Solving C++ 'target of assignment not really an lvalue' errors - c++

Question

Given this code:
void FrMemCopy(void *to, const void *from, size_t sz)
{
size_t sz8 = sz >> 3;
size_t sz1 = sz - (sz8 << 3);
while (sz8-- != 0) {
*((double *)to)++ = *((double *)from)++;
}
while (sz1-- != 0) {
*((char *)to)++ = *((char *)from)++;
}
}
I am receiving target of assignment not really an lvalue warnings on the 2 lines inside the while loops.
Can anyone break down those lines?
a cast then an increment?
What is a simplier way to write that?
What does the error mean?

Answer 1

It does not like the *((char*)to)++ statement.
Try this:
void FrMemCopy(void *to, const void *from, size_t sz)
{
size_t sz8 = sz >> 3;
size_t sz1 = sz - (sz8 << 3);
double * tod = (double *) to;
double * fromd = (double *) from;
while (sz8-- != 0) {
*(tod++) = *(fromd++);
}
char * toc = (char *) tod;
char * fromc = (char *) fromd;
while (sz1-- != 0) {
*(toc++) = *(fromc++);
}
}

Answer 2

You can't apply ++ to the result of a cast, only to an lvalue (a variable). So you need to create new variable with the appropriate types for the increments:
void FrMemCopy(void *to, const void *from, size_t sz)
{
size_t sz8 = sz >> 3;
size_t sz1 = sz - (sz8 << 3);
double *to1 = (double *)to;
double *from1 = (double *)from
while (sz8-- != 0) {
*to1++ = *from1++;
}
char *to2 = (char *)to1;
char *from2 = (char *)from1;
while (sz1-- != 0) {
*to2++ = *from2++;
}
}

Answer 3

I tried to rewrite it in a way that no warning appears:
void FrMemCopy(void *to, const void *from, size_t sz)
{
size_t sz8 = sz >> 3;
size_t sz1 = sz - (sz8 << 3);
double *xto = (double *)to;
double *xfrom = (double *)from;
while (sz8-- != 0) {
*xto++ = *xfrom++;
}
char *cto = (char *)to;
char *cfrom = (char *)from;
while (sz1-- != 0) {
*cto++ = *cfrom++;
}
}

Answer 4

The result of explicit type conversion is rvalue in this case - according to 5.4.1 of C++11 Standard. You cannot apply increment operator to rvalue, it shall be lvalue. See C++ value category for details.
Use temporary variables to obtain required effect:
double* to_dbl = static_cast<double*>(to);
double* from_dbl = static_cast<double*>(from);
while(sz8-- != 0)
{
*(to_dbl++) = *(from_dbl++);
}

Answer 5

You are performing an increment operation on the LValue (Left side value of the assignment operator). Logically and by definition, a LValue must always be a variable. It cannot be a constant. When you are performing an increment operation, it is leaving a constant value on the Left Side which is giving you the error.

Answer 6

First before answering let me just say: Don't try to out-micro-optimize your compiler/library. The compiler writers will win something like 99 times out of 100. Use std::copy or memcpy depending on the types you're copying and needs.
Other answers have noted that you can solve the immediate compilation errors with temporary variables.
I don't recommend this under any circumstances do the following, but I believe you can also accomplish this by casting to a reference type:
void FrMemCopy(void *to, const void *from, size_t sz)
{
size_t sz8 = sz >> 3;
size_t sz1 = sz - (sz8 << 3);
while (sz8-- != 0) {
*((double *&)to)++ = *((double *&)from)++;
}
while (sz1-- != 0) {
*((char *&)to)++ = *((char *&)from)++;
}
}