I have:
char message1[100];
char message2[100];
When I try to do message1 = message2, I get error:
incompatible types when assigning to type
‘char[100]’from type‘char *’
I have functions like
if(send(clntSocket, echoBuffer, recvMsgSize, 0) != recvMsgSize){
DieWithError("send() failed")
}
inbetween. Could these mess things up somehow? :(
I have a feeling maybe you can't do = on char arrays or something, but I looked around and couldn't find anything.
You can't assign anything to an array variable in C. It's not a 'modifiable lvalue'. From the spec, §6.3.2.1 Lvalues, arrays, and function designators:
An lvalue is an expression with an object type or an incomplete type other than
void; if an lvalue does not designate an object when it is evaluated, the behavior is undefined. When an object is said to have a particular type, the type is specified by the lvalue used to designate the object. A modifiable lvalue is an lvalue that does not have array type, does not have an incomplete type, does not have a const-qualified type, and if it is a structure or union, does not have any member (including, recursively, any member or element of all contained aggregates or unions) with a const-qualified type.
The error message you're getting is a bit confusing because the array on the right hand side of the expression decays into a pointer before the assignment. What you have is semantically equivalent to:
message1 = &message2[0];
Which gives the right side type char *, but since you still can't assign anything to message1 (it's an array, type char[100]), you're getting the compiler error that you see. You can solve your problem by using memcpy(3):
memcpy(message1, message2, sizeof message2);
If you really have your heart set on using = for some reason, you could use use arrays inside structures... that's not really a recommended way to go, though.
Your suspicions are correct. C (I'm assuming this is C) treats an array variable as a pointer.
You need to read the C FAQ about arrays and pointers: http://c-faq.com/aryptr/index.html
