![]() In the case of 9007199254740993, which is larger than the largest number that can be represented using 64 bits, the computer cannot represent it exactly and therefore it is rounded to the nearest representable number. This means that the largest number that can be represented using 64 bits is 2^64 - 1, which is approximately 1.8 x 10^19. In most modern computers, numbers are represented using a fixed number of bits, typically 32 or 64 bits. Regarding your second question, the reason why a computer can display 2^1023 (8.98846567431158e+307) but not 9007199254740993 is due to the way that numbers are represented in the computer's memory. However, in practice, these approximations are usually accurate enough for most applications. This means that for numbers that require an infinite number of bits to represent exactly, such as irrational numbers like pi, the computer can only represent an approximation of the number. Modern computers use finite-precision arithmetic to perform computations, which means that they can only represent numbers with a limited number of bits. I cound not answer ur problem and look around for a bit and found this: I also recommend this explanation of floating point representation: Try clicking the 1s or 0s to see what happens when they take on different values. It currently represents 0.75 in 64-bit floating point. I find it helpful to see what happens when you change bits in a representation. If it goes above that, then the exponent can be increased instead. The key thing here is that the mantissa only ever needs to be between 1.0 and 2.0 (excluding 2). In this case, there is a 1 in the first bit, so this mantissa is 1.5. The goal is for those bits to be able to represent values between 0 and 1, which then is considered a mantissa between 1 and 2. According to the floating point standard, the first bit represents 1/2 (0.5), the second bit represents 1/4, the third bit represents 1/8, etc. ![]() 1022-1023 is -1, which is indeed the exponent. According to the floating point standard, the exponent is calculated by subtracting 1023 from that value. The next 11 bits represents the exponent -1: The first bit represents the sign, where 0 is positive. The floating point representation of 0.750 in binary needs to include the sign (positive/negative), the mantissa, and the exponent. The number -1 is called the "exponent" (as per normal math term). The number 1.5 is called either the "mantissa" or the "significand". Let's step through how the computer actually represents that number in floating point representation and see if that helps. The operands are casted as ul (unsigned long) as a result, the compiler has populated the upper 16-bit of the 32-bit storage space with 0s and not by the copies of the MSbit (1) of the lower 16-bit data (D2F0).From the author: Apologies for the confusing explanation. Note: Can you figure out the reason(s) why the codes of Post#2 have worked with your uint32-t data type? Serial.println(currentSeconds) //shows: 4294955760Ĭonclusion: The compiler/MCU has not done any mistake it has exactly given what you have asked for. You have executed the following codes to show the value of Fig-2, which is an unsigned value, and you have have nicely got 4294955760- the decimal value of 0xFFFFD2F0. And, accordingly, you have the following variable (Fig-2) in memory with data: FFFFD2F0.Ħ. ![]() The compiler fills it up by the copies of the sign bit (MSBit) of the lower 16-bit of data which is 1. What will go at the upper 16-bit of Fig-1? There must be going something as you have asked for a 32-bit storage space. Compiler has created the following storage space (Fig-1) for your data where, the lower 16-bit has been filled up by your data of Step-1.ĥ. Now, as a matter of experiment you have declared the following unsigned long (32-bit) variable to hold your data of Step-1. You know the maximum size and polarity of your number and accordingly, you have declared data type as the following which is fine. But this doesn't explain why it worked with uint16_t and not with 32, strange behaviour.Ģ. ![]()
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