blocks|key|1328696|text|尝试将其作为函数定义void+myFunc(char+*a[])或void+myFunc(char+**a)，然后按如下方式使用：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1328697|char+*arr[20];
myFunc(arr);|code-block|syntax|javascript|1328698|entityMap^0|A|M|X|L|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

try this as a function definition <code>void myFunc(char *a[])</code> or <code>void myFunc(char **a)</code> then use it this way :

<pre><code>char *arr[20];
myFunc(arr);
</code></pre>

blocks|key|3781191|text|声明|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3781192|void+myFunc(char+&*a[]);++|code-block|syntax|javascript|3781193|不是有效的C语法。|3781194|若要传递字符指针数组的地址，请改用|3781195|void+myFunc(char+*(*a)[]);+++|3781196|上面函数中的*(*a)[]将a+声明为指向+char**s**.的指针数组的指针必须注意，a具有不兼容的类型。在[]中需要后缀才能使其完整。|offset|length|style|CODE|BOLD|3781197|entityMap^0|0|0|0|0|0|6|7|E|1|M|4|19|1|1K|2|E|1|G|5|M|6|0^^$0|@$1|2|3|4|5|6|7|V|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|W|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|X|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|Y|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|M|3|N|5|6|7|10|8|@$O|11|P|12|Q|R]|$O|13|P|14|Q|R]|$O|15|P|16|Q|R]|$O|17|P|18|Q|R]|$O|19|P|1A|Q|R]|$O|1B|P|1C|Q|S]|$O|1D|P|1E|Q|S]|$O|1F|P|1G|Q|S]]|9|@]|A|$]]|$1|T|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|U|$]]

Declaration

<pre><code>void myFunc(char &amp;*a[]); 
</code></pre>

is not a valid C syntax. 
To pass the address of character pointer arrays, use this instead 

<pre><code>void myFunc(char *(*a)[]); 
</code></pre>

<code>*(*a)[]</code> in the above function declares <code>a</code> as pointer to array of pointers to <code>char</code>s. Must note that <code>a</code> has an incompatible type. A suffix is needed in <code>[]</code> to make it complete.

blocks|key|1328722|text|好了，你几乎走在正确的道路上。void+myFunc(char+*a[]);|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1328723|示例|1328724|void+fun(char+*a[]){
++++printf("%25s",*a);++++//for+accessing+the+next+element+do+a%2B1
++++}

int+main(void)+{
++++char+*x[3];
++++x[0]="abcd";
++++fun(x);+++//+here+you+are+passing+the+address+first+array+element+
++++return+0;|code-block|syntax|javascript|1328725|1328726|entityMap^0|F|N|0|0|0|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|T|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|U|8|@]|D|@]|E|$]]|$1|N|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|O|$]]

Ok you are almost on the right path. <code>void myFunc(char *a[]);</code>

Example

<pre><code>void fun(char *a[]){
 printf("%s",*a); //for accessing the next element do a+1
 }

int main(void) {
 char *x[3];
 x[0]="abcd";
 fun(x); // here you are passing the address first array element 
 return 0;
</code></pre>

