Mixing
Java ternary operator datatype conversion based on the first value?
I came across this thing which I havent noticed before.
Here is a normal expression
int a = 5;
System.out.println(((a < 5) ? 0 : 9));
so this just prints 9 as an int. Now If I put the first value String instead of an int 0
int a = 5;
System.out.println(((a < 5) ? "asd" : 9));
Here the value 9 is printed as a string and not as an int. To confirm this just try to add this with another integer
int a = 5;
System.out.println((((a < 5) ? 0 : 9) + 4) );
Now this results in 13 but if you change the first value to string instead of an int 0 it gives a compile error
"The operator + is undefined for the argument type(s) Object&Serializable&Comparable<?>, int".
I am confused with this compile error. What is actually behind this ? Thanks for explanation
java operators
asked Jan 1 at 14:03
Naveen KumarNaveen Kumar
506
add a comment |
I came across this thing which I havent noticed before.
Here is a normal expression
int a = 5;
System.out.println(((a < 5) ? 0 : 9));
so this just prints 9 as an int. Now If I put the first value String instead of an int 0
int a = 5;
System.out.println(((a < 5) ? "asd" : 9));
Here the value 9 is printed as a string and not as an int. To confirm this just try to add this with another integer
int a = 5;
System.out.println((((a < 5) ? 0 : 9) + 4) );
Now this results in 13 but if you change the first value to string instead of an int 0 it gives a compile error
"The operator + is undefined for the argument type(s) Object&Serializable&Comparable<?>, int".
I am confused with this compile error. What is actually behind this ? Thanks for explanation
java operators
asked Jan 1 at 14:03
Naveen KumarNaveen Kumar
506
2
This is fully-described in the language spec, and absolutely not worth reproducing here. TL;DR: the result of the conditional operator is the least upper bound of the second and third operands.
– Andy Turner
Jan 1 at 14:04
I didnt know this so was confused, anyways thanks for clarifying.
– Naveen Kumar
Jan 1 at 14:09
if you think this is confusing, try to work out the type ofObject result = true ? Integer.valueOf(0) : Double.valueOf(0);.
– Andy Turner
Jan 1 at 14:11
add a comment |
I came across this thing which I havent noticed before.
Here is a normal expression
int a = 5;
System.out.println(((a < 5) ? 0 : 9));
so this just prints 9 as an int. Now If I put the first value String instead of an int 0
int a = 5;
System.out.println(((a < 5) ? "asd" : 9));
Here the value 9 is printed as a string and not as an int. To confirm this just try to add this with another integer
int a = 5;
System.out.println((((a < 5) ? 0 : 9) + 4) );
Now this results in 13 but if you change the first value to string instead of an int 0 it gives a compile error
"The operator + is undefined for the argument type(s) Object&Serializable&Comparable<?>, int".
I am confused with this compile error. What is actually behind this ? Thanks for explanation
java operators
asked Jan 1 at 14:03
Naveen KumarNaveen Kumar
506
I came across this thing which I havent noticed before.
Here is a normal expression
int a = 5;
System.out.println(((a < 5) ? 0 : 9));
so this just prints 9 as an int. Now If I put the first value String instead of an int 0
int a = 5;
System.out.println(((a < 5) ? "asd" : 9));
Here the value 9 is printed as a string and not as an int. To confirm this just try to add this with another integer
int a = 5;
System.out.println((((a < 5) ? 0 : 9) + 4) );
Now this results in 13 but if you change the first value to string instead of an int 0 it gives a compile error
"The operator + is undefined for the argument type(s) Object&Serializable&Comparable<?>, int".
I am confused with this compile error. What is actually behind this ? Thanks for explanation
java operators
java operators
asked Jan 1 at 14:03
Naveen KumarNaveen Kumar
506
asked Jan 1 at 14:03
Naveen KumarNaveen Kumar
506
asked Jan 1 at 14:03
Naveen KumarNaveen Kumar
506
asked Jan 1 at 14:03
Naveen KumarNaveen Kumar
506
asked Jan 1 at 14:03
Naveen KumarNaveen Kumar
506
506
2
This is fully-described in the language spec, and absolutely not worth reproducing here. TL;DR: the result of the conditional operator is the least upper bound of the second and third operands.
– Andy Turner
Jan 1 at 14:04
I didnt know this so was confused, anyways thanks for clarifying.
