blocks|key|586225|text|你正在尝试做的事情是不会成功的。首先，使用以下命令在linspace中分配堆栈上的内存|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|586226|double+line[c];|code-block|syntax|javascript|586227|您要么在调用前新分配内存并传递它，要么动态分配并返回它(并记得稍后释放它)。|586228|要动态分配，您可以执行以下操作：|586229|double+*+line+=+new+double[c];|586230|同样，当你完成它的时候，需要像这样释放它，否则你会有一个内存泄漏。|586231|delete+line[];|586232|此外，double+line[c];创建了一个双精度数组，line指向该数组。所以line是一个double+*。您将函数的返回类型指定为double|586233|entityMap^0|Q|8|0|0|0|0|0|0|0|3|F|T|4|1C|8|1X|6|0^^$0|@$1|2|3|4|5|6|7|Y|8|@$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|12|8|@]|D|@]|E|$]]|$1|M|3|N|5|6|7|13|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|15|8|@]|D|@]|E|$]]|$1|S|3|T|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|U|3|V|5|6|7|17|8|@$9|18|A|19|B|C]|$9|1A|A|1B|B|C]|$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|1G|8|@]|D|@]|E|$]]]|X|$]]

What you are trying to do what won't work. First off you are allocating the memory on the stack WITHIN <code>linspace</code> with

<pre><code>double line[c];
</code></pre>

You either new to allocate the memory before the call and pass it in, or allocate it dynamically and return it (and remember to free it later).

To allocate dynamically you can do the following:

<pre><code>double * line = new double[c];
</code></pre>

Again, this will need to be freed like so at some point later when you are finished with it, otherwise you will have a memory leak.

<pre><code>delete line[];
</code></pre>

Also <code>double line[c];</code> creates an array of doubles and <code>line</code> points to that. So line is a <code>double *</code>. You specified the return type of the function as <code>double</code>

blocks|key|587690|text|这样如何：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|587691|#include+<iostream>
#include+<vector>

template<typename+T>
std::vector<double>+linspace(T+start_in,+T+end_in,+int+num_in)
{

++std::vector<double>+linspaced;

++double+start+=+static_cast<double>(start_in);
++double+end+=+static_cast<double>(end_in);
++double+num+=+static_cast<double>(num_in);

++if+(num+==+0)+{+return+linspaced;+}
++if+(num+==+1)+
++++{
++++++linspaced.push_back(start);
++++++return+linspaced;
++++}

++double+delta+=+(end+-+start)+/+(num+-+1);

++for(int+i=0;+i+<+num-1;+%2B%2Bi)
++++{
++++++linspaced.push_back(start+%2B+delta+*+i);
++++}
++linspaced.push_back(end);+//+I+want+to+ensure+that+start+and+end
++++++++++++++++++++++++++++//+are+exactly+the+same+as+the+input
++return+linspaced;
}

void+print_vector(std::vector<double>+vec)
{
++std::cout+<<+"size:+"+<<+vec.size()+<<+std::endl;
++for+(double+d+:+vec)
++++std::cout+<<+d+<<+"+";
++std::cout+<<+std::endl;
}

int+main()
{
++std::vector<double>+vec_1+=+linspace(1,+10,+3);
++print_vector(vec_1);

++std::vector<double>+vec_2+=+linspace(6.0,+23.4,+5);
++print_vector(vec_2);

++std::vector<double>+vec_3+=+linspace(0.0,+2.0,+1);
++print_vector(vec_3);

++std::vector<double>+vec_4+=+linspace(0.0,+2.0,+0);
++print_vector(vec_4);


++return+0;
}|code-block|syntax|javascript|587692|C%2B%2B结果：|587693|size:+3
1+5.5+10+
size:+5
6+10.35+14.7+19.05+23.4+
size:+1
0+
size:+0|587694|Numpy结果：|587695|In+[14]:+np.linspace(1,+10,+3)
Out[14]:+array([++1.+,+++5.5,++10.+])

In+[15]:+np.linspace(6,+23.4,+5)
Out[15]:+array([++6.++,++10.35,++14.7+,++19.05,++23.4+])

In+[16]:+np.linspace(0.0,+2.0,+1)
Out[16]:+array([+0.])

In+[17]:+np.linspace(0.0,+2.0,+0)
Out[17]:+array([],+dtype=float64)|587696|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

How about something like this:

<pre><code>#include &lt;iostream&gt;
#include &lt;vector&gt;

template&lt;typename T&gt;
std::vector&lt;double&gt; linspace(T start_in, T end_in, int num_in)
{

 std::vector&lt;double&gt; linspaced;

 double start = static_cast&lt;double&gt;(start_in);
 double end = static_cast&lt;double&gt;(end_in);
 double num = static_cast&lt;double&gt;(num_in);

 if (num == 0) { return linspaced; }
 if (num == 1) 
 {
 linspaced.push_back(start);
 return linspaced;
 }

 double delta = (end - start) / (num - 1);

 for(int i=0; i &lt; num-1; ++i)
 {
 linspaced.push_back(start + delta * i);
 }
 linspaced.push_back(end); // I want to ensure that start and end
 // are exactly the same as the input
 return linspaced;
}

void print_vector(std::vector&lt;double&gt; vec)
{
 std::cout &lt;&lt; "size: " &lt;&lt; vec.size() &lt;&lt; std::endl;
 for (double d : vec)
 std::cout &lt;&lt; d &lt;&lt; " ";
 std::cout &lt;&lt; std::endl;
}

int main()
{
 std::vector&lt;double&gt; vec_1 = linspace(1, 10, 3);
 print_vector(vec_1);

 std::vector&lt;double&gt; vec_2 = linspace(6.0, 23.4, 5);
 print_vector(vec_2);

 std::vector&lt;double&gt; vec_3 = linspace(0.0, 2.0, 1);
 print_vector(vec_3);

 std::vector&lt;double&gt; vec_4 = linspace(0.0, 2.0, 0);
 print_vector(vec_4);


 return 0;
}
</code></pre>

C++ result:

<pre><code>size: 3
1 5.5 10 
size: 5
6 10.35 14.7 19.05 23.4 
size: 1
0 
size: 0
</code></pre>

Numpy result:

<pre><code>In [14]: np.linspace(1, 10, 3)
Out[14]: array([ 1. , 5.5, 10. ])

In [15]: np.linspace(6, 23.4, 5)
Out[15]: array([ 6. , 10.35, 14.7 , 19.05, 23.4 ])

In [16]: np.linspace(0.0, 2.0, 1)
Out[16]: array([ 0.])

In [17]: np.linspace(0.0, 2.0, 0)
Out[17]: array([], dtype=float64)
</code></pre>

im trying to write a similar version of python's numpy.linspace function.

<pre><code>double linspace(int a, int b, int c){
 double line[c];
 double delta =b-a/(c-1);
 for (int i=0; i&lt;c; ++i){
 line[i]=0 + (i*delta);
 }
 return line;
</code></pre>

with a and b being the first and the last component in the array, and c specifying the number of elements in the array. But when i compile this script it returns:

<pre><code>linspace.cpp: In function ‘double linspace(int, int, int)’:
linspace.cpp:11:9: error: cannot convert ‘double*’ to ‘double’ in return
 return line;
 ^
</code></pre>

would anyone happen to knw how to solve this problem?

python linspace in c++

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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我正在尝试编写一个类似版本的python的numpy.linspace函数。double linspace(int a, int b, int c){ double line[c]; double delta =b-a/(c-1); for (int i=0; i<c; ++i){ ...

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