<a href="http://ideone.com/0GJH3A" rel="nofollow">DEMO</a>

blocks|key|1465892|text|已更新已修改的问题陈述的答案|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1465893|考虑到您在注释中所说的内容，不需要传递指向数组的指针。您可以简单地传递一个指向数组第一个元素的指针。这样的指针就足够了，因为数组的其余元素显然位于第一个元素之后。|1465894|要编写将指针数组中的指针设置为指向char的函数，请执行以下操作：|offset|length|style|CODE|1465895|void+MyFunction(int+NumberToSet,+char+*Pointers[])
{
++++for+(int+i+=+0;+i+<+NumberToSet;+%2B%2Bi)
++++{
++++++++Pointers[i]+=+SomeString;
++++}
}|code-block|syntax|javascript|1465896|在上面的代码中，SomeString必须有“指向char的指针”类型。例如，这可以是一个字符串，如"Hello"，或char数组(自动转换为指向char的指针)，或某个已声明为char+*x+(并已初始化或赋值)的标识符x。|1465897|要使用此函数，请像这样调用它：|1465898|char+*MyArrayOfPointers[SomeNumber];
MyFunction(NumberOfPointersIWantToSet,+MyArrayOfPointers);|1465899|1465900|原始答案|1465901|在大多数情况下，要将指向char的指针数组传递给函数，只需传递第一个元素的地址即可。在这种情况下，您可以使用以下两种方法之一(它们等效)：|1465902|void+myFunc(char+**a)
void+myFunc(char+*a[])|1465903|如果你真的想要传递数组的地址，你可以使用：|1465904|void+myFunc(char+*(*a)[])|1465905|在这种情况下，a的类型是不完整的，因为缺少维度。根据您打算如何处理a，您可能需要在声明中提供维度。|1465906|当调用myFunc并传递一些声明为char+*array[N];的数组时，在前一种情况下，作为myFunc(array)传递，在后一种情况下，作为myFunc(&array)传递。|1465907|entityMap^0|0|0|H|4|0|0|8|A|O|4|1D|7|1M|4|20|4|2G|7|32|1|0|0|0|0|0|C|4|0|0|0|0|7|1|X|1|0|3|6|H|F|1B|D|21|E|0^^$0|@$1|2|3|4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|1C|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|1D|8|@$F|1E|G|1F|H|I]]|9|@]|A|$]]|$1|J|3|K|5|L|7|1G|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1H|8|@$F|1I|G|1J|H|I]|$F|1K|G|1L|H|I]|$F|1M|G|1N|H|I]|$F|1O|G|1P|H|I]|$F|1Q|G|1R|H|I]|$F|1S|G|1T|H|I]|$F|1U|G|1V|H|I]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1W|8|@]|9|@]|A|$]]|$1|S|3|T|5|L|7|1X|8|@]|9|@]|A|$M|N]]|$1|U|3|-4|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|V|3|W|5|6|7|1Z|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|20|8|@$F|21|G|22|H|I]]|9|@]|A|$]]|$1|Z|3|10|5|L|7|23|8|@]|9|@]|A|$M|N]]|$1|11|3|12|5|6|7|24|8|@]|9|@]|A|$]]|$1|13|3|14|5|L|7|25|8|@]|9|@]|A|$M|N]]|$1|15|3|16|5|6|7|26|8|@$F|27|G|28|H|I]|$F|29|G|2A|H|I]]|9|@]|A|$]]|$1|17|3|18|5|6|7|2B|8|@$F|2C|G|2D|H|I]|$F|2E|G|2F|H|I]|$F|2G|G|2H|H|I]|$F|2I|G|2J|H|I]]|9|@]|A|$]]|$1|19|3|-4|5|6|7|2K|8|@]|9|@]|A|$]]]|1A|$]]

<h3>Updated Answer For Modified Problem Statement</h3>

Given what you have said in comments, there is no need to pass a pointer to an array. You can simply pass a pointer to the first element of the array. Such a pointer suffices because the remaining elements of the array are obviously located after the first element.

To write a function that sets pointers in an array of pointers to <code>char</code>, do this:

<pre><code>void MyFunction(int NumberToSet, char *Pointers[])
{
 for (int i = 0; i &lt; NumberToSet; ++i)
 {
 Pointers[i] = SomeString;
 }
}
</code></pre>

In the above, <code>SomeString</code> must have type “pointer to <code>char</code>”. This could be a string, such as <code>"Hello"</code>, or an array of <code>char</code> (which is automatically converted to a pointer to <code>char</code>), or some identifier <code>x</code> that has been declared as <code>char *x</code> (and has been initialized or assigned), for example.

To use this function, call it like this:

<pre><code>char *MyArrayOfPointers[SomeNumber];
MyFunction(NumberOfPointersIWantToSet, MyArrayOfPointers);
</code></pre>

<hr>

<h3>Original Answer</h3>

In most cases, to pass an array of pointers to <code>char</code> to a function, it suffices to pass the address of the first element. In this case, you would use either of these (they are equivalent):

<pre><code>void myFunc(char **a)
void myFunc(char *a[])
</code></pre>

If you truly want to pass the address of the array, you would use:

<pre><code>void myFunc(char *(*a)[])
</code></pre>

In this case, the type of <code>a</code> is incomplete, since the dimension is missing. Depending on what you intend to do with <code>a</code>, you may need to provide the dimension in the declaration.

When calling <code>myFunc</code> and passing it some array declared as <code>char *array[N];</code>, you would pass it, in the former case, as <code>myFunc(array)</code> and, in the latter case, as <code>myFunc(&amp;array)</code>.

blocks|key|3781278|text|假设您有一个数组声明为|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3781279|char+*a[N];|code-block|syntax|javascript|3781280|并希望传递一个指向数组的指针，如下所示|3781281|foo(+&a+);|3781282|那么foo的原型需要是|offset|length|style|CODE|3781283|void+foo(+char+*(*aptr)[N]+);|3781284|请注意，在这种情况下，必须声明数组的大小；指向N元素数组的指针与指向M元素数组的指针的类型不同。|3781285|通常，您不希望这样做；相反，您通常会像这样传递数组表达式：|3781286|foo+(+a+);|3781287|相应的原型是：|3781288|void+foo+(+char+**aptr+);|3781289|除非是sizeof或一元&运算符的操作数，或者是用于在声明中初始化另一个数组的字符串文字，否则“T的N元素数组”类型的表达式将被转换(“衰减”)为“指向T的指针”类型的表达式，并且表达式的值将是数组的第一个元素的地址。|3781290|entityMap^0|0|0|0|0|2|3|0|0|0|0|0|0|0|3|6|C|1|1C|1|24|1|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|18|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1A|8|@$M|1B|N|1C|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|D|7|1D|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1E|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1F|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|1H|8|@]|9|@]|A|$]]|$1|10|3|11|5|D|7|1I|8|@]|9|@]|A|$E|F]]|$1|12|3|13|5|6|7|1J|8|@$M|1K|N|1L|O|P]|$M|1M|N|1N|O|P]|$M|1O|N|1P|O|P]|$M|1Q|N|1R|O|P]]|9|@]|A|$]]|$1|14|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]]|15|$]]

