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    <p>Usually you can use a cast to convert the operand to some type for which you know the right format.</p> <p>Your proposed solution:</p> <pre><code>time_t t = time(NULL); printf("%s", t); </code></pre> <p>clearly will <em>not</em> work, since <code>time_t</code> is a numeric type, not <code>char*</code>.</p> <p>We know, in general, that <code>time_t</code> is an arithmetic type. Something like this:</p> <pre><code> printf("%ld\n", (long)t); </code></pre> <p>is likely to work on most systems. It can fail (a) if <code>time_t</code> is an unsigned type no wider than <code>unsigned long</code> and the current value of <code>t</code> exceeds <code>LONG_MAX</code>, or (b) if <code>time_t</code> is a floating-point type.</p> <p>If you have C99 support, you can use <code>long long</code>, which is a little better:</p> <pre><code>printf("%lld\n", (long long)t); </code></pre> <p>If you really want to go overboard with portability, you can detect what kind of type <code>time_t</code> is:</p> <pre><code>if ((time_t)-1 &gt; 0) { // time_t is an unsigned type printf("%ju\n", (uintmax_t)t); } else if ((time_t)1 / 2 &gt; 0) { // time_t is a signed integer type printf("%jd\n", (intmax_t)t); } else { // time_t is a floating-point type (I've never seen this) printf("%Lg\n", (long double)t); } </code></pre> <p>You might want to tweak the <code>%Lg</code> format to something like <code>%Lf</code> or <code>%.10Lf</code>, depending on what output format you want.</p> <p>Again, this assumes C99 support -- and you'll need <code>#include &lt;stdint.h&gt;</code> to make <code>uintmax_t</code> and <code>intmax_t</code> visible.</p> <p><code>time_t</code> and <code>clock_t</code> are a bit unusual, in that the standard says <em>only</em> that they're arithmetic type capable of representing times. (In principle they could be complex types, but I'd say ignoring that possibility is worth the risk.)</p> <p>In most other cases, you'll probably know whether a given type is signed, unsigned, or floating-point, and you can just convert to the widest type of that kind.</p> <p>Note that if you don't know how <code>time_t</code> is represented, you probably won't understand the output of the <code>printf</code> (such as <code>1379375215</code>) either -- unless your goal is to figure that out.</p> <p>(If you were programming in C++ rather than C, <code>std::cout &lt;&lt; t &lt;&lt; "\n";</code> would automatically use the correct overloaded <code>operator&lt;&lt;</code>.)</p> <p>If you want <em>human-readable</em> output (like <code>Mon 2013-09-16 16:46:55 PDT</code>), you'll want to use one of the conversion functions declared in <code>&lt;time.h&gt;</code>, such as <code>asctime()</code> or <code>strftime()</code>.</p>
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