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    <p><a href="http://www.cppreference.com/wiki/keywords/sizeof" rel="noreferrer"><code>sizeof(x)</code></a> returns the amount of memory (in bytes) that the variable or type <code>x</code> occupies. It has nothing to do with the value of the variable.</p> <p>For example, if you have an array of some arbitrary type <code>T</code> then the distance between elements of that array is exactly <code>sizeof(T)</code>.</p> <pre><code>int a[10]; assert(&amp;(a[0]) + sizeof(int) == &amp;(a[1])); </code></pre> <p>When used on a variable, it is equivalent to using it on the type of that variable:</p> <pre><code>T x; assert(sizeof(T) == sizeof(x)); </code></pre> <p>As a rule-of-thumb, it is best to use the variable name where possible, just in case the type changes:</p> <pre><code>int x; std::cout &lt;&lt; "x uses " &lt;&lt; sizeof(x) &lt;&lt; " bytes." &lt;&lt; std::endl // If x is changed to a char, then the statement doesn't need to be changed. // If we used sizeof(int) instead, we would need to change 2 lines of code // instead of one. </code></pre> <p>When used on user-defined types, <code>sizeof</code> still returns the amount of memory used by instances of that type, but it's worth pointing out that this does not necessary equal the sum of its members.</p> <pre><code>struct Foo { int a; char b; }; </code></pre> <p>While <code>sizeof(int) + sizeof(char)</code> is typically <code>5</code>, on many machines, <code>sizeof(Foo)</code> may be <code>8</code> because the compiler needs to <a href="http://en.wikipedia.org/wiki/Data_structure_alignment" rel="noreferrer">pad out</a> the structure so that it lies on 4 byte boundaries. This is not always the case, and it's quite possible that on your machine <code>sizeof(Foo)</code> will be 5, but you can't depend on it.</p>
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    1. POWhat does sizeof do?
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    1. USPeter Alexander
    1. USPeter Alexander
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    1. COint means that if variable t is given t=100 than sizeof(t)= 1?
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      1. USdato datuashvili
    2. CObut it show me 4 why
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      1. USdato datuashvili
    3. CO`sizeof` has nothing to do with the *value* of the variable. It has to do with the type. It is telling you that on your system, all variables of type `int` require 4 bytes of storage.
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      1. USTyler McHenry
 

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