Assuming you have an array declared as 

<pre><code>char *a[N];
</code></pre>

and you want to pass a pointer to the array, as in

<pre><code>foo( &amp;a );
</code></pre>

then the prototype for <code>foo</code> needs to be

<pre><code>void foo( char *(*aptr)[N] );
</code></pre>

Note that in this case, the size of the array must be declared; a pointer to an N-element array is a different type from a pointer to an M-element array. 

Normally, you don't want to do this; instead, you would normally just pass the array expression like so:

<pre><code>foo ( a );
</code></pre>

and the corresponding prototype would be:

<pre><code>void foo ( char **aptr );
</code></pre>

Except when it is the operand of the<code>sizeof</code> or unary <code>&amp;</code> operators, or is a string literal being used to initialize another array in a declaration, an expression of type "N-element array of <code>T</code>" will be converted ("decay") to an expression of type "pointer to<code>T</code>", and the value of the expression will be the address of the first element of the array.

blocks|key|5249532|text|首先，C通常是一种“通过引用传递”的语言。一些数据项，如整数、浮点数和单个字符，可以通过值传递。但是，这些相同数据类型的数组始终是通过引用传递的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5249533|因此，当您说“我需要将字符指针数组的地址传递给函数”时，只需在函数原型中声明一个字符指针数组：|5249534|void+myFunc(char+*a[]);|code-block|syntax|javascript|5249535|因此，char+*声明了一个字符指针，而a[]定义了一个字符指针数组。要检查此声明，请参考：http://www.cdecl.org/，它将此表达式解析为“声明一个字符指针数组”。|offset|length|style|CODE|5249536|技术要点是*与char绑定而不是与a[]绑定。|5249537|因此，C将传递一个指向您已声明的数据结构的指针。关于这个话题的讨论可以深入到双指针，但对于这个问题，这样的讨论可能是离题的。|5249538|entityMap|0|LINK|mutability|MUTABLE|url|http://www.cdecl.org/^0|0|0|0|3|6|K|3|1A|L|0|0|5|1|7|4|H|3|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|11|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|12|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|13|8|@$K|14|L|15|M|N]|$K|16|L|17|M|N]]|9|@$K|18|L|19|1|1A]]|A|$]]|$1|O|3|P|5|6|7|1B|8|@$K|1C|L|1D|M|N]|$K|1E|L|1F|M|N]|$K|1G|L|1H|M|N]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|1I|8|@]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1J|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

First of all, C is in general a "pass by reference" language. Some data items such as integers, floats, and single characters can be passed by value. But, arrays of those same data types are ALWAYS passed by reference.

Thus, when you say "I need to pass into a function the address of a character pointer array" then simply declare an array of character pointers in your function prototype as:

<pre><code>void myFunc(char *a[]);
</code></pre>

Thus, <code>char *</code> declares a char pointer and <code>a[]</code> defines an array of them. To check this declaration refer to: <a href="http://www.cdecl.org/" rel="nofollow">http://www.cdecl.org/</a> which parses this expression as a "declare a as array of pointer to char". 

The technical point is that the <code>*</code> binds with <code>char</code> rather than with <code>a[]</code>.

So, C will pass a pointer to the data structure that you have declared. A discussion on this topic could delve into double pointers but for this question such a discussion is probably off topic.

I need to pass into a function the address of a character pointer array, what would be the correct function declaration/prototype syntax for that?

I tried:

<pre><code> void myFunc(char &amp;*a[]);
</code></pre>

But get an expected ; , or ) before &amp; error. 

I also tried:

<pre><code> void myFunc(char **a);
</code></pre>

Since the pointer I would be passing in is indeed a pointer to a pointer, but I get yet another error. This time the error has to do with: expected <code>char **</code>, but got <code>char *[]</code>

Or something like that. I have since attempted other solutions so I not remember exactly the error. 

Any suggestions?

Correct C syntax for "address of" character pointer array?

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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我需要将一个字符指针数组的地址传递给一个函数，正确的函数声明/原型语法是什么？我试过了：    void myFunc(char &*a[]);但是在& error之前得到一个预期的；，或者)。我也试过了：    void myFunc(char **a);因为我要传递的指针确实是一个指向指针的指针，但是我得到了另一个...

问"address of“字符指针数组的C语法正确吗？
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问"address of“字符指针数组的C语法正确吗？EN