如何将其中具有十六进制数据的char（或Byte）转换为整数（十进制值）

Question

I have

char tem;

Its value is shown as blow:

Printing description of tem:
(char) tem = '\xd1'

which should equals to 209 in decimal.

My question is how can I implement this conversion programmatically? That is I want to get a NSInteger that equals 209 in this case.

Answer 1

Maybe there's something I'm overlooking here, but given that both char and NSInteger are integral types, can't you just do

char tem = '\xd1';
NSInteger i = tem;

? Or perhaps, to avoid surprises from sign extension,

NSInteger i = tem & 0xff;

Answer 2

A char variable actually is a 8-bit integer. You don't have to convert it to a NSInteger to get its decimal value. Just explicitly tell the compiler to interpret it as an unsigned 8-bit integer, uint8_t :

char theChar = '\xd1';
NSLog(@"decimal: %d", (uint8_t)theChar);    //prints 'decimal: 209'

To convert it to a NSInteger :

NSInteger decimal = (uint8_t)theChar;

Answer 3

If yours char in ANSCII, do this:

char a = '3';//example char
int i = (int)(a - '0');