declaration

In this tutorial there is written: If you redeclare a JavaScript variable, it will not lose its value. Why should I redeclare a variable? Is it practical in some situations? thank you It's nothing more than a reminder that if you do this: var x=5; var x; alert(x); Result will be 5. If you re-declare variable in some other languages for example - result will be undefined, or NaN, but not in javascript. An example of redeclaring a variable can be found in Google Analytics . When the JavaScript tracking code is initiated by the Google Analytics script, it declares or redeclares _gaq in this way:

How do I interpret complex declarations like: int * (* (*fp1) (int) ) [10]; ---> declaration 1 int *( *( *[5])())(); --------> declaration 2 Is there any rule that should be followed to understand the above declarations? Here is a great article about how to read complex declarations in C: http://www.codeproject.com/KB/cpp/complex_declarations.aspx It helped me a lot! Especially - You should read "The right rule" section. Here quote: int * (* (*fp1) (int) ) [10]; This can be interpreted as follows: Start from the variable name -------------------------- fp1 Nothing to right but ) so go left to

I was just going through some code on the Internet and found this: float * (*(*foo())[SIZE][SIZE])() How do I read this declaration? Is there a specific set of rules for reading such complex declarations? Kos I haven't done this in a while! Start with foo and go right. float * (*(* foo() )[SIZE][SIZE])() foo is a function with no arguments... Can't go right since there's a closing parenthesis. Go left: float * (*( * foo() )[SIZE][SIZE])() foo is a function with no arguments returning a pointer Can't go left further, so let's cross the parentheses and go right again float * (* (* foo()) [SIZE]

How exactly does the standard define that, for example, float (*(*(&e)[10])())[5] declares a variable of type "reference to array of 10 pointer to function of () returning pointer to array of 5 float "? Inspired by discussion with @DanNissenbaum Joseph Mansfield I refer to the C++11 standard in this post Declarations Declarations of the type we're concerned with are known as simple-declaration s in the grammar of C++, which are of one of the following two forms (§7/1): decl-specifier-seq opt init-declarator-list opt ; attribute-specifier-seq decl-specifier-seq opt init-declarator-list ; The

I've been looking at some code and am having a hard time working out variable declaration in php classes. Specifically it appears that the code i'm looking at doesn't declare the class variables before it uses them. Now this may be expected but I can't find any info that states that it is possible. So would you expect this: class Example { public function __construct() { $this->data = array(); $this->var = 'something'; } } to work? and does this create these variables on the class instance to be used hereafter? This works the same as a normal variable declaration would work: $foo = 'bar'; //

In the following code, why is the variable i not assigned the value 1 ? #include <stdio.h> int main(void) { int val = 0; switch (val) { int i = 1; //i is defined here case 0: printf("value: %d\n", i); break; default: printf("value: %d\n", i); break; } return 0; } When I compile, I get a warning about i not being initialized despite int i = 1; that clearly initializes it $ gcc -Wall test.c warning: ‘i’ is used uninitialized in this function [-Wuninitialized] printf("value %d\n", i); ^ If val = 0 , then the output is 0 . If val = 1 or anything else, then the output is also 0. Please explain to