如何判断函数何时访问局部变量或外部变量？

Question

I am currently attempting to understand this code I came across a C certification test. The correct output of the code is 12, 13, 13. I tried printing out the values, before it calls the 3 final outputs and i noticed the reason being is that there's a scope difference between the external x declared at the top and the one inside the function. My question is, how do I know which of those functions are accessing which object throughout the code?

#include <stdio.h>
#include <stdlib.h>

int x; 
int modifyvalue() 
{ 
    return(x+=10); 
}
int changevalue(int x) 
{ 
    return(x+=1); 
}


int main(){
    int x=10; 
    x++; 
    printf("[1] %d \n\n", x);  
    changevalue(x); 
    printf("[2] %d \n\n", x);
    x++;
    printf("[3] %d \n\n", x); 
    modifyvalue(); 
    printf("First output:%d \n\n\n",x);
    x++; 
    changevalue(x); 
    printf("Second output:%d \n\n\n",x); 
    modifyvalue(); 
    printf("Third output:%dn \n\n\n",x);
}

Answer 1

Always take the variable from nearest scope.
Whenever the printf("...", x) is called, it takes the x from the function main() .

The function modifyvalue() always operates on the x outside of all functions.

The function changevalue(int x) always operates on the parameter x - which is a copy of the variable passed in.

So in your case, both functions essentially do nothing to the x in main() .

Answer 2

The rule is simple, in every { ... } (block) you can reference either a variable that is define in that block or any "parent" of it. For example in your code:

• modifyvalue: no x, x is in parent, "global"
• changevalue: x is a parameter, increment the parameter , change not reflected on the caller (because the parameter is a copy)
• main: x is local

Notice that you are printing always "the main's x" that is incremented only in the main. As noted by @Jens, the parent scope in C is called enclosing block and the global scope is called program scope (file scope + external)