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Int32 means you have 32 bits available to store your number. The highest bit is the sign-bit, this indicates if the number is positive or negative. So you have 2^31 bits for positive and negative numbers. With zero being a positive number you get the logical range of (mentioned before) +2147483647 to -2147483648. If you think that is to small ...
Thank you for your answer. But the reason a compiler has to deal with different int-sized is not about the year 2038 problem. A compiler just compiles as given. And the solution to the Year 2038 is to use a larger int size indeed, but neither got the 64-bit integer invented because that that, nor does it solve the problem on 32-bit hardware. –
A 32-bit unsigned int has a range from 0 to 4,294,967,295. 0 to 65535 would be a 16-bit unsigned. An unsigned long long (and, on a 64-bit implementation, possibly also ulong and possibly uint as well) have a range (at least) from 0 to 18,446,744,073,709,551,615 (2 64-1). In theory it could be greater than that, but at least for now that's rare ...
First off, neither IEEE-754-2008 nor -1985 have 16-bit floats; but it is a proposed addition with a 5-bit exponent and 10-bit fraction. IEE-754 uses a dedicated sign bit, so the positive and negative range is the same. Also, the fraction has an implied 1 in front, so you get an extra bit.
The most used size of an integer is 32 bits. The last bit is used to distinguish positive and negative numbers. If the last bit is NOT set, then the number is positive. Therefore, the maximal positive number is 0x7FFFFFFF = (1<<31)-1=2147483647 (the last bit is not set). For the negative numbers, two's complement notation is widely used.
the max 32-bit integer I believe is (2^31) which is 2,147,483,647. This is so that negative values can be stored as well (-2^31) being the 32 bit limit (this is what "signed" means). So any number higher than that, you can return 0 for the sake of your program. If the prompt asked you for "unsigned", the range would be 0 to 2^32 as you ...
The values for the float data type come from having 32 bits in total to represent the number which are allocated like this: 1 bit: sign bit. 8 bits: exponent p. 23 bits: mantissa. The exponent is stored as p + BIAS where the BIAS is 127, the mantissa has 23 bits and a 24th hidden bit that is assumed 1. This hidden bit is the most significant ...
Integers are 32 bits wide even on 64-bit platforms. The Java language specification states quite clearly (slightly paraphrased to make more readable, in my opinion): The integral types are byte, short, int, and long, whose values are respectively 8-bit, 16-bit, 32-bit and 64-bit signed two's-complement integers, and char, whose values are 16 ...
The right most bit (number) when set to 1 represents 2 0, the next bit 2 1, then 2 2 and so on until we get to the left most bit which if the number is unsigned represents 2 7. So the number represents 2 6 + 2 3 = 64 + 8 = 72, because the 4th bit from the right and the 7th bit right the left are set. If we set all values to 1: 11111111
You can use long, which is guaranteed to be at least 32-bits wide as a result of the minimum range requirements specified by the standard. If you would rather use the smallest integer type to fit a 32-bit number, then you can use preprocessor statements like the following with the macros defined in <limits.h>: