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up vote 25 down vote favorite

This code compiles, but I have a run time error in Visual Studio:

Run-time check failure #3 - the variable 'x' is being used without being initialized...

int x = 15;
int main()
{
    int x = x;
    return 0;
}

I don't understand that behavior... in the error box when I click continue the program resumes and x has a corrupted content (like -8556328 instead of 15).

Why does this code work without a problem, and the int array is well declared?

const int x = 5;
int main()
{
     int x[x] = {1,2,3,4};
     return 0;
}
share|improve this question
13  
You tagged both C and C++. Which did you compile? –  underscore_d 23 hours ago
2  
Some interesting facts: gcc 4.8.4, compiles and this program can be run with -Wall -Wextra -pedantic turned on. clang 7.0.0 compiles it, and can be run as is. However if printf("%d\n", x); is added after int x=x; (I guess any actual usage of x), the compiler gives the more friendly warning: warning: variable 'x' is uninitialized when used within its own initialization [-Wuninitialized]. gcc still compiles and runs it even with the printf and printed 0. However running the program through valgrind gives Conditional jump or move depends on uninitialised value(s) –  Joakim 20 hours ago
    
@Joakim: Interesting; thanks for the results. Are GCC and Clang within their rights, i.e. is this canonically undefined behaviour? –  underscore_d 12 hours ago
1  
@underscore_d - C++ doesn't require any diagnostics on uninitialized variables. And the compiler is free to optimize away a variable altogether, especially if it is never really used after the assignment. In fact, "undefined behavior" means that the compiler can do whatever it wishes. –  Jirka Hanika 6 hours ago
    
Yup, I know what UB means, just wanted to check that the standard defined (or rather, omits to define) these particular cases as UB. Thanks for the info –  underscore_d 6 hours ago

4 Answers 4

up vote 29 down vote accepted

When you declare a new variable, its name becomes visible right here

int x =
//     ^- there

because it is at that point the variable is fully declared, and as such; its name means something. At this point in time any other (previously declared variable) in a surrounding scope will be hidden.

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up vote 34 down vote

x is defined at the left of =.

so in x[x], [x] refer to the global one,

whereas in x = x;, x hides the global x and initializes from itself -> UB.

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up vote 4 down vote

There is no scope resolution operator in C, so you may not be able to use 

int x = x;

in your program.

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6  
The OP doesn't seem to know whether they want an answer for C or for C++. The latter does support scope resolution. –  underscore_d 23 hours ago
    
I think that it compiles I checked it in ideone.com with the C compiler... –  Aminos 23 hours ago
    
@underscore_d, yes C++ supports and C doesn't. –  Adi 23 hours ago
up vote -10 down vote

You didn't describe your code correctly: probably your main function is static. A static function of any class can't use the non-static members of that class

class Program
{
    int x = 4;
    static int y = 5;

    static void Main(string[] args)
    {
        Console.WriteLine(x.ToString()); // <- compiler error: x not static
        Console.WriteLine(y.ToString()); // <- Ok, y is also static
    }
}
share|improve this answer
11  
This doesn't look like C or C++ to me... –  Stig Hemmer 11 hours ago
3  
I do not see the C# tag on the question, which this answer appears to assume. –  Snowman 8 hours ago

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