Bitwise operation not concatenating with string in print() in Java
17
This code
int a = 6;
System.out.print("The result is " + a*a);
works just fine, but this one
int a = 6;
System.out.print("The result is " + a^a);
produces an exception:
Exception in thread "main" java.lang.RuntimeException: Uncompilable
source code - Erroneous tree type: at
pkg1.pkg4.taking.input.TakingInput.main(TakingInput.java:11)
Why so?
The question arose when I was trying to print the results of several bitwise operations in one swoop, like so:
System.out.print(a&b + "n" + a|b + "n" + a^b);
I looked up the description of the print() method and several topics on bitwise operators and printing to the console on SO including the recommended topics when composing the question, but couldn't find an answer.
java printing bitwise-operators operator-precedence
asked Dec 4 at 5:34
John Allison
382216
add a comment |
17
This code
int a = 6;
System.out.print("The result is " + a*a);
works just fine, but this one
int a = 6;
System.out.print("The result is " + a^a);
produces an exception:
Exception in thread "main" java.lang.RuntimeException: Uncompilable
source code - Erroneous tree type: at
pkg1.pkg4.taking.input.TakingInput.main(TakingInput.java:11)
Why so?
The question arose when I was trying to print the results of several bitwise operations in one swoop, like so:
System.out.print(a&b + "n" + a|b + "n" + a^b);
I looked up the description of the print() method and several topics on bitwise operators and printing to the console on SO including the recommended topics when composing the question, but couldn't find an answer.
java printing bitwise-operators operator-precedence
asked Dec 4 at 5:34
John Allison
382216
I wonder what Java version are you compiling this with, it doesn't compile with java-11 for sure, forgetRuntimeException!
– nullpointer
Dec 4 at 5:46
That's what the error says:uncompilable source code. Another problem was I was using single quotes instead of double quotes withn.
– John Allison
Dec 4 at 5:50
add a comment |
17
17
17
1
This code
int a = 6;
System.out.print("The result is " + a*a);
works just fine, but this one
int a = 6;
System.out.print("The result is " + a^a);
produces an exception:
Exception in thread "main" java.lang.RuntimeException: Uncompilable
source code - Erroneous tree type: at
pkg1.pkg4.taking.input.TakingInput.main(TakingInput.java:11)
Why so?
The question arose when I was trying to print the results of several bitwise operations in one swoop, like so:
System.out.print(a&b + "n" + a|b + "n" + a^b);
I looked up the description of the print() method and several topics on bitwise operators and printing to the console on SO including the recommended topics when composing the question, but couldn't find an answer.
java printing bitwise-operators operator-precedence
asked Dec 4 at 5:34
John Allison
382216
This code
int a = 6;
System.out.print("The result is " + a*a);
works just fine, but this one
int a = 6;
System.out.print("The result is " + a^a);
produces an exception:
Exception in thread "main" java.lang.RuntimeException: Uncompilable
source code - Erroneous tree type: at
pkg1.pkg4.taking.input.TakingInput.main(TakingInput.java:11)
Why so?
The question arose when I was trying to print the results of several bitwise operations in one swoop, like so:
System.out.print(a&b + "n" + a|b + "n" + a^b);
I looked up the description of the print() method and several topics on bitwise operators and printing to the console on SO including the recommended topics when composing the question, but couldn't find an answer.
java printing bitwise-operators operator-precedence
java printing bitwise-operators operator-precedence
asked Dec 4 at 5:34
John Allison
382216
asked Dec 4 at 5:34
John Allison
382216
edited Dec 4 at 5:54
asked Dec 4 at 5:34
John Allison
382216
asked Dec 4 at 5:34
John Allison
382216
asked Dec 4 at 5:34
John Allison
382216
382216
I wonder what Java version are you compiling this with, it doesn't compile with java-11 for sure, forgetRuntimeException!
– nullpointer
Dec 4 at 5:46
That's what the error says:uncompilable source code. Another problem was I was using single quotes instead of double quotes withn.
– John Allison
Dec 4 at 5:50
add a comment |
I wonder what Java version are you compiling this with, it doesn't compile with java-11 for sure, forgetRuntimeException!
– nullpointer
Dec 4 at 5:46
That's what the error says:uncompilable source code. Another problem was I was using single quotes instead of double quotes withn.
– John Allison
Dec 4 at 5:50
I wonder what Java version are you compiling this with, it doesn't compile with java-11 for sure, forget
RuntimeException!– nullpointer
Dec 4 at 5:46
I wonder what Java version are you compiling this with, it doesn't compile with java-11 for sure, forget
RuntimeException!– nullpointer
Dec 4 at 5:46
That's what the error says:
uncompilable source code. Another problem was I was using single quotes instead of double quotes with n.– John Allison
Dec 4 at 5:50
That's what the error says:
uncompilable source code. Another problem was I was using single quotes instead of double quotes with n.– John Allison
Dec 4 at 5:50
add a comment |
1 Answer
1
active
oldest
votes
22
This is because the + has higher precedence than the ^ so it compiles to:
("The result is " + a) ^ a
Which obviously will not work. Put parenthesis around it:
System.out.print("The result is " + (a^a));
Or as Holger mentioned, you can eliminate this problem by using printf:
System.out.printf("The result is %d", a^a);
answered Dec 4 at 5:37
GBlodgett
9,10641632
@JohnAllison If you used an IDE, you would probably know about this before you actually compile and run it.
