Thread: redefinition of an int

The definition of a variable declaration is to have the type of variable name you need followed by a name for the variable of that type. By doing what you are showing in the above code, you are telling the compiler that you want to create a new instance of an integer variable and call it the same thing as something you have already defined. YOU CAN'T DO THIS.

Removing the keyword "int" from in front of the variable "x" in the second use leaves you with just "x = 5;" which in the syntax of C tells the compiler you are using a variable called "x" and storing the value 5 in that variable. This is how it should be in this case.

I think this whole issue is simply a problem you may be having with the syntax of C. Other languages have different syntax and it may be perfectly acceptable in those languages to do what you have above, you just aren't allowed to do it using C.

Sure the compiler could use the same name for another variable. But that doesn't make the code any better if your variables have different meanings in different places. The maintainability would not be easy either.

just out of interest why can you not redifine a data type

Well, if you think about it, why would you want to? Defining a variable happens when you specify it's type, just as hk_mp5kpdw said. if you want to change a variable's type half way through a function that's using it then there's something wrong with your code.

When defining your variables, you're telling the compiler that you need space to store a particular piece of information. Does it really make sense to tell it AGAIN once it's been defined? I don't think it does.

If you do want to do this kind of thing, then just use different scopes... like so:

Code:

#include<iostream>
using namespace std;

int main(void)
{
    { // define a scope
        int x=5;
        cout << x << endl;
    } // end scope, therefore x doesn't exist any more

    { // define another scope
        int x=2;  // this is now fine since the above x no longer exists
        cout << x << endl;
    } // end scope again

    return(0);
}

hope that helps.

U.