blocks|key|1554914|text|据我所见，这似乎没有什么不同。也许问题在于评论者所说的，将字符串数组更改为字符数组。无论如何，这段代码适用于您的场景：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1554915|int+main()+{

++std::string+str+=+"1%2B1";
++
++char+Operators[4]+=+{'%2B','-','*','/'};+//+using+the+commenters+suggestion,+char+array+is+better

++for+(int+i+=+0;+i+<+4;+i%2B%2B)+{
++++for+(int+j+=+0;+j+<+str.length();+j%2B%2B)+{
++++++if+(Operators[i]+==+str[j])
++++++++std::cout+<<+"A+'"+<<+Operators[i]+<<+"'+is+present"+<<+std::endl;
++++}
++}

++return+0;

}|code-block|syntax|javascript|1554916|正如另一位评论者所言，您也可以使用迭代器。|1554917|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

From what I can see, it does not seem different. Maybe the issue was what the commenter said, changing your array of strings to an array of chars. Regardless this code works for your scenario:
<pre><code>int main() {

 std::string str = &quot;1+1&quot;;
 
 char Operators[4] = {'+','-','*','/'}; // using the commenters suggestion, char array is better

 for (int i = 0; i &lt; 4; i++) {
 for (int j = 0; j &lt; str.length(); j++) {
 if (Operators[i] == str[j])
 std::cout &lt;&lt; &quot;A '&quot; &lt;&lt; Operators[i] &lt;&lt; &quot;' is present&quot; &lt;&lt; std::endl;
 }
 }

 return 0;

}
</code></pre>
As another commenter said, you could also use an iterator.

blocks|key|771033|text|使用find_first_of|type|unstyled|depth|inlineStyleRanges|entityRanges|data|771034|+std::string+Equation+=+"1%2B1";

+std::size_t+found+=+Equation.find_first_of("%2B-/*");|code-block|syntax|javascript|771035|如果找到char，“found”将是偏移量(在本例中为1)。如果未找到，则“found”将等于std::string::npos|771036|看见|771037|of/+of/|offset|length|771038|要知道找到哪个操作员，请执行|771039|+++char+op+=+Equation[found];|771040|entityMap|0|LINK|mutability|MUTABLE|url|https://www.cplusplus.com/reference/string/string/find_first_of/^0|0|0|0|0|0|7|0|0|0|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|13|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|14|8|@]|9|@$M|15|N|16|1|17]]|A|$]]|$1|O|3|P|5|6|7|18|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$U|$5|V|W|X|A|$Y|Z]]]]

use find_first_of
<pre><code> std::string Equation = &quot;1+1&quot;;

 std::size_t found = Equation.find_first_of(&quot;+-/*&quot;);
</code></pre>
if char is found 'found' will be the offset (1 in this case). If not found then 'found' will equal std::string::npos
See
<a href="https://www.cplusplus.com/reference/string/string/find_first_of/" rel="nofollow noreferrer">https://www.cplusplus.com/reference/string/string/find_first_of/</a>
To know which operator was found do
<pre><code> char op = Equation[found];
</code></pre>

blocks|key|1555020|text|通过std::string类的成员函数，特别是，为了您的目的，检查搜索成员函数。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|1555021|下面是使用find类的std::string成员函数在方程中寻找操作符的一个例子|CODE|1555022|#include+<string>
#include+<iostream>
+
int+main()
{
++++std::string+Operators[4]+=+{
++++++"%2B",
++++++"-",
++++++"*",
++++++"/"
++++};
++++
++++std::string+Equation+=+"1%2B1";
+
++++for+(std::size_t+i+=+0;+i+<+sizeof(Operators)/sizeof(Operators[0]);+%2B%2Bi)+{
++++++++std::string::size_type+n+=+Equation.find(Operators[i]);

++++++++std::cout+<<+Operators[i]+<<+"+:+";
++++++++if+(n+==+std::string::npos)+{
++++++++++++std::cout+<<+"not+found\n";
++++++++}+else+{
++++++++++++std::cout+<<+"found+at+position+:+"+<<+n+<<+'\n';
++++++++}
++++}

