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  1. PO
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    <p>Well, you are trying to return a vector as strings. This won't work because they are different types and have no conversion defined from one to the other. Your function has the return type <code>string</code>.</p> <p><strong>Solution 1</strong></p> <p>In your case you could append the lines to a string instead of adding them to a vector? You are using the result as a string anyhow.</p> <p>You could change seqs to <code>string</code> and append data to it with the <code>+=</code> operator. </p> <p><strong>Solution 2</strong></p> <p>You could also change the return type to <code>vector&lt;string&gt;</code> but you would need to loop over the items and print them instead in your <code>main</code>. </p> <pre><code>vector&lt;string&gt; getseq(char * db_file) { ... return seqs; } </code></pre> <p><em>Caveat Lector: this will copy all the items. If you want to avoid this pass the vector as a reference to the function and add to it.</em></p> <p>Looping is quite easy using iterators:</p> <pre><code>// Get the strings as a vector. vector&lt;string&gt; seqs = getseq(argv[1]); // This is just a matter of taste, we create an alias for the vector&lt;string&gt; iterator type. typedef vector&lt;string&gt;:iterator_t string_iter; // Loop till we hit the end of the vector. for (string_iter i = seqs.begin(); i != seqs.end(); i++) { cout &lt;&lt; *i; // you could add endlines, commas here etc. } </code></pre> <hr> <p>If you want to avoid copying a vector and all the strings make <code>getseq</code> take a reference to a <code>vector&lt;string&gt;</code>.</p> <pre><code>void getseq(char * db_file, vector&lt;string&gt; &amp;seqs) { ... // vector&lt;string&gt; seqs; this line is not needed anymore. ... // we don't need to return anything anymore } </code></pre> <p>You would then need to create the <code>vector&lt;string&gt;</code> in your main instead, making my above code:</p> <pre><code>// Get the strings as a vector. vector&lt;string&gt; seqs; // Holds our strings. getseq(argv[1], seqs); // We don't return anything. // This is just a matter of taste, we create an alias for the vector&lt;string&gt; iterator type. typedef vector&lt;string&gt;:iterator_t string_iter; // Print prelude. cout &lt;&lt; "Sekwencje: \n"; // Loop till we hit the end of the vector. for (string_iter i = seqs.begin(); i != seqs.end(); i++) { cout &lt;&lt; *i &lt;&lt; " "; // Do processing, add endlines, commas here etc. } cout &lt;&lt; endl; </code></pre> <hr> <p><em>Edit after comments</em></p> <pre><code>int main(int argc, char * argv[1]) { // This is what you need, sorry for the confusion. // This copies the vector returned to seqs vector&lt;string&gt; seqs = getseq(argv[1]); // This is just a matter of taste, we create an alias for the vector&lt;string&gt; iterator type. typedef vector&lt;string&gt;::iterator string_iter; // Print prelude. cout &lt;&lt; "Sekwencje: \n"; // Loop till we hit the end of the vector. for (string_iter i = seqs.begin(); i != seqs.end(); i++) { cout &lt;&lt; *i &lt;&lt; " "; // Do processing, add endlines, commas here etc. } cout &lt;&lt; endl; } </code></pre>
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    1. POvector in function - how to do return
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    1. USSkurmedel
    1. USSkurmedel
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    1. COThanks for this advice! I want to pront it only to check if it read lines wright. In program I want to have this vector because I want to iterate and do some operations on vector items - it will be some biological sequences.
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      1. USMateusz Korycinski
    2. COMateusz K: Ok :) I added a bit more flesh to my answer.
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      1. USSkurmedel
    3. COThanks. :) Unfortunatelly after I've compiled it, and execute (./fpars seqs.fasta) it returns some signs (like in binary file) and statement: Violation of memory security. I have this statement in polish so I do not know english analog. :/
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      1. USMateusz Korycinski
 

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