blocks|key|540934|text|double+readScores(double+scoreArray)
{
++++double+scoreArray[maxStudents];
++++...
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|540935|编译错误的发生是因为您使用与形参double+scoreArray同名的名称scoreArray声明了一个局部变量。更改其中一个的名称。|unstyled|offset|length|style|CODE|540936|entityMap^0|0|G|H|12|A|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|S|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>double readScores(double scoreArray)
{
 double scoreArray[maxStudents];
 ...
}
</code></pre>

The compile error happens because you declared a local variable with the name <code>scoreArray</code>, which is the same name as that of the formal parameter <code>double scoreArray</code>. Change the name of one of them.

blocks|key|1561557|text|你就快成功了：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1561558|double+readScores(double+scoreArray[])+{
++++double+*scorePTR;
++++*scorePTR+=+scoreArray;

++++for(int+count+=+0;+count+<+maxStudents;+count%2B%2B)
++++{
++++++++cout<<"Please+enter+score+for+student+"<<count%2B1<<"+or+-999+to+end.\n";
++++++++cin>>*(scorePTR%2Bcount);
++++++++if(*(scorePTR%2Bcount)+==+-999)
++++++++break;
++++}
}|code-block|syntax|javascript|1561559|我还建议在访问数组时切换到“直接”数组语法，而不是依赖数组即指针语义，即在表达式中使用scoreArray[count]而不是*(scoreArray%2Bcount)：|offset|length|style|CODE|1561560|double+readScores(double+scoreArray[])+{
++++for(int+count+=+0;+count+<+maxStudents;+count%2B%2B)
++++{
++++++++cout<<"Please+enter+score+for+student+"<<count%2B1<<"+or+-999+to+end.\n";
++++++++cin+>>+scoreArray[count];
++++++++if(scoreArray[count]+==+-999)
++++++++break;
++++}
}|1561561|附注:我假设您还没有开始学习STL；使用标准C%2B%2B库可以找到更合适的解决方案。|1561562|entityMap^0|0|0|17|H|1R|J|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@$I|V|J|W|K|L]|$I|X|J|Y|K|L]]|9|@]|A|$]]|$1|M|3|N|5|D|7|Z|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|10|8|@]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|11|8|@]|9|@]|A|$]]]|R|$]]

You're almost there:

<pre><code>double readScores(double scoreArray[]) {
 double *scorePTR;
 *scorePTR = scoreArray;

 for(int count = 0; count &lt; maxStudents; count++)
 {
 cout&lt;&lt;"Please enter score for student "&lt;&lt;count+1&lt;&lt;" or -999 to end.\n";
 cin&gt;&gt;*(scorePTR+count);
 if(*(scorePTR+count) == -999)
 break;
 }
}
</code></pre>

I also suggest switching to "straight" array syntax when accessing the array, rather than relying on array-as-pointer semantic, i.e. use <code>scoreArray[count]</code> rather than <code>*(scoreArray+count)</code> in your expressions:

<pre><code>double readScores(double scoreArray[]) {
 for(int count = 0; count &lt; maxStudents; count++)
 {
 cout&lt;&lt;"Please enter score for student "&lt;&lt;count+1&lt;&lt;" or -999 to end.\n";
 cin &gt;&gt; scoreArray[count];
 if(scoreArray[count] == -999)
 break;
 }
}
</code></pre>

P.S.: I assume that you have not started learning about STL; more appropriate solutions exist with the use of standard C++ library.

