Are these operators in C ?

[AmberJain]

I found these OPERATORS ("PROBABLY this is right name for them") mentioned in an online tutorial for C--------->
__________________________________________________ _______________
1. |=
2. ^=
3. &=
4. <<=
5. >>=
__________________________________________________ _______________
Are these operators in C? Somebody please explain them in a bit detail.... I could not find them mentioned in the book I'm referring to.

NEWBIE HERE......PLZ HELP

THANKS TO EVERYONE IN ADVANCE................

[weaknessforcats]

What does your compiler have to say about this?

Write a code snippet and see.

[Ganon11]

The first three look like bitwise operators to me. The last two look like bit shift operators. But what's with the equals sign?

I also don't get the equals sign in +=, -=, *=, /=,...hmm, maybe there's a connection...

[Parul Bagadia]

Even i dont know about those operators; but i can tell you about these;
+=, *=,/=, -=;
these all are used for postincrementing the operands,
for ex.
i+=1;
means; i=i+1;
these goes for rest of the above mentioned operators; so may be its the same thing for all those u hav mentioned. But i dont know about them; but m sure regarding this.

[AmberJain]

I found this in my compiler's (BORLAND Turbo C++ 3.0) help manual and thought it would be nice to post it here. This solves my problem also------------->

Assignment operators ------------------>

There are 11 assignment operators in Turbo C++.

The = operator is the simple assignment operator; the other 10 are known as
compound assignment operators.

This is the complete set of assignment operators:

=
*= /= %= += -=
<<= >>= &= ^= |=

Syntax:
unary-expr assignment-op assignment-expr

In the expression E1 = E2, E1 must be a modifiable lvalue.

The value of E2, after conversion to the type of E1, is stored in the object
designated by E1 (replacing E1's previous value).

The value of the assignment expression is the value of E1 after the
assignment.

The assignment expression is not itself an lvalue.

For both simple and compound assignment, the operands E1 and E2 must obey
one of the following sets of rules:

1. E1 is of qualified or unqualified arith-
metic type and E2 is of 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. One of 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.

The compound assignments are op=, where op can be any one of the ten
operator symbols

* / % + -
<< >> & ^ |

The expression

E1 op= E2

has the same effect as

E1 = E1 op E2

except that the lvalue E1 is evaluated only once. For example, E1 += E2 is
the same as E1 = E1 + E2.