– Jai
Dec 4 at 5:43
4
@Jai The OP surely used an IDE as without, he wouldn’t even try to run a program which didn’t compile. Besides that, here, the original Java designers are to blame. Using+for string concatenation was a big mistake. It not only has an unfortunate precedence, the usual expectation of commutativity for an addition does not hold when at least one of the operands of+is a string. But even Java itself has an alternative,System.out.printf("The result is %d", a^a);…
– Holger
Dec 4 at 7:44
@Holger Hmm, my Eclipse is showing a compile time error when I do this. I guess it's the same whichever way - you can always miss the error messages and run the application...
– Jai
Dec 4 at 7:55
@Jai the standardjavaccommand line tool does not generate.classfiles when there are compile errors, hence, it is impossible to run the application afterwards. It is a special (dubious imho) option of IDEs to still generate classes on compiler errors, which will throw an exception at runtime then. You can turn this feature off and never miss an error, but the default is on for most IDEs.
– Holger
Dec 4 at 8:06
add a comment |
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oldest
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oldest
votes
22
This is because the + has higher precedence than the ^ so it compiles to:
("The result is " + a) ^ a
Which obviously will not work. Put parenthesis around it:
System.out.print("The result is " + (a^a));
Or as Holger mentioned, you can eliminate this problem by using printf:
System.out.printf("The result is %d", a^a);
answered Dec 4 at 5:37
GBlodgett
9,10641632
@JohnAllison If you used an IDE, you would probably know about this before you actually compile and run it.
– Jai
Dec 4 at 5:43
4
@Jai The OP surely used an IDE as without, he wouldn’t even try to run a program which didn’t compile. Besides that, here, the original Java designers are to blame. Using+for string concatenation was a big mistake. It not only has an unfortunate precedence, the usual expectation of commutativity for an addition does not hold when at least one of the operands of+is a string. But even Java itself has an alternative,System.out.printf("The result is %d", a^a);…
– Holger
Dec 4 at 7:44
@Holger Hmm, my Eclipse is showing a compile time error when I do this. I guess it's the same whichever way - you can always miss the error messages and run the application...
– Jai
Dec 4 at 7:55
@Jai the standardjavaccommand line tool does not generate.classfiles when there are compile errors, hence, it is impossible to run the application afterwards. It is a special (dubious imho) option of IDEs to still generate classes on compiler errors, which will throw an exception at runtime then. You can turn this feature off and never miss an error, but the default is on for most IDEs.
– Holger
Dec 4 at 8:06
add a comment |
22
This is because the + has higher precedence than the ^ so it compiles to:
("The result is " + a) ^ a
Which obviously will not work. Put parenthesis around it:
System.out.print("The result is " + (a^a));
Or as Holger mentioned, you can eliminate this problem by using printf:
System.out.printf("The result is %d", a^a);
answered Dec 4 at 5:37
GBlodgett
9,10641632
@JohnAllison If you used an IDE, you would probably know about this before you actually compile and run it.
– Jai
Dec 4 at 5:43
4
@Jai The OP surely used an IDE as without, he wouldn’t even try to run a program which didn’t compile. Besides that, here, the original Java designers are to blame. Using+for string concatenation was a big mistake. It not only has an unfortunate precedence, the usual expectation of commutativity for an addition does not hold when at least one of the operands of+is a string. But even Java itself has an alternative,System.out.printf("The result is %d", a^a);…
– Holger
Dec 4 at 7:44
@Holger Hmm, my Eclipse is showing a compile time error when I do this. I guess it's the same whichever way - you can always miss the error messages and run the application...
– Jai
Dec 4 at 7:55
@Jai the standardjavaccommand line tool does not generate.classfiles when there are compile errors, hence, it is impossible to run the application afterwards. It is a special (dubious imho) option of IDEs to still generate classes on compiler errors, which will throw an exception at runtime then. You can turn this feature off and never miss an error, but the default is on for most IDEs.
– Holger
Dec 4 at 8:06
add a comment |
22
22
22
This is because the + has higher precedence than the ^ so it compiles to:
("The result is " + a) ^ a
Which obviously will not work. Put parenthesis around it:
System.out.print("The result is " + (a^a));
Or as Holger mentioned, you can eliminate this problem by using printf:
System.out.printf("The result is %d", a^a);
answered Dec 4 at 5:37
GBlodgett
9,10641632
This is because the + has higher precedence than the ^ so it compiles to:
("The result is " + a) ^ a
Which obviously will not work. Put parenthesis around it:
System.out.print("The result is " + (a^a));
Or as Holger mentioned, you can eliminate this problem by using printf:
System.out.printf("The result is %d", a^a);
answered Dec 4 at 5:37
GBlodgett
9,10641632
edited Dec 5 at 0:05
answered Dec 4 at 5:37
GBlodgett
9,10641632
answered Dec 4 at 5:37
GBlodgett
9,10641632
answered Dec 4 at 5:37
GBlodgett
9,10641632
9,10641632
@JohnAllison If you used an IDE, you would probably know about this before you actually compile and run it.
