sum:                sum:                                                   
                        pushl   %ebp                                    
                        movl    %esp, %ebp      
                        subl    $16, %esp       
 reg1 = b               movl    12(%ebp), %eax                 
 add a, reg1            addl    8(%ebp), %eax   
 mov reg1, ans          movl    %eax, -4(%ebp)                 
 mov ans, reg1          movl    -4(%ebp), %eax                 
 return                 leave                             
                        ret                     

• One of the operands of mov must be a register or an immediate value, not a memory location.

• Moving the value in %eax to memory location for ans and then moving it back to a register (actually the same register), would be avoided with optimization level 2.