需使用std::string即可。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2461496|struct+myStruct+{

++++int+nCode;
++++std::string+str1;
++++myStruct+()+:+str1(STRING_)
++++{
++++++++nLangCode+=+0x0409;
++++}
}+;|code-block|syntax|javascript|2461497|这样你就不必担心复制构造函数、赋值运算符和析构函数--这是其他两个答案都遗漏的。|2461498|entityMap^0|19|B|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

If you only know the size at runtime, there's no way to declare your member as an array, because variable length arrays aren't a C++ feature. Just use a <code>std::string</code>.

<pre><code>struct myStruct {

 int nCode;
 std::string str1;
 myStruct () : str1(STRING_)
 {
 nLangCode = 0x0409;
 }
} ;
</code></pre>

This way you don't have to worry about copy constructors, assignment operators and destructors - which is what both other answers missed.

blocks|key|4939312|text|您应该使用标准类std::string|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4939313|例如|4939314|#include+<string>
struct+myStruct+{

++++int+nLangCode;
++++std::string+str1;
++++myStruct+(+const+char+*s,+int+code+)+:+nLangCode(+code+),+str1(+s+)++
++++{
++++}
}+;|code-block|syntax|javascript|4939315|否则，您需要使用运算符new动态分配字符数组。在这种情况下，您还必须显式定义复制构造函数、析构函数和复制赋值运算符。|4939316|在这种情况下，构造函数可能如下所示|4939317|#include+<cstring>

struct+myStruct+{

++++int+nLangCode;
++++char+*str;
++++myStruct+(+const+char+*s,+int+code+)+:+nLangCode(+code+)++
++++{
++++++++str+=+new+char[std::strlen(+s+)+%2B+1];
++++++++std::strcpy(+str,+s+);
++++}
++++//+Other+special+functions...
}+;|4939318|entityMap^0|8|B|0|0|0|B|3|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|X|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|12|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|13|8|@]|D|@]|E|$K|L]]|$1|S|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|T|$]]

You should use standard class <code>std::string</code>

For example

<pre><code>#include &lt;string&gt;
struct myStruct {

 int nLangCode;
 std::string str1;
 myStruct ( const char *s, int code ) : nLangCode( code ), str1( s ) 
 {
 }
} ;
</code></pre>

otherwise you need yourself dynamically allocate a character array using operator <code>new</code>. In this case you also have to define explicitly a copy constructor, destructor and the copy assignment operator.

In this case the constructor could look the following way

<pre><code>#include &lt;cstring&gt;

struct myStruct {

 int nLangCode;
 char *str;
 myStruct ( const char *s, int code ) : nLangCode( code ) 
 {
 str = new char[std::strlen( s ) + 1];
 std::strcpy( str, s );
 }
 // Other special functions...
} ;
</code></pre>

blocks|key|2457363|text|如果不需要保存字符串的副本，也不需要修改字符串，那么一开始就不要使用数组，而是使用像char+const+*+const+str1;这样的原始指针，然后在构造函数中像str1(+STRING_+)一样初始化它，比如|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2457364|#define+_STRING+"myString"

struct+myStruct+{
++++const+char+*+const+str1;
++++myStruct()+:+str1(+_STRING+)+{+}
};|code-block|syntax|javascript|2457365|如果您不需要字符串的副本，但确实需要修改它，请直接将其存储在数组中，并让编译器计算出正确的大小：|2457366|#define+_STRING+"myString"

struct+myStruct+{
++++static+char+str1[];
++++myStruct()+{}
};

const+myStruct::str1[]+=+_STRING;|2457367|如果您确实需要一个副本，并且确实需要修改该字符串，请使用Luchian+Grigore的答案中所示的普通std::string。|2457368|entityMap^0|16|O|2B|F|0|0|0|0|1G|B|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@$9|11|A|12|B|C]]|D|@]|E|$]]|$1|Q|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|R|$]]

If there's no need to hold a copy of the string and you don't need to modify the string, just don't use an array in the first place but just a raw pointer like <code>char const * const str1;</code> which you then initialize like <code>str1( STRING_ )</code> in the constructor, like

<pre><code>#define _STRING "myString"

struct myStruct {
 const char * const str1;
 myStruct() : str1( _STRING ) { }
};
</code></pre>

If you don't need a copy of the string but you do need to modify it, store it in an array directly and let the compiler figure out the correct size:

<pre><code>#define _STRING "myString"

struct myStruct {
 static char str1[];
 myStruct() {}
};

const myStruct::str1[] = _STRING;
</code></pre>

If you do need a copy and you do need to modify the string, use a plain <code>std::string</code> as shown in the answer by Luchian Grigore.

I have a struct with a char array and a constructor that initializes the array with a defined String.
I want to avoid the <code>#define</code> and instead pass a C++ string to the Constructor. But then again, the size of the char array is not known at compile time. 
What would be a good approach to this?

<pre><code>#define STRING_ "myString"

 struct myStruct {

 int nCode;
 char str1[sizeof(STRING_)];
 myStruct () 
 {
 nLangCode = 0x0409;
 strcpy(str1, STRING_ );
 }
 } ;
</code></pre>

Initialize struct having a char array member with a string of dynamic length

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个带有char数组的结构和一个用定义的字符串初始化数组的构造函数。我希望避免使用#define，而是将一个C++字符串传递给构造函数。但话又说回来，char数组的大小在编译时是未知的。什么是解决这个问题的好方法？#define STRING_ "myString"    struct myStruct {    ...

问使用动态长度字符串初始化具有char数组成员的结构
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问使用动态长度字符串初始化具有char数组成员的结构EN