– Naveen Kumar
Jan 1 at 14:09
if you think this is confusing, try to work out the type ofObject result = true ? Integer.valueOf(0) : Double.valueOf(0);.
– Andy Turner
Jan 1 at 14:11
add a comment |
2
This is fully-described in the language spec, and absolutely not worth reproducing here. TL;DR: the result of the conditional operator is the least upper bound of the second and third operands.
– Andy Turner
Jan 1 at 14:04
I didnt know this so was confused, anyways thanks for clarifying.
– Naveen Kumar
Jan 1 at 14:09
if you think this is confusing, try to work out the type ofObject result = true ? Integer.valueOf(0) : Double.valueOf(0);.
– Andy Turner
Jan 1 at 14:11
2
2
This is fully-described in the language spec, and absolutely not worth reproducing here. TL;DR: the result of the conditional operator is the least upper bound of the second and third operands.
– Andy Turner
Jan 1 at 14:04
This is fully-described in the language spec, and absolutely not worth reproducing here. TL;DR: the result of the conditional operator is the least upper bound of the second and third operands.
– Andy Turner
Jan 1 at 14:04
I didnt know this so was confused, anyways thanks for clarifying.
– Naveen Kumar
Jan 1 at 14:09
I didnt know this so was confused, anyways thanks for clarifying.
– Naveen Kumar
Jan 1 at 14:09
if you think this is confusing, try to work out the type of
Object result = true ? Integer.valueOf(0) : Double.valueOf(0);.– Andy Turner
Jan 1 at 14:11
if you think this is confusing, try to work out the type of
Object result = true ? Integer.valueOf(0) : Double.valueOf(0);.– Andy Turner
Jan 1 at 14:11
add a comment |
2 Answers
2
active
oldest
votes
The type of
(a < 5) ? "asd" : 9
is
Object&Serializable&Comparable<?>
You can see this in the compiler error later in the question. The int 9 is boxed to Integer and then the common type between that and String is found. So you are actually calling the println(Object) overload not println(String) or println(int). println(Object) calls toString on its argument.
If you try to apply + to a Object&Serializable&Comparable<?> then neither string concatenation or arithmetic is applicable, hence the compiler error.
answered Jan 1 at 14:39
Tom Hawtin - tacklineTom Hawtin - tackline
127k28183273
add a comment |
The type of the expression (a < 5) ? "asd" : 9 depends on the types of the 2nd and 3rd operands - "asd" and 9. The only common type to those two operands is Object (if the int is boxed to Integer).
Therefore the type of the expression is Object, Java has no + operator that accepts Object and int as operands. Hence the compilation error.
In
System.out.println((((a < 5) ? 0 : 9) + 4));
the type of the ternary conditional expression is int, and int and int are acceptable operands of the + operator.
edited Jan 1 at 14:10
Andy Turner
83.4k983141
answered Jan 1 at 14:05
EranEran
288k37467557
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The type of
(a < 5) ? "asd" : 9
is
Object&Serializable&Comparable<?>
You can see this in the compiler error later in the question. The int 9 is boxed to Integer and then the common type between that and String is found. So you are actually calling the println(Object) overload not println(String) or println(int). println(Object) calls toString on its argument.
If you try to apply + to a Object&Serializable&Comparable<?> then neither string concatenation or arithmetic is applicable, hence the compiler error.
answered Jan 1 at 14:39
Tom Hawtin - tacklineTom Hawtin - tackline
127k28183273
add a comment |
The type of
(a < 5) ? "asd" : 9
is
Object&Serializable&Comparable<?>
You can see this in the compiler error later in the question. The int 9 is boxed to Integer and then the common type between that and String is found. So you are actually calling the println(Object) overload not println(String) or println(int). println(Object) calls toString on its argument.
If you try to apply + to a Object&Serializable&Comparable<?> then neither string concatenation or arithmetic is applicable, hence the compiler error.
answered Jan 1 at 14:39
Tom Hawtin - tacklineTom Hawtin - tackline
127k28183273
add a comment |
The type of
(a < 5) ? "asd" : 9
is
Object&Serializable&Comparable<?>
You can see this in the compiler error later in the question. The int 9 is boxed to Integer and then the common type between that and String is found. So you are actually calling the println(Object) overload not println(String) or println(int). println(Object) calls toString on its argument.
