Static scoping in C/C++

2018-06-29 13:16:23

In the following code, 2 is printed.

int x = 1;
int f(int y)
{
    return x;
}

int main() {
    x = 2;
    printf("%d", f(0));
}

How is it happening if we have static scoping in C? Why isn't 1 printed?

Printing 2 in this case isn't a dynamic scoping, is it?

I thought that in static scoping it should take the nearest x to the function definition.

It does take the nearest x , but since you only have one x it doesn't really matter.

If you change the code to

int x = 1;
int f(int y)
  {
    return x ;
  }

int main() {
    int x=2;       
    printf("%d", f(0));
}

so you have 2 x , the global one and the local one in main you will see 1 getting printed.

The scoping is moot here since you have not declared an x locally that would have otherwise shadowed the global x .

2 is printed.

x is assigned in main to 2 immediately before f is called with the parameter 0 .

(Conceptually int x = 1; is ran before main is entered.)

It is the way the compiler generates the assembly/machine code.

first global variable X is stored in memory location "abc"

next main is executed: global variable X at "abc" is changed to 2

Now function f() is called:

function f returns the value of global variable X at "abc": being 2

the return value of f() is printed.

So if you wanted a different X in the main-function scope, you should have made a new object, like in the answer by nwp.

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