Division by base and remainder method (conversion of integer part)

Multiplication base rounding method (conversion of decimal part)

Think in conjunction with the numerical representation formulas of r-base numbers instead of memorizing them by rote.

The sign bit of positive numbers is represented by 0, the sign bit of negative numbers is represented by 1, and the value bit remains unchanged.

Two ways to find the complement of a negative number

The sign bit is the same as the original code. When the true value is a positive number, the complement code is the same as the original code; when the true value is a negative number, the value bit of the complement code is the inversion of the true value bit.

The frameshift and complement of a true value differ by only one sign bit.

internal principles

Single bus structure operator

Dual bus structure operator

Three-bus structure operator

The object of arithmetic shift is a signed number, and the sign bit remains unchanged during the shift process.

The overflow judgment condition of the arithmetic shift operation is: the original code arithmetic left shift shifts by 1, or the complement arithmetic left shift shifts the bit with the opposite sign.

Logical shifts treat operands as unsigned numbers.

Shifting rules: When the logic is shifted to the left, the high bit is shifted and the low bit is added with 0; when the logic is shifted to the right, the low bit is shifted and the high bit is added with 0.

Overflow detection Overflow is only possible when adding two numbers with the same sign.

C language operation overflow example

Fixed-point multiplication

Fixed-point division operation

The bit values ​​are kept unchanged, only the way the bits are interpreted changes.

When a large word length variable is forced to type into a small word length variable, the system directly truncates the redundant high bits and assigns the low bits directly.

When converting from short word length to long word length, not only the corresponding bit values ​​must be equalized, but also the high-order part must be expanded.

The machine number of an int type variable i is 01 23 45 67H, its most significant byte MSB=01H, and its least significant byte LSB=67H.

border alignment

Borders not aligned

There is no boundary alignment problem for data of one byte length

Left-hand rule: When the highest digit of the mantissa of the operation result is not a significant digit, that is, in the form of ±0.0…0x…x, a left-hand rule is required. In left-hand rule, every time the mantissa is shifted one position to the left, the exponent code is decreased by 1 (when the base is 2). The left rule may need to be performed multiple times.

Right-hand calculation: When the significant digit of the mantissa of the operation result goes before the decimal point, right-hand calculation is required. Shift the mantissa right by one position and add 1 to the exponent (when the base is 2). When you need to adjust the right direction, you only need to do it once.

"The small steps should be aligned with the big steps"

If the mantissa is in the form of 1......., it is a standardized mantissa.

If the mantissa is in the form of 1×.…, it needs to be normalized to the right once, that is, the mantissa is shifted to the right by one position and the exponent code is increased by 1.

If the mantissa is in the form of 0..., you need to normalize it to the left, shift the mantissa to the left, and shift the mantissa to the left by 1 bit each time, and at the same time reduce the exponent by 1, until the mantissa is in the form of 1...

The last bit is always set to 1 method: as long as one of the bits lost due to the shift is 1, the lowest bit of the operation result is set to 1, regardless of whether the lowest bit was originally 0 or 1.

0-1 rounding method: When the highest digit of the missing digit is 1, add 1 to the last digit of the mantissa, similar to the rounding of decimal numbers.

Truncation method: directly intercept the required number of digits and discard all subsequent digits. This rounding process is the simplest.

The overflow of floating point operations can be judged by the overflow of the exponent code. For IEEE754 single-precision floating-point numbers, normalization overflow occurs when normalizing to the right causes the exponent to be all 1 (that is, 11111111, the true value is 128). Normalization underflow occurs when normalizing to the left causes the exponent to be all zeros.

Operations that may cause overflow

Index overflow and index underflow

double→float

float/double→int

float→double

int→float

int→double