Is `long` guaranteed to be at least 32 bits?

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C++ uses the limits defined in the C standard (C++: 18.3.2 (c.limits), C: 5.2.4.2.1): 

LONG_MIN -2147483647 // -(2^31 - 1)
LONG_MAX +2147483647 //   2^31 - 1

So you are guaranteed that a long is at least 32 bits.

And if you want to follow the long circuitous route to whether LONG_MIN/LONG_MAX are representable by a long, you have to look at 18.3.1.2 (numeric.limits.members) in the C++ standard:

static constexpr T min() throw(); // Equivalent to CHAR_MIN, SHRT_MIN, FLT_MIN, DBL_MIN, etc.
static constexpr T max() throw(); // Equivalent to CHAR_MAX, SHRT_MAX, FLT_MAX, DBL_MAX, etc.

I moved the footnotes into the comment, so it’s not exactly what appears in the standard. But it basically implies that std::numeric_limits<long>::min()==LONG_MIN==(long)LONG_MIN and std::numeric_limits<long>::max()==LONG_MAX==(long)LONG_MAX.

So, even though the C++ standard does not specify the bitwise representation of (signed) negative numbers, it has to either be twos-complement and require 32-bits of storage in total, or it has an explicit sign bit which means that it has 32-bits of storage also.

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