1) 源代码 

long long fact(long long n) {if (n < 1) return 1;else return (n*fact(n - 1));
}

2) 汇编代码 

fact:addi sp, sp, -16       // adjust stack for 2 itemssd x1, 8(sp)           // save the return addresssd x10, 0(sp)          // save the argument naddi x5, x10, -1       // x5 = n - 1bge x5, x0, L1         // if (n - 1) >= 0, go to L1addi x10, x0, 1        // return 1addi sp, sp, 16        // pop 2 items off stackjalr x0, 0(x1)         // return to callerL1:addi x10, x10, -1      // n >= 1: argument gets (n - 1)jal x1, fact           // call fact with (n - 1)ld x6, 0(sp)           // restore argument nld x1, 8(sp)           // restore the return addressaddi sp, sp, 16        // adjust stack pointer to pop 2 itemsmul x10, x10, x6       // return n * fact(n - 1)jalr x0, 0(x1)         // return to the caller