NumberFormatException - with String of an Hex Value
I got a stream of byte this stream is a bytearray ( byte stream ) now i want parse this byte first to an hex value, e.g. 80 = "0x50" then transform this to an int
But for some values like 0xB8 i got an `java.lang.NumberFormatException: For input string: "0xB8"
How could i manage this exception? Maybe with an other type of parse or different type data?
byte k = stream[i];
int b = 0;
try {
b = Integer.parseInt(String.format("0x%02X",k),16);
} catch (NumberFormatException e) {
e.printStackTrace();
}
java exception
asked Nov 15 '18 at 17:21
Gianluca BenucciGianluca Benucci
779
add a comment |
I got a stream of byte this stream is a bytearray ( byte stream ) now i want parse this byte first to an hex value, e.g. 80 = "0x50" then transform this to an int
But for some values like 0xB8 i got an `java.lang.NumberFormatException: For input string: "0xB8"
How could i manage this exception? Maybe with an other type of parse or different type data?
byte k = stream[i];
int b = 0;
try {
b = Integer.parseInt(String.format("0x%02X",k),16);
} catch (NumberFormatException e) {
e.printStackTrace();
}
java exception
asked Nov 15 '18 at 17:21
Gianluca BenucciGianluca Benucci
779
Check this stackoverflow.com/questions/11377944/…
– mettleap
Nov 15 '18 at 17:31
add a comment |
I got a stream of byte this stream is a bytearray ( byte stream ) now i want parse this byte first to an hex value, e.g. 80 = "0x50" then transform this to an int
But for some values like 0xB8 i got an `java.lang.NumberFormatException: For input string: "0xB8"
How could i manage this exception? Maybe with an other type of parse or different type data?
byte k = stream[i];
int b = 0;
try {
b = Integer.parseInt(String.format("0x%02X",k),16);
} catch (NumberFormatException e) {
e.printStackTrace();
}
java exception
asked Nov 15 '18 at 17:21
Gianluca BenucciGianluca Benucci
779
I got a stream of byte this stream is a bytearray ( byte stream ) now i want parse this byte first to an hex value, e.g. 80 = "0x50" then transform this to an int
But for some values like 0xB8 i got an `java.lang.NumberFormatException: For input string: "0xB8"
How could i manage this exception? Maybe with an other type of parse or different type data?
byte k = stream[i];
int b = 0;
try {
b = Integer.parseInt(String.format("0x%02X",k),16);
} catch (NumberFormatException e) {
e.printStackTrace();
}
java exception
java exception
asked Nov 15 '18 at 17:21
Gianluca BenucciGianluca Benucci
779
asked Nov 15 '18 at 17:21
Gianluca BenucciGianluca Benucci
779
edited Nov 16 '18 at 7:45
Gianluca Benucci
asked Nov 15 '18 at 17:21
Gianluca BenucciGianluca Benucci
779
asked Nov 15 '18 at 17:21
Gianluca BenucciGianluca Benucci
779
asked Nov 15 '18 at 17:21
Gianluca BenucciGianluca Benucci
779
779
Check this stackoverflow.com/questions/11377944/…
– mettleap
Nov 15 '18 at 17:31
add a comment |
Check this stackoverflow.com/questions/11377944/…
– mettleap
Nov 15 '18 at 17:31
Check this stackoverflow.com/questions/11377944/…
– mettleap
Nov 15 '18 at 17:31
Check this stackoverflow.com/questions/11377944/…
– mettleap
Nov 15 '18 at 17:31
add a comment |
3 Answers
3
active
oldest
votes
You must use overloaded parseInt method with radix.
For hex value an example:
Integer.parseInt("0x84B",16);
answered Nov 15 '18 at 17:42
TixTix
616
Always get same error. I've updated my code.
– Gianluca Benucci
Nov 16 '18 at 7:45
@GianlucaBenucci Check what it returns String.format("0x%02X",k), it can return a non-hex string.
– Tix
Nov 16 '18 at 7:58
1
You'll need to remove the leading0xfor this to work.Integer.parseInt("84B",16)returns 2123, but with0x84Byou get aNumberFormatException.
– jsheeran
Nov 16 '18 at 8:03
add a comment |
Try this:
Integer.parseInt(String.format("0b%02X",k),16);
Binary values are written using "0bxx" syntax
answered Nov 16 '18 at 7:59
NickNick
5118
add a comment |
I've resolved my problem with this question
byte k = stream[i];
int b = 0;
try {
String hexStringNumber = String.format("0x%02X",k);
b = Integer.decode(hexStringNumber);
} catch (NumberFormatException e) {
e.printStackTrace();
}
answered Nov 16 '18 at 8:06
Gianluca BenucciGianluca Benucci
779
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
You must use overloaded parseInt method with radix.
For hex value an example:
Integer.parseInt("0x84B",16);
answered Nov 15 '18 at 17:42
TixTix
616
Always get same error. I've updated my code.
– Gianluca Benucci
Nov 16 '18 at 7:45
@GianlucaBenucci Check what it returns String.format("0x%02X",k), it can return a non-hex string.
– Tix
Nov 16 '18 at 7:58
1
You'll need to remove the leading0xfor this to work.Integer.parseInt("84B",16)returns 2123, but with0x84Byou get aNumberFormatException.
– jsheeran
Nov 16 '18 at 8:03
add a comment |
You must use overloaded parseInt method with radix.
For hex value an example:
Integer.parseInt("0x84B",16);
answered Nov 15 '18 at 17:42
TixTix
616
Always get same error. I've updated my code.
