blocks|key|4808061|text|orderFiles是一个const+char+**，你不能修改它(数组指针本身)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4808062|您可以尝试分配数组成员(例如orderFiles+=+argv等等)。|4808063|entityMap^0|D|D|0|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|M|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|N|8|@]|D|@]|E|$]]]|I|$]]

orderFiles is a <code>const char **</code>, you can't modify it (the array pointer itself).

You could try assigning the array members (i.e. orderFiles[0] = argv[0] and so on).

blocks|key|771160|text|问题是，在没有长度的情况下初始化的数组和使用长度初始化的数组之间存在差异。从全局变量声明中删除10，然后它就可以工作了|type|unstyled|depth|inlineStyleRanges|entityRanges|data|771161|这样做的原因是，argv实际上只是一个字符**，而orderFiles是一个包含10个字符*的数组。|771162|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

The problem is that there is a difference between arrays initialized without a length, and those initialized with one. Remove the 10 from the global variables declaration, and then it should work

The reason for this is that argv is really just a char**, but orderFiles is an array of 10 char*.

blocks|key|617150|text|对于orderFiles是的，有一个隐式的char**，但是它是恒定的，因为您通过指定大小[10]将它初始化为一个链接时分配的内存块。您应该创建一个非常数char**，或者从argv到数组的memcpy。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|617151|entityMap^0|2|A|L|6|19|4|25|6|2F|4|2N|6|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|G|$]]

There is an implicit <code>char**</code> for <code>orderFiles</code> yes, but it's constant because you initialized it to a link time allocated block of memory by specifying the size <code>[10]</code>. You should create a non-constant <code>char**</code> or maybe <code>memcpy</code> from <code>argv</code> to your array.

blocks|key|771245|text|它会给你一个错误，因为char+*x[10]给了你一个十个字符指针的数组，这个数组是不可修改的。换句话说，您不能将其赋值给x，也不能以任何方式更改它。等效的可变版本是char+**orderFiles+-您可以将argv分配给它。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|771246|另外，您可以将单个参数传递到数组中，如下所示：|771247|for+(i+=+0;+i+<=+argc+&&+i+<+sizeof(orderFiles)/(sizeof(*orderFiles);+i%2B%2B)
++++orderFiles[i]+=+argv[i];|code-block|syntax|javascript|771248|但这似乎相当复杂。它要么用第一个N参数填充orderFiles，要么部分填充它，使下一个参数为空。|771249|如果您的目的只是简单地将参数隐藏到全局变量中，以便可以在任何地方引用它们，那么您应该这样做：|771250|#include+<stdio.h>

char+**orderFiles;
int+orderCount;

static+void+someFn+(void)+{
++++int+i;
++++printf+("Count+=+%25d\n",+orderCount);
++++for+(i+=+0;+i+<+orderCount;+i%2B%2B)
++++++++printf+("%253d:+[%25s]\n",+i,+orderFiles[i]);
++++//+or,+without+orderCount:
++++//++++for+(i+=+0;+orderFiles[i]+!=+NULL;+i%2B%2B)
++++//++++++++printf+("%253d:+[%25s]\n",+i,+orderFiles[i]);
++++//++++printf+("Count+was+%25d\n",+i);

}

int+main+(int+argc,+char+**argv)+{
++++orderCount+=+argc;
++++orderFiles+=+argv;
++++someFn();
++++return+0;
}|771251|该代码将参数保存到全局变量中，以便可以在不同的函数中访问它们。|771252|如果您还想使用argc，则应该将这两个参数保存到main，尽管从技术上讲，这并不是必需的，因为对于托管环境，argv[argc]保证为NULL+-您可以用它来检测参数数组的结尾。|771253|entityMap^0|B|B|1P|1|2B|H|2Y|4|0|0|0|G|1|L|A|0|0|0|0|7|4|O|4|1I|A|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]|$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|17|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|18|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|19|8|@$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]]|D|@]|E|$]]|$1|O|3|P|5|6|7|1E|8|@]|D|@]|E|$]]|$1|Q|3|R|5|J|7|1F|8|@]|D|@]|E|$K|L]]|$1|S|3|T|5|6|7|1G|8|@]|D|@]|E|$]]|$1|U|3|V|5|6|7|1H|8|@$9|1I|A|1J|B|C]|$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1O|8|@]|D|@]|E|$]]]|X|$]]

It's giving you an error because <code>char *x[10]</code> gives you an array of ten char pointers which is non-modifiable. In other words, you cannot assign to <code>x</code>, nor change it in any way. The equivalent changeable version would be <code>char **orderFiles</code> - you can assign <code>argv</code> to that just fine.

As an aside, you could transfer individual arguments to your array thus:

<pre><code>for (i = 0; i &lt;= argc &amp;&amp; i &lt; sizeof(orderFiles)/(sizeof(*orderFiles); i++)
 orderFiles[i] = argv[i];
</code></pre>

but that seems rather convoluted. It will either fill up <code>orderFiles</code> with the first <code>N</code> arguments or partially fill it, making the next one NULL.

If your intent is simply to stash away the arguments into a global so that you can reference them anywhere, you should do something like:

<pre><code>#include &lt;stdio.h&gt;

char **orderFiles;
int orderCount;

static void someFn (void) {
 int i;
 printf ("Count = %d\n", orderCount);
 for (i = 0; i &lt; orderCount; i++)
 printf ("%3d: [%s]\n", i, orderFiles[i]);
 // or, without orderCount:
 // for (i = 0; orderFiles[i] != NULL; i++)
 // printf ("%3d: [%s]\n", i, orderFiles[i]);
 // printf ("Count was %d\n", i);

}

int main (int argc, char **argv) {
 orderCount = argc;
 orderFiles = argv;
 someFn();
 return 0;
}
</code></pre>

That code saves the arguments into globals so they can be accessed in a different function.

You should save both arguments to <code>main</code> if you want to use <code>argc</code> as well although, technically, it's not necessary since <code>argv[argc]</code> is guaranteed to be NULL for hosted environments - you could use that to detect the end of the argument array.

blocks|key|1638220|text|就像他们都说的两种不同的数据类型。换句话说，可以这样想:+argv是一个c-string数组，您的orderFiles被声明为一个c-string。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1638221|因此，如何分配orderFiles取决于您想要做什么。我通常会遍历argv以获取传递给应用程序的参数。请注意，argv是应用程序名称。|1638222|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|F|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|G|8|@]|9|@]|A|$]]|$1|D|3|-4|5|6|7|H|8|@]|9|@]|A|$]]]|E|$]]

Like they all said two different data types. In other terms think of it this way: argv is an array of c-strings, and your orderFiles is declared as a single c-string.

So how to assign orderFiles depends on what you're trying to do. I typically iterate through argv to get the arguments passed to the application. Note that argv[0] is the application name.

I have the following bit of code:

As a global variable:

<pre><code>char *orderFiles[10];
</code></pre>

And then my main method:

<pre><code>int main(int argc, char *argv[])
{
 orderFiles = argv;
}
</code></pre>

However it keeps giving me an error. What am I doing wrong?

Incompatible types in assignment char ** to char *

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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我有以下几段代码：作为全局变量：char *orderFiles[10];然后是我的主要方法：int main(int argc, char *argv[]){    orderFiles = argv;}然而，它总是给我一个错误。我做错了什么？

问赋值char ** to char *中的类型不兼容
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