1,000,000,000,000,000,
Sorry, I really don't understand what values does the compiler give you...-32768..32767 is the value range of a 16-bit-long signed binary number (short in C)
8x8x8x8x8 equals 32,768. This can be calculated by multiplying 8 by itself five times, which is equivalent to 8 raised to the power of 5. In mathematical terms, this can be represented as 8^5 = 32,768.
(-32)3 = -32768(-32)3 = -32768(-32)3 = -32768(-32)3 = -32768
No and Yes... No because as such the two numbers are different and convention will define what the bit pattern should represent. However, with 16 bits, the UNSIGNED bit pattern for 32768 is the same as the SIGNED pattern for -32768 BUT the two numbers are being represented in different ways (ie in different number systems). If the binary representation is a SIGNED number then the top bit will be set if it is negative and with 16 bits the possible range of numbers is -32768 to +32767 and so it is IMPOSSIBLE to represent +32768. With an UNSIGNED number of 16 bits the top bit has no such special meaning and the range possible is 0 to +65535, so it is IMPOSSIBLE to represent -32768. ie it is IMPOSSIBLE with 16 bits to represent -32768 and +32768 in the same number system as they have the same bit pattern, BUT in different number systems the bit pattern can be used to represent the two numbers as the different number systems cannot represent BOTH -32678 and +32768.
The binary equivalent of Hexidecimal B3 = 1011 0011
The binary equivalent of the decimal number 47 is 101111.
it is itself in a binary form :) but if its in decimal form.. then its binary equivalent will be..(10011100011011) if it is in binry form.. its equivalent will be..(19)
If 1101 is a decimal number, then its binary equivalent is 10001001101. If 1101 is a binary number, then its decimal equivalent is 13.
100100 is 36 in binary.
51 in binary is... 110011
178 in binary is 10110010.
The binary equivalent of the decimal number 131 is 10000011.