[英]code in mips with use of “push” without “pop”
C中的代码:
int recursive_power (int base, int power)
{
if (power <=0)
return (-1);
else if (power ==1)
return (base);
else
return (base*recursive_power (base, power-1)); }
技巧:
recursive_power: addi $sp, $sp, -8 #Assign stack space
sw $ra, 0($sp) #Store current value of $ra
sw $a1, 4($sp) #Store current value of power
slti $t0, $a1, 1 #if power<1, $t0==1, else $t0==0;
beq $t0, $0, next #power>0, continue ($t0==0)
addi $v0, $0, -1 #power<=0, return -1
j return2
next: addi $t1, $0, 1 #put ‘1’ in $t1
beq $a1, $t1, return1 #if power==1, return(base)
addi $a1, $a1, -1 #power=power-1
jal recursive_power #call recursive_power (base,power-1)
lw $ra, 0($sp) #Restore original return address
lw $a1, 4($sp) #Restore original value of power
mul $v0, $a0, $v0 #return value is $v0*base
j return2
return1: addi $v0,$a0, 0 #$v0=base
return2: addi $sp, $sp, 8 #Restore stack
jr $ra #Return to caller address
既然在代码中出现“ return1”和“ return2”,这是正确的,因为不会有“ lw”的推荐吗? 使用栈-“推”是否还需要“弹出”? 您可以帮助改善mips代码吗?
您将流行音乐放置在错误的位置。 每次进入函数时,您要将两个寄存器压入堆栈,因此,每次从函数返回时,都应将它们弹出:
jal recursive_power #call recursive_power (base,power-1)
mul $v0, $a0, $v0 #return value is $v0*base
j return2
return1: addi $v0,$a0, 0 #$v0=base
return2:
lw $ra, 0($sp) #Restore original return address
lw $a1, 4($sp) #Restore original value of power
addi $sp, $sp, 8 #Restore stack
jr $ra #Return to caller address
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