global variable initialization

Question

In principle, a variable defined outside any function (that is, global, namespace, and class static variables) is initialized before main() is invoked. Such nonlocal variables in a translation unit are initialized in their declaration order

Above are the lines from the class notes given by my lecturer.

#include <iostream>

using namespace std;
int a=99;
int main(int argc, char *argv[]) 
{
  cout<<a<<endl;
  cout<<b<<endl;
  return 0;
}
int b=100;

There is an error while I run this. Isn't it true that b assigned to 100 before main() is called?

You're getting confused between initialisation order and compilation order - they are not the same thing. — Paul R, Aug 07 '12 at 08:52
So,is it like,Compilation is done line by line in order.When compilation reaches cout< — tez, Aug 07 '12 at 09:03
@tez: yes, that's more or less correct - you need to declare `b` before you use it so that the compiler "sees" its declaration first. — Paul R, Aug 07 '12 at 09:04
@tez: that is still not correct - you need `extern int b;` above `main` and `int b = 100;` below `main`. — Paul R, Aug 07 '12 at 10:09
Because [expression statements are not allowed in namespace scope](https://stackoverflow.com/a/8470186/5376789). — xskxzr, Jun 10 '18 at 07:04
The question is largely change by the 4th edit (by OP), I'd suggest rollback it. — apple apple, Apr 27 '22 at 07:53
I edited the question to match most of the answers (and the accepted one). — apple apple, Apr 27 '22 at 08:00

score 5 · Accepted Answer · answered Aug 07 '12 at 08:56

The problem here is not initialisation order: b is indeed initialised before main starts running.

The problem is the "visibility" of b. At the point where main is being compiled, there is no b.

You can fix it by either moving the definition/initialisation of b to before main:

#include <iostream>

using namespace std;
int a = 99;
int b = 100;
int main (int argc, char *argv[]) {
    cout << a << '\n';
    cout << b << '\n';
    return 0;
}

or simply indicate that b exists:

#include <iostream>

using namespace std;
int a = 99;
extern int b;
int main (int argc, char *argv[]) {
    cout << a << '\n';
    cout << b << '\n';
    return 0;
}
int b = 100;

Neither of those two solutions change when b is created or initialised at run-time, they simply make b available within main.

score 3 · Answer 2 · answered Aug 07 '12 at 08:56

Your lecturer is wrong; global variables are initialised in order of definition, not declaration.

For example,

#include <iostream>
struct S { S(const char *s) { std::cout << s << '\n'; } };
extern S a;    // declaration
extern S b;    // declaration
int main() { }
S b("b");      // definition
S a("a");      // definition

will print

b
a

The code you posted doesn't work because b is not even declared at the point of use. A declaration (for example, extern int b), is required because C++ (like C) was originally designed as a one-pass compiler.

score 2 · Answer 3 · answered Aug 07 '12 at 08:50

2

The problem is here: cout<<b<<endl;

You can't access the variable before it's declaration.

answered Aug 07 '12 at 08:50

Andrew

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Heisenbug · Answer 4 · 2012-08-07T08:58:15.427

2

Read carefully.

A variable defined outside any function (B) is initialized before main is invoked (not compiled).

To be compiled correctly B should be defined(and declared) before it's first use (and that's your error: B is used before been declared anywhere).

edited Aug 07 '12 at 08:58

answered Aug 07 '12 at 08:52

Heisenbug

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score 0 · Answer 5 · edited Jul 13 '21 at 13:17

The problem is the statement b=100;: you can't place a statement in a global scope (except for variable initialization).

If this line remains as is, the code won't compile regardless of b declarations/definitions or the use of b within main or anywhere.

Without this line, the code is correct and works, with b value equals 0, since uninitialized global variables are initialized to 0 (or the line b=100; could be moved into any function scope).

global variable initialization

5 Answers5

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