Example 1

此示例在相同大小的数据类型之间转换：

TypeCast（UINT8（255），'INT8'）ANS = -1键盘（INT16（-1），'UINT16'）ANS = 65535

Example 2

Set X to a 1-by-3 vector of 32-bit integers, then cast it to an 8-bit integer type:

X = uint32([1 255 256]) X = 1 255 256

Running this on a little-endian system produces the following results. Each 32-bit value is divided up into four 8-bit segments:

Y = typecast(X, 'uint8') Y = 1 0 0 0 255 0 0 0 0 1 0 0

The third element ofX, 256, exceeds the 8 bits that it is being converted to inY（9）因此溢出到Y（10）：

Y(9:12) ans = 0 1 0 0

不te thatlength(Y)is equal to4.*length(X). Also note the difference between the output oftypecastversus that ofcast:

Z = cast(X, 'uint8') Z = 1 255 255

Example 3

此示例投用了较小的数据类型（uint8）进入一个更大的人（uint16). Displaying the numbers in hexadecimal format makes it easier to see just how the data is being rearranged:

十六进制格式X = uint8 ([44 55 66 77]) X = 2c 37 42 4d

The firsttypecastis done on a big-endian system. The four 8-bit segments of the input data are combined to produce two 16-bit segments:

Y = typecast(X, 'uint16') Y = 2c37 424d

The second is done on a little-endian system. Note the difference in byte ordering:

Y = typecast(X, 'uint16') Y = 372c 4d42

You can format the little-endian output into big-endian (and vice versa) using theswapbytesfunction:

y = swapbytes（typecast（x，'uint16'））y = 2c37 424d

Example 4

This example attempts to make a 32-bit value from a vector of three 8-bit values. MATLAB issues an error because there are an insufficient number of bytes in the input:

format hex typecast(uint8([120 86 52]), 'uint32') Error using typecast Too few input values to make output type.

Repeat the example, but with a vector of four 8-bit values, and it returns the expected answer:

TypeCast（UINT8（[120 86 52 18]），'UINT32'）ANS = 12345678