– Jai
Dec 4 at 5:43
4
@Jai The OP surely used an IDE as without, he wouldn’t even try to run a program which didn’t compile. Besides that, here, the original Java designers are to blame. Using+for string concatenation was a big mistake. It not only has an unfortunate precedence, the usual expectation of commutativity for an addition does not hold when at least one of the operands of+is a string. But even Java itself has an alternative,System.out.printf("The result is %d", a^a);…
– Holger
Dec 4 at 7:44
@Holger Hmm, my Eclipse is showing a compile time error when I do this. I guess it's the same whichever way - you can always miss the error messages and run the application...
– Jai
Dec 4 at 7:55
@Jai the standardjavaccommand line tool does not generate.classfiles when there are compile errors, hence, it is impossible to run the application afterwards. It is a special (dubious imho) option of IDEs to still generate classes on compiler errors, which will throw an exception at runtime then. You can turn this feature off and never miss an error, but the default is on for most IDEs.
– Holger
Dec 4 at 8:06
add a comment |
@JohnAllison If you used an IDE, you would probably know about this before you actually compile and run it.
– Jai
Dec 4 at 5:43
4
@Jai The OP surely used an IDE as without, he wouldn’t even try to run a program which didn’t compile. Besides that, here, the original Java designers are to blame. Using+for string concatenation was a big mistake. It not only has an unfortunate precedence, the usual expectation of commutativity for an addition does not hold when at least one of the operands of+is a string. But even Java itself has an alternative,System.out.printf("The result is %d", a^a);…
– Holger
Dec 4 at 7:44
@Holger Hmm, my Eclipse is showing a compile time error when I do this. I guess it's the same whichever way - you can always miss the error messages and run the application...
– Jai
Dec 4 at 7:55
@Jai the standardjavaccommand line tool does not generate.classfiles when there are compile errors, hence, it is impossible to run the application afterwards. It is a special (dubious imho) option of IDEs to still generate classes on compiler errors, which will throw an exception at runtime then. You can turn this feature off and never miss an error, but the default is on for most IDEs.
– Holger
Dec 4 at 8:06
@JohnAllison If you used an IDE, you would probably know about this before you actually compile and run it.
– Jai
Dec 4 at 5:43
@JohnAllison If you used an IDE, you would probably know about this before you actually compile and run it.
– Jai
Dec 4 at 5:43
4
4
@Jai The OP surely used an IDE as without, he wouldn’t even try to run a program which didn’t compile. Besides that, here, the original Java designers are to blame. Using
+ for string concatenation was a big mistake. It not only has an unfortunate precedence, the usual expectation of commutativity for an addition does not hold when at least one of the operands of + is a string. But even Java itself has an alternative, System.out.printf("The result is %d", a^a);…– Holger
Dec 4 at 7:44
@Jai The OP surely used an IDE as without, he wouldn’t even try to run a program which didn’t compile. Besides that, here, the original Java designers are to blame. Using
+ for string concatenation was a big mistake. It not only has an unfortunate precedence, the usual expectation of commutativity for an addition does not hold when at least one of the operands of + is a string. But even Java itself has an alternative, System.out.printf("The result is %d", a^a);…– Holger
Dec 4 at 7:44
@Holger Hmm, my Eclipse is showing a compile time error when I do this. I guess it's the same whichever way - you can always miss the error messages and run the application...
– Jai
Dec 4 at 7:55
@Holger Hmm, my Eclipse is showing a compile time error when I do this. I guess it's the same whichever way - you can always miss the error messages and run the application...
– Jai
Dec 4 at 7:55
@Jai the standard
javac command line tool does not generate .class files when there are compile errors, hence, it is impossible to run the application afterwards. It is a special (dubious imho) option of IDEs to still generate classes on compiler errors, which will throw an exception at runtime then. You can turn this feature off and never miss an error, but the default is on for most IDEs.– Holger
Dec 4 at 8:06
@Jai the standard
javac command line tool does not generate .class files when there are compile errors, hence, it is impossible to run the application afterwards. It is a special (dubious imho) option of IDEs to still generate classes on compiler errors, which will throw an exception at runtime then. You can turn this feature off and never miss an error, but the default is on for most IDEs.– Holger
Dec 4 at 8:06
add a comment |
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I wonder what Java version are you compiling this with, it doesn't compile with java-11 for sure, forget
RuntimeException!– nullpointer
Dec 4 at 5:46
That's what the error says:
uncompilable source code. Another problem was I was using single quotes instead of double quotes withn.– John Allison
Dec 4 at 5:50