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    <p>As shown <a href="http://www.cplusplus.com/reference/string/string/substr/" rel="noreferrer">here</a>, the second argument to <code>substr</code> is the <em>length</em>, not the ending position:</p> <blockquote> <p><code>string substr ( size_t pos = 0, size_t n = npos ) const;</code></p> <p><em>Generate substring</em></p> <p>Returns a string object with its contents initialized to a substring of the current object. This substring is the character sequence that starts at character position <code>pos</code> and has a length of <code>n</code> characters.</p> </blockquote> <p>Your line <code>b = a.substr(i,i+1);</code> will generate, for values of <code>i</code>:</p> <pre><code>substr(0,1) = 1 substr(1,2) = 23 substr(2,3) = 345 substr(3,4) = 45 (since your string stops there). </code></pre> <p>What you need is <code>b = a.substr(i,2);</code></p> <p>You should also be aware that your output will look funny for a number like 12045. You'll get <code>12 20 4 45</code> due to the fact that you're using <code>atoi()</code> on the string section and outputting that integer. You might want to try just outputing the string itself which <em>will</em> be two characters long:</p> <pre><code>b = a.substr(i,2); cout &lt;&lt; b &lt;&lt; " "; </code></pre> <p>In fact, the entire thing could be more simply written as:</p> <pre><code>#include &lt;iostream&gt; #include &lt;string&gt; using namespace std; int main(void) { string a; cin &gt;&gt; a; for (int i = 0; i &lt; a.size() - 1; i++) cout &lt;&lt; a.substr(i,2) &lt;&lt; " "; cout &lt;&lt; endl; return 0; } </code></pre>
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