blocks|key|618823|text|您得到的错误是因为您命名了一个与函数参数相同的局部变量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|618824|double+readScores(double+scoreArray+/*+parameter+named+scoreArray+*/)
{
++++double+scoreArray[maxStudents];+//+this+cannot+have+the+same+name!|code-block|syntax|javascript|618825|您所显示的代码揭示了很多混淆。我会试着解释其中的错误。|618826|cout<<scoreArray[1],scoreArray[2];|618827|这不会输出元素1和2。要这样做，它应该是cout+<<+scoreArray[1]+<<+scoreArray[2];。还要注意，C%2B%2B中的数组是从0开始的，因此第一个元素是scoreArray[0]，第二个是scoreArray[1]，第三个是scoreArray[2]，依此类推。|offset|length|style|CODE|618828|该代码声明了一个名为readScores的函数，该函数接受指向double的指针并返回一个double：|618829|double+readScores(double[]);|618830|但它只定义了一个接受double的函数|618831|double+readScores(double+scoreArray)|618832|我确信这不是我们想要的，一旦编译器错误被修复，它将产生一个链接器错误，因为没有对double+readScores(double[])函数的定义。|618833|现在，数组。|618834|当您将参数编写为double+x[]时，它不是一个数组。它是一个指向double的指针。它与double*+x完全相同。是的，语法具有很强的误导性。这就是为什么你不应该使用它。它播下了混乱的种子。最好是显式地将指针称为指针。|618835|在C%2B%2B中不能直接返回数组，在C%2B%2B中也不能直接传递数组参数。您可以传递对数组的引用，也可以返回对数组的引用，但这里似乎有使用指针的说明，因此我不再赘述。|618836|代码似乎没有使用从readScores返回的值，所以只使用void返回类型可能更好。无论如何，代码都会将值直接写入数组。|618837|void+readScores(double*+scorePtr)|618838|语法*(scorePTR%2Bcount)与scorePTR[count]完全相同，所以使用它没有任何好处。无论如何，代码都是使用索引(count)进行迭代的。|618839|我认为赋值对这个限制的意义实际上是有用的，即使只是略微有用:使用指针而不是索引进行迭代。为此，您需要找到三件事:如何开始循环，如何移动到下一个元素，以及如何结束循环。|618840|如何启动循环？您应该从指向第一个元素的指针开始。幸运的是，这样的指针正是作为参数传递给函数的。|618841|for(double*+ptr+=+scorePtr;+/*+...+*/;+/*+...+*/)+|618842|如何移动到下一个元素？如果递增指针，它将移动到下一个元素。|618843|for(double*+ptr+=+scorePtr;+/*+...+*/;+%2B%2Bptr)|618844|如何结束循环？当指针经过整个数组时，循环结束。该数组具有maxStudents元素，因此当指针远离开始处的maxStudents元素时，循环结束：|618845|double*+end+=+scorePtr+%2B+maxStudents;
for(double*+ptr+=+scorePtr;+ptr+!=+end;+%2B%2Bptr)|618846|如何在循环中使用这个指针？正常情况下，使用取消引用运算符。|618847|for(double*+ptr+=+scorePtr;+ptr+!=+end;+%2B%2Bptr)
{
++++cout<<"Please+enter+score+for+student+"<<+/*+(1)+*/+<<"+or+-999+to+end.\n";
++++cin>>*ptr;
++++if(*ptr+==+-999)
++++break;
}

//+(1)+left+as+an+exercise+for+the+reader|618848|entityMap^0|0|0|0|0|K|13|2F|D|2X|D|3F|D|0|A|A|V|6|19|6|0|0|A|6|0|0|14|R|0|0|8|A|Y|6|1A|9|0|0|9|A|T|4|0|0|2|H|K|F|1U|5|0|0|0|0|0|0|S|B|1H|B|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1W|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1X|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1Y|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|1Z|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|20|8|@$M|21|N|22|O|P]|$M|23|N|24|O|P]|$M|25|N|26|O|P]|$M|27|N|28|O|P]]|9|@]|A|$]]|$1|Q|3|R|5|6|7|29|8|@$M|2A|N|2B|O|P]|$M|2C|N|2D|O|P]|$M|2E|N|2F|O|P]]|9|@]|A|$]]|$1|S|3|T|5|D|7|2G|8|@]|9|@]|A|$E|F]]|$1|U|3|V|5|6|7|2H|8|@$M|2I|N|2J|O|P]]|9|@]|A|$]]|$1|W|3|X|5|D|7|2K|8|@]|9|@]|A|$E|F]]|$1|Y|3|Z|5|6|7|2L|8|@$M|2M|N|2N|O|P]]|9|@]|A|$]]|$1|10|3|11|5|6|7|2O|8|@]|9|@]|A|$]]|$1|12|3|13|5|6|7|2P|8|@$M|2Q|N|2R|O|P]|$M|2S|N|2T|O|P]|$M|2U|N|2V|O|P]]|9|@]|A|$]]|$1|14|3|15|5|6|7|2W|8|@]|9|@]|A|$]]|$1|16|3|17|5|6|7|2X|8|@$M|2Y|N|2Z|O|P]|$M|30|N|31|O|P]]|9|@]|A|$]]|$1|18|3|19|5|D|7|32|8|@]|9|@]|A|$E|F]]|$1|1A|3|1B|5|6|7|33|8|@$M|34|N|35|O|P]|$M|36|N|37|O|P]|$M|38|N|39|O|P]]|9|@]|A|$]]|$1|1C|3|1D|5|6|7|3A|8|@]|9|@]|A|$]]|$1|1E|3|1F|5|6|7|3B|8|@]|9|@]|A|$]]|$1|1G|3|1H|5|D|7|3C|8|@]|9|@]|A|$E|F]]|$1|1I|3|1J|5|6|7|3D|8|@]|9|@]|A|$]]|$1|1K|3|1L|5|D|7|3E|8|@]|9|@]|A|$E|F]]|$1|1M|3|1N|5|6|7|3F|8|@$M|3G|N|3H|O|P]|$M|3I|N|3J|O|P]]|9|@]|A|$]]|$1|1O|3|1P|5|D|7|3K|8|@]|9|@]|A|$E|F]]|$1|1Q|3|1R|5|6|7|3L|8|@]|9|@]|A|$]]|$1|1S|3|1T|5|D|7|3M|8|@]|9|@]|A|$E|F]]|$1|1U|3|-4|5|6|7|3N|8|@]|9|@]|A|$]]]|1V|$]]

The error you get is because you named a local variable the same as the function's argument.

<pre><code>double readScores(double scoreArray /* parameter named scoreArray */)
{
 double scoreArray[maxStudents]; // this cannot have the same name!
</code></pre>

The code you've shown reveals a lot of confusion. I'll try to explain the mistakes in it.

