Bonjour,
Il existe plusieurs opérateur logique : ++, --, += ...
Mais quelqun pourrait-il me confirmer ce que signifie ^=
Est ce un complément ?
Version imprimable
Bonjour,
Il existe plusieurs opérateur logique : ++, --, += ...
Mais quelqun pourrait-il me confirmer ce que signifie ^=
Est ce un complément ?
Salut,
il me semble que l'opérateur ^= est le OU EXCLUSIF (XOR)...
Regardes dans l'aide, ça y est aussi...
L'aide est un peu juste.
Citation:
Syntax
unary-expr assignment-op assignment-expr
Remarks
The assignment operators are:
= *= /= %= += -=
<<= >>= &= ^= |=
The = operator is the only simple assignment operator, the others are compound assignment operators.
In the expression E1 = E2, E1 must be a modifiable lvalue. The assignment expression itself is not an lvalue.
The expression
E1 op= E2
has the same effect as
E1 = E1 op E2
except the lvalue E1 is evaluated only once. For example, E1 += E2 is the same as E1 = E1 + E2.
The expression's value is E1 after the expression evaluates.
For both simple and compound assignment, the operands E1 and E2 must obey one of the following rules:
1. E1 is a qualified or unqualified arithmetic type and E2 is an arithmetic type.
2. E1 has a qualified or unqualified version of a structure or union type compatible with the type of E2.
3. E1 and E2 are pointers to qualified or unqualified versions of compatible types, and the type pointed to by the left has all the qualifiers of the type pointed to by the right.
4. Either E1 or E2 is a pointer to an object or incomplete type and the other is a pointer to a qualified or unqualified version of void. The type pointed to by the left has all the qualifiers of the type pointed to by the right.
5. E1 is a pointer and E2 is a null pointer constant.
Note: Spaces separating compound operators (+<space>=) will generate errors.
Note: There are certain conditions where assignment operators are not supported when used with properties.
J'ai vu ca, il donne tous les opérateurs, mais pas la signification.
Si tu as 2 nombres en hexa par exemple h1 et h2,
faire h1 ^= h2 revient à dire :
h1 = h1 XOR h2
XOR est la fonction OU EXCLUSIF dont je te donne la table de vérité :
h1 h2 (h1 XOR h2)
0 0 0
0 1 1
1 0 1
1 1 0
Voilà, ça devrait t'aider...
J'ai trouve.
pour la recherche il faut ecrire operator ^=Citation:
Returns this Variant after performing a bitwise xor operation.
Variant& __fastcall operator ^=(const Variant& rhs);
Description
The actual types of the Variants determine the meaning of this operation. If the Variants can be converted to types that make sense for the given operation, the assignment is performed. If any of these operations is not possible because of an invalid variant type conversion, an EVariantError exception is thrown.
Impec, Merci bien les gars ;)