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copied!<p>If you mean "how large is an int" then <code>sizeof(int)</code> is the answer.</p> <p>If you mean "how small a type can I use to store values of this magnitude" then that's a bit more complex. If you already have the value in integer form, then presumably it fits in 4, 3, 2, or 1 bytes. For unsigned values, if it's 16777216 or over you need 4 bytes, 65536-16777216 requires 3 bytes, 256-65535 needs 2, and 0-255 fits in 1 byte. The formula for this comes from the fact that each byte can hold 8 bits, and each bit holds 2 digits, so 1 byte holds 2^8 values, ie. 256 (but starting at 0, so 0-255). 2 bytes therefore holds 2^16 values, ie. 65536, and so on.</p> <p>You can generalise that beyond the normal 4 bytes used for a typical int if you like. If you need to accommodate signed integers as well as unsigned, bear in mind that 1 bit is effectively used to store whether it is positive or negative, so the magnitude is 1 power of 2 less.</p> <p>You can calculate the number of bits you need iteratively from an integer by dividing it by two and discarding the remainder. Each division you can make and still have a non-zero value means you have one more bit of data in use - and every 8 bits you're using means 1 byte.</p> <p>A quick way of calculating this is to use the shift right function and compare the result against zero.</p> <pre><code>int value = 23534; // or whatever int bits = 0; while (value) { value >> 1; ++bits; } std::cout << "Bits used = " << bits << std::endl; std::cout << "Bytes used = " << (bits / 8) + 1 << std::endl; </code></pre>

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