在C中跟踪与局部变量同名的全局变量

Question

I'm having a few issues tracking the values of global variables when there are local variables that exist with the same name.

This is the code I'm working with:

#include <stdio.h>

void func(int);

int x=6, y=7, z=10;

int
main(int argc, char **argv) {
    int z=5;
    printf("main: x=%2d, y=%2d, z=%2d\n", x, y, z);
    func(x);
    printf("main: x=%2d, y=%2d, z=%2d\n", x, y, z);
    func(y);
    printf("main: x=%2d, y=%2d, z=%2d\n", x, y, z);
    func(z);
    printf("main: x=%2d, y=%2d, z=%2d\n", x, y, z);
    return 0;
    }

void
func(int x) {
    x = x+1;
    y = y+1;
    printf("func: x=%2d, y=%2d, z=%2d\n", x, y, z);
}

Note the globals have the same name as the locals in func(), except for z. When I run the program. I get the following output:

main: x= 6, y= 7, z= 5
func: x= 7, y= 8, z=10
main: x= 6, y= 8, z= 5
func: x= 9, y= 9, z=10
main: x= 6, y= 9, z= 5
func: x= 6, y=10, z=10
main: x= 6, y=10, z= 5

I can understand where the first line comes from. It's just the values of the global variables, but main is using 5 rather than 10 because the global variable shadows the local variable. I can also understand line 2. global variable x is passed into func, giving a 7. Global variable y is used also, giving an 8.

Line 3 is where I lose track of values. Why has the value of global variable y stayed as 8? Did the func() function call change its value permanently? I thought this was not possible since y in func() is just a local variable. I understand where the x = 6 comes from in line 3 though.

Line 4 I lose track even more. How can the value of x jump from 6 to 9?

It would be great if someone could walk me through this output, and perhaps give me a quick explanation on scope and also on shadowing.

Answer 1

To answer your first question

Why has the value of global variable y stayed as 8? Did the func() function call change its value permanently?

Yes it did change the global variable y permanently, because it is global and not local (it would be local if it was declared as int y; )

How can the value of x jump from 6 to 9?

You pass the global variable y , which is changed to 8 in the first function call, to the function. This value is locally stored in the function as x . So x + 1 will become 9.

This is the danger of global variables. Once you name local variables the same you WILL lose track of what value is contains what value.

A little example to explain some details:

int x = 3;

int main(){
    int x = 8;
    printf("%d", x);
}

This will print 8 because the code will grab the local value of x .

Furthermore, when you change a global variable this will be remembered for the lifetime of the program.

Answer 2

y is not a local variable inside func() . In order for it to be a local variable (which would shadow the global y variable), you would have to declare another y variable inside func() , eg as in the example below. (This version will always print "y=1", and will not modify the global y variable.)

void
func(int x) {
    int y = 0;

    x = x+1;
    y = y+1;
    printf("func: x=%2d, y=%2d, z=%2d\n", x, y, z);
}

Since y is not a local variable inside the original func() , it's incrementing the global y .

Note that x is local to func() too, and is passed in by value. That is, a call func(foo) will copy the value of foo into the x parameter, and changing the x parameter inside func() will not change foo .

Line 4 comes from the func(y) call. It will set the (local) x parameter inside func() to the current value of the global variable y . y happens to have the value 8 at that point. (It was incremented from 7 to 8 by the previous call to func() .) The assignment x = x+1 inside func() bumps that to 9, which is then printed.