Contents
V6044. Postfix increment/decrement is senseless because this variable is overwritten.
Analyzer detected a likely error that has to do with using a postfix increment or decrement in an assignment to the same variable.
Consider the following example:
int i = 5;
// Some code
i = i++;The increment operation here will not affect the expression result and the 'i' variable will be assigned the value 5 after executing this code.
This is explained by the fact that postfix increment and decrement operations are executed after evaluating the right operand of the assignment operator, while the result of the assignment is temporarily cached and is assigned later to the left part of the expression after the increment/decrement operation has executed. Therefore, the result of the increment/decrement is overwritten with the result of the whole expression.
As for the correct version of this code, it can look differently depending on the intended behavior.
This error may be a typo and the programmer unintentionally wrote variable 'i' twice in the assignment statement. Then the correct version could look as follows:
int i = 5;
// Some code
q = i++;Another scenario is that the programmer did not know that the postfix increment operator adds one to the value of the variable but returns its initial value. Then the assignment statement is redundant and the fixed code could look like this:
int i = 5;
// Some code
i++;This diagnostic is classified as:
|
Was this page helpful?
