If sizeof(int) == sizeof(long), then is INT_MIN == LONG_MIN && INT_MAX == LONG_MAX always true?

Question

If sizeof(int) == sizeof(long), then is INT_MIN == LONG_MIN && INT_MAX == LONG_MAX always true?

Is there any real existing cases demonstrating "not true"?

UPD. The similar question: Is there any hosted C implementations which have CHAR_BIT > 8?.

Answer 1

It need not be true. C11 6.2.6.2p2:

2. For signed integer types, the bits of the object representation shall be divided into three groups: value bits, padding bits, and the sign bit. There need not be any padding bits; signed char shall not have any padding bits.There shall be exactly one sign bit. Each bit that is a value bit shall have the same value as the same bit in the object representation of the corresponding unsigned type (if there are M value bits in the signed type and N in the unsigned type, then M <= N ). If the sign bit is zero, it shall not affect the resulting value. If the sign bit is one, the value shall be modified in one of the following ways:

   • the corresponding value with sign bit 0 is negated (sign and magnitude);
   • the sign bit has the value -(2^M) (two's complement);
   • the sign bit has the value -(2^M- 1) (ones' complement).

   Which of these applies is implementation-defined, as is whether the value with sign bit 1 and all value bits zero (for the first two), or with sign bit and all value bits 1 (for ones' complement), is a trap representation or a normal value. In the case of sign and magnitude and ones' complement, if this representation is a normal value it is called a negative zero.

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Now, the question is "is there any implementation that has different amount of padding bits" or, even as stark mentioned, different representations for different types of integers - it is very hard to prove that there is no such implementation currently in use. But I believe it is very unlikely that one would come across a system like this in real life.