– Gianluca Benucci
Nov 16 '18 at 7:45
@GianlucaBenucci Check what it returns String.format("0x%02X",k), it can return a non-hex string.
– Tix
Nov 16 '18 at 7:58
1
You'll need to remove the leading0xfor this to work.Integer.parseInt("84B",16)returns 2123, but with0x84Byou get aNumberFormatException.
– jsheeran
Nov 16 '18 at 8:03
add a comment |
You must use overloaded parseInt method with radix.
For hex value an example:
Integer.parseInt("0x84B",16);
answered Nov 15 '18 at 17:42
TixTix
616
You must use overloaded parseInt method with radix.
For hex value an example:
Integer.parseInt("0x84B",16);
answered Nov 15 '18 at 17:42
TixTix
616
answered Nov 15 '18 at 17:42
TixTix
616
answered Nov 15 '18 at 17:42
TixTix
616
answered Nov 15 '18 at 17:42
TixTix
616
616
Always get same error. I've updated my code.
– Gianluca Benucci
Nov 16 '18 at 7:45
@GianlucaBenucci Check what it returns String.format("0x%02X",k), it can return a non-hex string.
– Tix
Nov 16 '18 at 7:58
1
You'll need to remove the leading0xfor this to work.Integer.parseInt("84B",16)returns 2123, but with0x84Byou get aNumberFormatException.
– jsheeran
Nov 16 '18 at 8:03
add a comment |
Always get same error. I've updated my code.
– Gianluca Benucci
Nov 16 '18 at 7:45
@GianlucaBenucci Check what it returns String.format("0x%02X",k), it can return a non-hex string.
– Tix
Nov 16 '18 at 7:58
1
You'll need to remove the leading0xfor this to work.Integer.parseInt("84B",16)returns 2123, but with0x84Byou get aNumberFormatException.
– jsheeran
Nov 16 '18 at 8:03
Always get same error. I've updated my code.
– Gianluca Benucci
Nov 16 '18 at 7:45
Always get same error. I've updated my code.
– Gianluca Benucci
Nov 16 '18 at 7:45
@GianlucaBenucci Check what it returns String.format("0x%02X",k), it can return a non-hex string.
– Tix
Nov 16 '18 at 7:58
@GianlucaBenucci Check what it returns String.format("0x%02X",k), it can return a non-hex string.
– Tix
Nov 16 '18 at 7:58
1
1
You'll need to remove the leading
0x for this to work. Integer.parseInt("84B",16) returns 2123, but with 0x84B you get a NumberFormatException.– jsheeran
Nov 16 '18 at 8:03
You'll need to remove the leading
0x for this to work. Integer.parseInt("84B",16) returns 2123, but with 0x84B you get a NumberFormatException.– jsheeran
Nov 16 '18 at 8:03
add a comment |
Try this:
Integer.parseInt(String.format("0b%02X",k),16);
Binary values are written using "0bxx" syntax
answered Nov 16 '18 at 7:59
NickNick
5118
add a comment |
Try this:
Integer.parseInt(String.format("0b%02X",k),16);
Binary values are written using "0bxx" syntax
answered Nov 16 '18 at 7:59
NickNick
5118
add a comment |
Try this:
Integer.parseInt(String.format("0b%02X",k),16);
Binary values are written using "0bxx" syntax
answered Nov 16 '18 at 7:59
NickNick
5118
Try this:
Integer.parseInt(String.format("0b%02X",k),16);
Binary values are written using "0bxx" syntax
answered Nov 16 '18 at 7:59
NickNick
5118
answered Nov 16 '18 at 7:59
NickNick
5118
answered Nov 16 '18 at 7:59
NickNick
5118
answered Nov 16 '18 at 7:59
NickNick
5118
5118
add a comment |
add a comment |
I've resolved my problem with this question
byte k = stream[i];
int b = 0;
try {
String hexStringNumber = String.format("0x%02X",k);
b = Integer.decode(hexStringNumber);
} catch (NumberFormatException e) {
e.printStackTrace();
}
answered Nov 16 '18 at 8:06
Gianluca BenucciGianluca Benucci
779
add a comment |
I've resolved my problem with this question
byte k = stream[i];
int b = 0;
try {
String hexStringNumber = String.format("0x%02X",k);
b = Integer.decode(hexStringNumber);
} catch (NumberFormatException e) {
e.printStackTrace();
}
answered Nov 16 '18 at 8:06
Gianluca BenucciGianluca Benucci
779
add a comment |
I've resolved my problem with this question
byte k = stream[i];
int b = 0;
try {
String hexStringNumber = String.format("0x%02X",k);
b = Integer.decode(hexStringNumber);
} catch (NumberFormatException e) {
e.printStackTrace();
}
answered Nov 16 '18 at 8:06
Gianluca BenucciGianluca Benucci
779
I've resolved my problem with this question
byte k = stream[i];
int b = 0;
try {
String hexStringNumber = String.format("0x%02X",k);
b = Integer.decode(hexStringNumber);
} catch (NumberFormatException e) {
e.printStackTrace();
}
answered Nov 16 '18 at 8:06
Gianluca BenucciGianluca Benucci
779
answered Nov 16 '18 at 8:06
Gianluca BenucciGianluca Benucci
779
answered Nov 16 '18 at 8:06
Gianluca BenucciGianluca Benucci
779
answered Nov 16 '18 at 8:06
Gianluca BenucciGianluca Benucci
779
779
add a comment |
add a comment |
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Check this stackoverflow.com/questions/11377944/…
– mettleap
Nov 15 '18 at 17:31