<pre><code>cout&lt;&lt;scoreArray[1],scoreArray[2];
</code></pre>

This will not output the elements 1 and 2. To do that, it should be <code>cout &lt;&lt; scoreArray[1] &lt;&lt; scoreArray[2];</code>. Also, note that arrays in C++ are 0-based, so the first element will be <code>scoreArray[0]</code>, the second <code>scoreArray[1]</code>, the third <code>scoreArray[2]</code>, and so on.

The code declares a function named <code>readScores</code> taking a pointer to <code>double</code> and returning a <code>double</code>: 

<pre><code>double readScores(double[]);
</code></pre>

But it only defines a function taking a <code>double</code>:

<pre><code>double readScores(double scoreArray)
</code></pre>

I'm sure that's not what was intended, and once the compiler error is fixed, it will produce a linker error because there's no definition for the <code>double readScores(double[])</code> function.

Now, the arrays.

When you write a parameter as <code>double x[]</code>, it's not an array. It's a pointer to a <code>double</code>. It's exactly the same as <code>double* x</code>. Yes, the syntax is highly misleading. That's why you shouldn't use it. It sows confusion. It's better to be explicit and call a pointer a pointer.

You cannot directly return arrays in C++ and you cannot directly pass array arguments in C++. You can pass references to arrays and return references to arrays, but here it seems you have instructions to use pointers, so I won't go there.

The code doesn't seem to be making use of the value returned from <code>readScores</code>, so it's probably better to just use a <code>void</code> return type. The code writes the values directly into the array anyway.

<pre><code>void readScores(double* scorePtr)
</code></pre>

The syntax <code>*(scorePTR+count)</code> is exactly the same as <code>scorePTR[count]</code>, so there's no gain in using it. And the code is iterating with an index (<code>count</code>) anyway.

I suppose what the assignment meant with that restriction was something actually useful, even if only marginally: iterate using a pointer, not an index. In order to do that, you need to find three things: how to start the loop, how to move to the next element and how to end the loop.

How to start the loop? You should start with a pointer to the first element. Luckily, such a pointer is exactly what gets passed as an argument to the function.

<pre><code>for(double* ptr = scorePtr; /* ... */; /* ... */) 
</code></pre>

How to move to the next element? If you increment the pointer, it moves to the next element.

<pre><code>for(double* ptr = scorePtr; /* ... */; ++ptr)
</code></pre>

How to end the loop? The loop ends when the pointer has passed over the entire array. The array has <code>maxStudents</code> elements, so the loop ends when the pointer is <code>maxStudents</code> elements away from the start:

<pre><code>double* end = scorePtr + maxStudents;
for(double* ptr = scorePtr; ptr != end; ++ptr)
</code></pre>

And how do you use this pointer in the loop? As normal, with the dereferencing operator.

<pre><code>for(double* ptr = scorePtr; ptr != end; ++ptr)
{
 cout&lt;&lt;"Please enter score for student "&lt;&lt; /* (1) */ &lt;&lt;" or -999 to end.\n";
 cin&gt;&gt;*ptr;
 if(*ptr == -999)
 break;
}

// (1) left as an exercise for the reader
</code></pre>

I'm learning about pointers in class but I never really understood arrays 100%

I have a assignment where I have to read in an array of student scores from a function. Each element of the array would hold a different score and the element identifies what student it is.

I'm trying to read the scores into a array but when it comes to returning it I'm lost. I've tried google but just get more confused because people talking about dynamic memory and using the new and delete operand which I haven't gotten to yet, and I don't want to do the 
assignment wrong.

I have to user pointer instead of index to access elements in the array.
 The code below gives me the error "redefinition of formal parameter 'scoreArray'"

<pre><code>#include &lt;iostream&gt;
using namespace std;

const int maxStudents = 30;
double readScores(double[]);

int main()
{
 double scoreArray[maxStudents];
 readScores(scoreArray);

 cout&lt;&lt;scoreArray[1],scoreArray[2];

 system ("PAUSE");
 return 0;
}

double readScores(double scoreArray)
{
 double scoreArray[maxStudents];
 double *scorePTR;
 *scorePTR = scoreArray;

 for(int count = 0; count &lt; maxStudents; count++)
 {
 cout&lt;&lt;"Please enter score for student "&lt;&lt;count+1&lt;&lt;" or -999 to end.\n";
 cin&gt;&gt;*(scorePTR+count);
 if(*(scorePTR+count) == -999)
 break;
 }
}
</code></pre>

How do I return a array from a function?

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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我在课堂上学习指针，但我从来没有100%理解数组。我有一个作业，我必须从一个函数中读取学生成绩的数组。数组中的每个元素都有不同的分数，元素标识它是哪个学生。我试图将分数读入数组，但当返回它时，我就迷失了方向。我试过google，但我更困惑了，因为人们都在谈论动态内存和使用新的和delete操作数，而我还没有做到这一点，...

问如何从函数返回数组？
EN

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问如何从函数返回数组？EN