++++return+0;+
}|code-block|syntax|javascript|1555023|输出：|1555024|#+./a.out
%2B+:+found+at+position+:+1
-+:+not+found
*+:+not+found
/+:+not+found|1555025|注意，在上面的程序中，如果将Operators数组的类型更改为|1555026|char+Operators[4]+=+{
++'%2B',
++'-',
++'*',
++'/'
};|1555027|如果不进行任何其他更改，它将非常好地工作，因为find成员函数有一个过载，它以char作为参数。|1555028|entityMap|0|LINK|mutability|MUTABLE|url|https://en.cppreference.com/w/cpp/string/basic_string^0|X|2|2|C|0|0|5|4|B|B|0|0|0|0|E|9|0|0|N|4|13|4|0^^$0|@$1|2|3|4|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|F|3|G|5|6|7|1B|8|@$9|1C|A|1D|B|H]|$9|1E|A|1F|B|H]]|D|@]|E|$]]|$1|I|3|J|5|K|7|1G|8|@]|D|@]|E|$L|M]]|$1|N|3|O|5|6|7|1H|8|@]|D|@]|E|$]]|$1|P|3|Q|5|K|7|1I|8|@]|D|@]|E|$L|M]]|$1|R|3|S|5|6|7|1J|8|@$9|1K|A|1L|B|H]]|D|@]|E|$]]|$1|T|3|U|5|K|7|1M|8|@]|D|@]|E|$L|M]]|$1|V|3|W|5|6|7|1N|8|@$9|1O|A|1P|B|H]|$9|1Q|A|1R|B|H]]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1S|8|@]|D|@]|E|$]]]|Y|$Z|$5|10|11|12|E|$13|14]]]]

Go through the member functions of <a href="https://en.cppreference.com/w/cpp/string/basic_string" rel="nofollow noreferrer">std::string class</a> and specifically, for your purpose, check the Search member functions.
Below is an example of using <code>find</code> member function of <code>std::string</code> class for finding Operators in Equation:
<pre><code>#include &lt;string&gt;
#include &lt;iostream&gt;
 
int main()
{
 std::string Operators[4] = {
 &quot;+&quot;,
 &quot;-&quot;,
 &quot;*&quot;,
 &quot;/&quot;
 };
 
 std::string Equation = &quot;1+1&quot;;
 
 for (std::size_t i = 0; i &lt; sizeof(Operators)/sizeof(Operators[0]); ++i) {
 std::string::size_type n = Equation.find(Operators[i]);

 std::cout &lt;&lt; Operators[i] &lt;&lt; &quot; : &quot;;
 if (n == std::string::npos) {
 std::cout &lt;&lt; &quot;not found\n&quot;;
 } else {
 std::cout &lt;&lt; &quot;found at position : &quot; &lt;&lt; n &lt;&lt; '\n';
 }
 }

 return 0; 
}
</code></pre>
Output:
<pre><code># ./a.out
+ : found at position : 1
- : not found
* : not found
/ : not found
</code></pre>
Note that, in the above program, if you change the type of <code>Operators</code> array to
<pre><code>char Operators[4] = {
 '+',
 '-',
 '*',
 '/'
};
</code></pre>
it will work absolutely fine without any other change because the <code>find</code> member function has an overload which takes <code>char</code> as argument.

So I have an array that looks as such:
<pre><code>std::string Operators[4] = {
 &quot;+&quot;,
 &quot;-&quot;,
 &quot;*&quot;,
 &quot;/&quot;
};
</code></pre>
And I have another string, that looks like this
<pre><code>std::string Equation = &quot;1+1&quot;;
</code></pre>
I'm trying to make the code check if the plus from the array is in the equation string. Is it no different than finding a substring in a string? Any help is greatly appreciated.

How can I find object in array in a string?

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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所以我有一个数组，看起来是这样的：std::string Operators[4] = {  "+",  "-",  "*",  "/"};我还有一根绳子，看上去像这样std::string Equation = "1+1";我试图让代码检查来自数组的加号是否在等式字符串中。这和在字符串中找到子字符串没有什么不同吗？任...

问如何在数组中找到字符串中的对象？
EN

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问如何在数组中找到字符串中的对象？EN