What is the difference between int8(0x98), uint8(0x98), typecast(uint8(0x98), 'int8')

>> A=int8(0x98)

A =

int8

127

>> B=uint8(0x98)

B =

uint8

152

>> C=typecast(uint8(0x98), 'int8')

C =

int8

-104

I want 0x98 to be handled as signed Integer. Expectation is a negative values (-104) - two's complement.

Why is the calculation A=int8(0x98) leading to 127 and not to -104.

[Dyuman Joshi]

"Why is the calculation A=int8(0x98) leading to 127 and not to -104."

From the documentation of int8 - "Y = int8(X) converts the values in X to type int8. Values outside the range [-2^7,2^7-1] map to the nearest endpoint."

Additionally, from the documentation of typecast - "Y = typecast(X,type) converts the bit patterns of X to the data type specified by type without changing the underlying data."

format hex
A=int8(0x98)

A = int8
   7f

B=uint8(0x98)

B = uint8
   98

C=typecast(B, 'int8')

C = int8
   98

If you use cast, the result is same as the behaviour of the data type of the converted class.

C0=cast(B,'int8')

C0 = int8
   7f

[Henry]

All of above is understood.

The ind8 value range is specified to be [-2^7,2^7-1].

How do I get the negative values?

[Dyuman Joshi]

I don't understand, is using typecast() not satisfactory?

Or am I misunderstanding what you want to obtain? In that case, please specify.

A=int8(0x98)

A = int8 127

B=uint8(0x98)

B = uint8 152

C=typecast(B, 'int8')

C = int8 -104

Answer 1

[Steven Lord]

If you use the following syntax, you create a uint8 value and then cast that value to int8. Since the uint8 value is larger than intmax for int8, it saturates at intmax.

y1 = 0x98

y1 = uint8 152

z = int8(y1)

z = int8 127

Why uint8 not int8? From that documentation page I linked above: "By default, MATLAB stores the number as the smallest unsigned integer type that can accommodate it."

Instead tell MATLAB you want an int8 (a signed 8-bit integer) value from the start. To do this add the "s8" suffix to the hex value.

y2 = 0x98s8

y2 = int8 -104

The typecast function works because it converts not the value stored in the input but the bit pattern of the input to the new type.

format hex % Show the hex digits of the bit pattern
y1

y1 = uint8
   98

y2

y2 = int8
   98

Answer 2

[Swastik Sarkar]

In MATLAB, the number 0x98 will be represented as 127 in int8 due to overflow.

It will map to the nearest endpoint of the range. This is how MATLAB handles the overflow instead of setting the sign bit, like you would see in C.

However while typcasting a uint to int, it does not truncate and keeps the last bit. Hence you get -104

Read the description in this page to understand it better: int8

[Henry]

int8 explains the following:

Y = int8(X) converts the values in X to type int8. Values outside the range [-2^7,2^7-1] map to the nearest endpoint.

Integers: explains that matlab distiguishes between unsigned (uint) and signed (int).

To all what I experience is that Y = int8(X) ignores the MSB and maps to 2^7-1 in case x>=127.

Expected behaviour would have been that the MSB triggers the 2th-complement calculation to deliver the negative values down to -2^7.

I have the impression that in my version MATLAB R2022b the Y = int8(X) is not implemented as described in int8. Is this a known limitation/bug?

Please describe why not using the defined range [-2^7,2^7-1] is intended behavior.

[Dyuman Joshi]

"I have the impression that in my version MATLAB R2022b the Y = int8(X) is not implemented as described in int8."

I doubt there is a difference in the behaviour of int8 in your version and the one ran above in the comment I posted (R2023a).

"Expected behaviour would have been that the MSB triggers the 2th-complement calculation to deliver the negative values down to -2^7."

Please take a look at @Steven Lord's (informative) answer. It explains the behaviour of MATLAB w.r.t the issue of discussion.

[Steven Lord]

To all what I experience is that Y = int8(X) ignores the MSB and maps to 2^7-1 in case x>=127.

It sounds like you expected that conversion to wrap. It does not. It saturates at intmax.

To forestall what I expect will be your next question no, there is no way to change how MATLAB performs integer conversion or arithmetic to make it wrap instead of saturate.

[Henry]

my takeaaway:

y=int8(x) is not implemented as one might understand from the documentation int8: signed integer with number range [-2^7 .. 2^7-1]

• The number range is limited to [0 .. 2^7-1]
• The defined number range [-2^7 .. -1] cannot be achieved: value saturates at 2^7-1 instead
• negative sign is not provided

Several workarounds are present to generate a signed interger among which:

• y=typecast(uint8(x), 'int8')
• "s8"-suffix to HEX-number
• ...

I consider this topic closed.

[Steven Lord]

The number range is limited to [0 .. 2^7-1]

The range of values that can be represented in the int8 data type is:

[intmin('int8'), intmax('int8')]

ans = 1×2

-128 127

where 2^7-1 is:

2^7-1

ans = 127

So your statement is incorrect.

The defined number range [-2^7 .. -1] cannot be achieved: value saturates at 2^7-1 instead

Incorrect. If you try to cast a uint8 value to int8, anything greater than intmax of int8 becomes intmax of int8.

y = intmax('uint8')

y = uint8 255

z1 = int8(y)

z1 = int8 127

If you try to typecast a uint8 bit pattern to int8 then you can easily obtain a negative int8 value.

z2 = typecast(y, 'int8')

z2 = int8 -1

But you can certainly create negative int8 values directly

y = int8(-1)

y = int8 -1

negative sign is not provided

Incorrect, as shown above.