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This group of errors results from code that is syntactically correct but violates the semantics of Java, in particular the rules relating to type checking.
An instance variable declared in a class has a default initial value. A class is a template that is used to create or instantiate instances called objects . Since memory is allocated separately for each object, these variables are called instance variables . However, a local variable declared within a method does not, so you are required to initialize it before use, either in its declaration or in anassignment statement:
void m(int n) { // n is initialized from the actual parameter
int i, j; i = 2; // i is initialized by the assignment
int k = 1; // k is initialized in its declaration if (i == n) // OK
k = j; // Error, j is not initialized else
j = k;}
The variable must be initialized on every path from declaration to use even if the semantics of the program ensure that the path cannot be taken:
void m(int n) {
int i; if (n == n) // Always true
i = n; else
n = i; // Error, although never executed!!}
Eclipse: The local variable ... may not have been initialized
Note: If the expression in the
if -statement can be computed at compile-time:
if (true) // OK
if ('n' == 'n') // OK
the error will not occur.
This error message is very common and results from an incompatibility between a method call and the method's declaration:
void m(int i) { ... }
m(5.5); // Error, the literal is of type double, not int
Check the declaration and call carefully. You can also get the error message cannot find symbol if the declaration of the method is in the API. The Application Programming Interface (API) describes how to use the library of classes supplied as part of the Java system. Byextension, it is also used as a name for the library itself.
Eclipse: The method ... in the type ... is not applicable for the arguments ...
Operators are only defined for certain types, although implicit type conversion is allowed between certain numeric types:
int a = 5;
boolean b = true;int c = a + b; // Error, can't add a boolean value
double d = a + 1.4; // OK, int is implicitly converted to double
Eclipse: The operator + is undefined for the argument type(s) int, boolean
This error arises when trying to assign
a value of higher precision to a variable of lower precision without anexplicit type cast. Surprisingly, perhaps, floating point literals are of
type
double and you will get this message if you try to assign one to a variable
of type
float :
float sum = 0.0; // Error, literal is not of type float
The correct way to do this is to use a literal of type
float or
an explicit type cast:
float sum = 0.0f; // OK
float sum = (float) 0.0; // OK
Eclipse: Type mismatch: cannot convert from double to float
Java checks that the type of an expression is compatible with the type of the variable in an assignment statement and this error will result ifthey are incompatible:
boolean b = true;
int a = b; // Error, can't assign boolean to int
Eclipse: Type mismatch: cannot convert from boolean to int
Important note
In the C language it is (unfortunately!) legal to write
if (a = b) using the assignment operator instead of
if (a == b) using the equality operator.
The meaning is to execute the assignment, convert the value to an integerand use its value to make the decision for the
if -statement (zero is false
and non-zero is true). In Java, the assignment is legal and results in a value,but (unless
a is of type
boolean !) this error message will result
because the expression in an
if -statement must be of type
boolean , not
int . Writing
== instead of
= becomes a simple compile-time error in
Java, whereas in C this error leads to runtime errors that are extremelydifficult to find.
Not every type conversion is legal:
boolean b = true;
int x = (int) b; // Error, can't convert boolean to int
Eclipse: Type mismatch: cannot convert from ... to ...
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