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Q67040: /Ol Causes register Variables to Be Allocated on the Stack

Article: Q67040
Product(s): See article
Version(s): 6.00 6.00a | 6.00 6.00a
Operating System(s): MS-DOS | OS/2
Keyword(s): ENDUSER | buglist6.00 buglist6.00a | mspl13_c
Last Modified: 28-NOV-1990

In most cases, a variable declared with the register storage-class
will not be allocated to a register if loop optimization (/Ol) is
enabled. Instead, the compiler will place the variable on the stack.

Only if the function contains an inline assembly (_asm) block will the
register storage be allocated, due to the fact that inline assembly
takes precedence over optimization. This is reflected by the following
warning message, which is generated when /Ol is used on a function
with an _asm block:

   warning C4204: in-line assembler precludes global optimizations

The sample program below demonstrates both of the situations described
above. If the program is compiled as is with /Ol, the register
declared variable "i" will not be put in a register; it will be
handled exactly like "j", which is not declared with register
storage-class.

If the line at the end of main() with the _asm block is uncommented,
then "i" will be allocated storage in a register and the C4204 warning
will be displayed if the warning level is set at 3 or 4 (/W3 or /W4).
To see the difference, you can generate an assembly listing with /Fa,
or you can compile and link the program with CodeView information and
then view the mixed source and assembly listing in the debugger.

Microsoft has confirmed this to be a problem in C versions 6.00 and
6.00a. We are researching this problem and will post new information
here as it becomes available.

Sample Code
-----------

int foo(void);

void main(void)
{
    register int i = 7;
    int          j = 9;

    while ( i < 10 ) {
       i += foo();
       j += foo();
    }

    /* Uncomment the following line to have i put in a register */
    /* _asm xor i,i */
}

int foo(void)
{
   return (1);
}

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