Problem 2. (20 points): Condider the following assembly code for a C for lop loop: pushl %ebp movi %esp,%ebp movl 8 (%ebp), %ecx movl 12 (%ebp), %edx xor1 %eax, %eax cmp1 %edx, %ecx le .L4 : decl incl incl 1 %ecx %edx %eax %edx, %ecx L4: incl movl popl ret %eax %ebp, %esp %ebp Based on the assembly code above, fill in the blanks below in its corresponding C source code. Notes: You may only use the symbolicvariables x, y, and result in your expressions below- do not use register names Some of the code above is about stack discipline. x ends up in %ecx and y ends up in %edx int loop (int x, int y) int result; for ( 2.6 ; resulttt return result; Page 3 of 3