I'm converting assembly code to C code and I'm confused about part of the codes. I knew that after imul, the higher and lower part of tmp1*tmp2 will be stored in %edx and %eax separately. However, I'm confused that
1, why it stores tmp1 into %eax (line 132d) again? I think it will erase the value of tmp3in %eax.
2, In line 1332, the instruction uses %edx. I know it should be the upper part of tmp3, but how can I represent it in C code?
3, Why there is nop in line 1339?
//prologue
1306: f3 0f 1e fb endbr32
130a: 55 push %ebp
130b: 89 e5 mov %esp,%ebp
//body
130d: 8b 45 0c mov 0xc(%ebp),%eax
1310: 8b 00 mov (%eax),%eax
1312: 8b 55 08 mov 0x8(%ebp),%edx //%edx<-para1
1315: c1 e2 03 shl $0x3,%edx //para1<<3
1318: 01 c2 add %eax,%edx //para1+=*para2
131a: 8b 45 0c mov 0xc(%ebp),%eax //%eax<-para2
131d: 89 10 mov %edx,(%eax)
131f: 8b 45 0c mov 0xc(%ebp),%eax //
1322: 8b 08 mov (%eax),%ecx
1324: ba 56 55 55 55 mov $0x55555556,%edx
1329: 89 c8 mov %ecx,%eax //%eax<-tmp1
132b: f7 ea imul %edx //long tmp3=tmp1*tmp2
132d: 89 c8 mov %ecx,%eax //%eax<-tmp1
132f: c1 f8 1f sar $0x1f,%eax //tmp1>>31
1332: 29 c2 sub %eax,%edx //?
1334: 8b 45 0c mov 0xc(%ebp),%eax
1337: 89 10 mov %edx,(%eax)
1339: 90 nop
133a: 5d pop %ebp
133b: c3 ret
Thank you so much~
