Difference between global and $ GLOBALS [] in PHP

In PHP, global and $ GLOBALS [] are different. many people think that the difference between global and $ GLOBALS [] is just the difference in writing the above code. According to the official explanation, 1. $ GLOBALS [var] is the external global variable itself. 2. global $ var is the reference or pointer of the external $ var with the same name. For example: <? Php ?? $ Var1 ?? 1 ;?? Difference between global and $ GLOBALS [] in $ PHP

Many people think that the difference between global and $ GLOBALS [] is actually not the same.

According to the official explanation

1. $ GLOBALS ['var'] is an external global variable.

2. global $ var is an external reference or pointer with the same name as $ var.

For example:

1. <? Php ??
2. $ Var1? =? 1 ;??
3. $ Var2? =? 2 ;??
4. Function? Test (){??
5. ???? $ GLOBALS ['var2']? =? & $ GLOBALS ['var1'];?
6. }??
7. Test ();??
8. Echo? $ Var2 ;??
9. ?> ??

The normal print result is 1.

1. <? Php ??
2. $ Var1? =? 1 ;??
3. $ Var2? =? 2 ;??
4. Function? Test (){??
5. ???? Global? $ Var1, $ var2 ;??
6. ???? $ Var2? =? & $ Var1 ;??
7. }??
8. Test ();??
9. Echo? $ Var2 ;??
10. ?> ??

The unexpected result is 2.

Why is the result 2 printed? In fact, it is because the $ var1 reference points to the reference address of $ var2. The actual value is not changed.

Let's look at an example.

1. <? Php ??
2. $ Var1? =? 1 ;??
3. Function? Test (){??
4. ???? Unset ($ GLOBALS ['var1']);?
5. }??
6. Test ();??
7. Echo? $ Var1 ;??
8. ?> ??

Because $ var1 is deleted, nothing is printed.

1. <? Php ??
2. $ Var1? =? 1 ;??
3. Function? Test (){??
4. ???? Global ?? $ Var1 ;??
5. ???? Unset ($ var1 );??
6. }??
7. Test ();??
8. Echo? $ Var1 ;??
9. ?> ??

Accidentally printed 1. It indicates that only the alias is deleted. | the referenced value is not changed.

Do you understand?

That is to say, global $ var is actually $ var = & $ GLOBALS ['var']. Call an alias for an external variable.