With reference to this situation who can explain the next

Code Snippet 1:

$array _a= Array (0 = ' a '), $array _b= & $array _a; $array _c= & $array _b; $array _d= $array _b; $array _c[0]= ' B '; echo ' Variable A: '; Var_dump ($array _a); Echo ' variable B: '; Var_dump ($array _b); Echo ' variable c: '; Var_dump ($array _c); Echo ' variable d: '; Var_dump ($ Array_d);

Output Result:

Variable A:array (size=1)  0 = = String ' B ' (length=1) variable B:array (size=1)  0 = string ' B ' (length=1) variable C:array (size=1 )  0 = string ' B ' (length=1) variable D:array (size=1)  0 = String ' a ' (length=1)

Here the variable D value is not changed, well understood, but change the reference position of the variable C, as follows:
Code Snippet 2:

$array _a= Array (0 = ' a '), $array _b= & $array _a; $array _c= & $array _b[0]; $array _d= $array _b; $array _c= ' B '; echo ' Variable A: '; Var_dump ($array _a); Echo ' variable B: '; Var_dump ($array _b); Echo ' variable c: '; Var_dump ($array _c); Echo ' variable d: '; Var_dump ($ Array_d);

Output Result:

Variable A:array (size=1)  0 = &string ' B ' (length=1) variable B:array (size=1)  0 = &string ' B ' (length=1) variable C: String ' B ' (length=1) variable D:array (size=1)  0 = &string ' B ' (length=1)

Here the variable d also changed, and also became a reference, do not understand why
Ask the Great God for advice

Reply to discussion (solution)

The manual says: If an array with a reference is copied, its value is not dereferenced. This is true for arrays that pass values to functions.

$array _c = & $array _b[0]; This allows $array_b to have a reference, C and b[0] pointing to the same memory

above example:

