Forum : ST7/STM8
Original Post
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November 4, 2010 - 8:28pm
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The compiler does not correctly apply post increment operator on these instructions:
int i = 0;
long l = 0x0102;
void *addri = (void*)&i;
*((int*)addri++) = l;
printf ("%s works\r\n", (&i + 1 == addri) ? "" : "not");
This code compiles but it behaves wrongly because the post increment adds just 1 to addri because the compiler doesn't know the actual size of addri. It shouldn't compile rather than produce a wrong result. When you try to cast the addri pointer to the correct size it doesn't compile:
int i = 0;
long l = 0x0102;
void *addri = (void*)&i;
*(((int*)addri)++) = l;
printf ("%s works\r\n", (&i + 1 == addri) ? "" : "not");
It says the "operand is an lvalue". Other compiler works and produce the correct result. Regards. |
Hi,
The ANSI/ISO C Standard mandates that "A cast does not yield an lvalue. Thus, a cast to a qualified type has the same effect as a cast to the unqualified version of the type."
In other words, you cannot cast the left-side of an assignment or of a address-of operator. As the postincrement operator "++" is an affectation operator, the conversion is not allowed.
So our compiler is perfectly right and MUST reject your "*((int*)addri++) = l;" statement.
The best workaround is to use the proper pointer type instead of "void *". Alternatively, you can assign a "char *" pointer to your pointer's value and manipulate it freely.
Regards,
Bruno,
ok, if the compiler is strictly ANSI compliant you're right. But the rejected statement is
not this:
Regards.
Right, the problem comes from the postincrement operator (++), which requires its operand being an lvalue.
Regards,