Write char array to specific memory-address

Question

I have a huge byte array (char array) that I am using to write to another address. e.g.

char myBytes[] = { 0x43, 0x31, 0x63 };//just an example
(char*)0x123456 = &myBytes;

but the problem is that I get an error saying

"expression must be a modifiable lvalue"

I've tried some other stuff like

char myBytes[] = { 0x43, 0x31, 0x63 };//just an example
*(char*)0x123456 = myBytes;

But I get the same error. What am i doing wrong and what can I do get what I want accomplished?

memcpy((char*)0x123456, myBytes, sizeof(myBytes)); Bad idea, though... — Eugene Sh.
– Eugene Sh., Commented Jan 13, 2017 at 20:48
(char*)0x123456 is a constant of type char *. You cannot assign to it any more than you can assign to the value 1. — John Bollinger
– John Bollinger, Commented Jan 13, 2017 at 20:48
@SergeyA Working with explicit addresses in C or C++ is justified in a very special cases only, while I am not quite sure it is the case for the OP. And even in these cases there are often other ways. — Eugene Sh.
– Eugene Sh., Commented Jan 13, 2017 at 20:54
@EugeneSh., apparently, you never worked with MCUs, did you? — SergeyA
– SergeyA, Commented Jan 17, 2017 at 14:37

Barmar · Accepted Answer · 2017-01-13 20:50:01Z

3

You can't assign arrays with = in C++. The array name decays to a pointer when used like that, so you're trying to write the address of the array into the memory location. Use memcpy() to copy memory.

memcpy((void*)0x123456, (void*)mBytes, sizeof(mBytes));

answered Jan 13, 2017 at 20:50

Barmar

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8 Comments

John Bollinger Over a year ago

Not only does the array name decay to a pointer, any array value decays to a pointer when evaluated. Thus, you cannot even do tricks such as *(char(*)[3]) myBytes to get a handle on an array -- the array value resulting from that dereference just decays back to a pointer.

M.M Over a year ago

@JohnBollinger the "decay" does not happen on every evaluation. It only happens when the context demands a decay (which is most contexts, but not all). For example in &myBytes there is no decay, the address of the array results. And in your example (which is the same as *&myBytes), the & operator could be applied to the result.

John Bollinger Over a year ago

@M.M, it is a fine point, to be sure, but evaluating the expression &myBytes, does not involve evaluating the array designated by myBytes, therefore that array does not decay to a pointer. This (and my earlier comment) are consistent with what you've said. I like this way of thinking about it because it has the advantage of requiring fewer exceptions, but it is admittedly a bit subtle.

M.M Over a year ago

@JohnBollinger & evaluates its operand. Every operand is evaluated unless the standard specifically says it is an unevaluated operand. Which is true for sizeof but not for &. Consider the code &foo() ; if & did not evaluate its operand then the function would not be called.

John Bollinger Over a year ago

@M.M, good point, and I see that this is an area where C++ and C differ. In C++, the result of evaluating a function call expression is an lvalue under some (but by no means all) circumstances. As far as I can determine, the result of evaluating a C function call is never an lvalue. When it is not an lvalue, the result of a function call expression is not an allowed operand for the & operator in either language.

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