If you try to apply + to a Object&Serializable&Comparable<?> then neither string concatenation or arithmetic is applicable, hence the compiler error.
answered Jan 1 at 14:39
Tom Hawtin - tacklineTom Hawtin - tackline
127k28183273
The type of
(a < 5) ? "asd" : 9
is
Object&Serializable&Comparable<?>
You can see this in the compiler error later in the question. The int 9 is boxed to Integer and then the common type between that and String is found. So you are actually calling the println(Object) overload not println(String) or println(int). println(Object) calls toString on its argument.
If you try to apply + to a Object&Serializable&Comparable<?> then neither string concatenation or arithmetic is applicable, hence the compiler error.
answered Jan 1 at 14:39
Tom Hawtin - tacklineTom Hawtin - tackline
127k28183273
answered Jan 1 at 14:39
Tom Hawtin - tacklineTom Hawtin - tackline
127k28183273
answered Jan 1 at 14:39
Tom Hawtin - tacklineTom Hawtin - tackline
127k28183273
answered Jan 1 at 14:39
Tom Hawtin - tacklineTom Hawtin - tackline
127k28183273
127k28183273
add a comment |
add a comment |
The type of the expression (a < 5) ? "asd" : 9 depends on the types of the 2nd and 3rd operands - "asd" and 9. The only common type to those two operands is Object (if the int is boxed to Integer).
Therefore the type of the expression is Object, Java has no + operator that accepts Object and int as operands. Hence the compilation error.
In
System.out.println((((a < 5) ? 0 : 9) + 4));
the type of the ternary conditional expression is int, and int and int are acceptable operands of the + operator.
edited Jan 1 at 14:10
Andy Turner
83.4k983141
answered Jan 1 at 14:05
EranEran
288k37467557
add a comment |
The type of the expression (a < 5) ? "asd" : 9 depends on the types of the 2nd and 3rd operands - "asd" and 9. The only common type to those two operands is Object (if the int is boxed to Integer).
Therefore the type of the expression is Object, Java has no + operator that accepts Object and int as operands. Hence the compilation error.
In
System.out.println((((a < 5) ? 0 : 9) + 4));
the type of the ternary conditional expression is int, and int and int are acceptable operands of the + operator.
edited Jan 1 at 14:10
Andy Turner
83.4k983141
answered Jan 1 at 14:05
EranEran
288k37467557
add a comment |
The type of the expression (a < 5) ? "asd" : 9 depends on the types of the 2nd and 3rd operands - "asd" and 9. The only common type to those two operands is Object (if the int is boxed to Integer).
Therefore the type of the expression is Object, Java has no + operator that accepts Object and int as operands. Hence the compilation error.
In
System.out.println((((a < 5) ? 0 : 9) + 4));
the type of the ternary conditional expression is int, and int and int are acceptable operands of the + operator.
edited Jan 1 at 14:10
Andy Turner
83.4k983141
answered Jan 1 at 14:05
EranEran
288k37467557
The type of the expression (a < 5) ? "asd" : 9 depends on the types of the 2nd and 3rd operands - "asd" and 9. The only common type to those two operands is Object (if the int is boxed to Integer).
Therefore the type of the expression is Object, Java has no + operator that accepts Object and int as operands. Hence the compilation error.
In
System.out.println((((a < 5) ? 0 : 9) + 4));
the type of the ternary conditional expression is int, and int and int are acceptable operands of the + operator.
edited Jan 1 at 14:10
Andy Turner
83.4k983141
answered Jan 1 at 14:05
EranEran
288k37467557
edited Jan 1 at 14:10
Andy Turner
83.4k983141
edited Jan 1 at 14:10
Andy Turner
83.4k983141
edited Jan 1 at 14:10
Andy Turner
83.4k983141
83.4k983141
answered Jan 1 at 14:05
EranEran
288k37467557
answered Jan 1 at 14:05
EranEran
288k37467557
answered Jan 1 at 14:05
EranEran
288k37467557
288k37467557
add a comment |
add a comment |
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2
This is fully-described in the language spec, and absolutely not worth reproducing here. TL;DR: the result of the conditional operator is the least upper bound of the second and third operands.
– Andy Turner
Jan 1 at 14:04
I didnt know this so was confused, anyways thanks for clarifying.
– Naveen Kumar
Jan 1 at 14:09
if you think this is confusing, try to work out the type of
Object result = true ? Integer.valueOf(0) : Double.valueOf(0);.– Andy Turner
Jan 1 at 14:11