我正在尝试将此c ++函数转换为mips。我认为我在循环中遇到问题,因为当我运行它时,它给了我13..1.17.5 ..但我的输出应该是两个ip地址:130.52.0.10和171.9.50.186
C ++函数代码:
void IPtoDD(int arg0, char *arg1)
{
int temp, numChar, shift = 24;
for (int i=0; i<4; i++) {
temp = arg0 >> shift;
temp = temp & 0x000000ff;
numChar = byteToDec(temp,arg1);
arg1 += numChar;
*arg1++ = '.';
shift -= 8;
}
arg1--;
*arg1 = 0;
return;
}
MIPS代码:
IPtoDD: addi $sp, $sp, -20
sw $ra, ($sp)
sw $s0, 4($sp)
sw $s1, 8($sp)
sw $s2, 12($sp)
sw $s3, 16($sp)
move $s0, $a0
move $s1, $a1
li $s3, 24 #s3=shift
li $s2, 0 #s2=i
li $t5, 0 #t5=temp
li $t3, 0
move $s1, $a1 #s1=*arg1
loop: srl $t5, $s0, $s3 #t3= numChar
and $t5, $t5, 0xff #t4= (*arg1)
move $a0, $t5
move $a1, $s1
jal byteToDec
move $t3, $v0
add $s1, $s1, $t3
li $t5, '.'
sb $t5, ($a1)
addi $a1, $a1, 1
addi $s3, $s3, -8
addi $s2, $s2, 1
blt $s2, 4, loop
addi $s1, $s1, -1
sb $0, ($a1)
lw $s3, 16($sp)
ra: lw $s2, 12($sp)
lw $s1, 8($sp)
lw $s0, 4($sp)
lw $ra, ($sp)
addi $sp, $sp, 20
jr $ra
请在这里帮忙。我尝试了很多,但没能使它正常运行。
编辑: byteToDec的C ++函数
int byteToDec(int arg0, char *arg1)
{
int temp, flag = 0, count = 0;
if (arg0==0) {
*arg1 = '0';
return 1;
}
else {
temp = arg0/100;
if (temp != 0) {
*arg1++ = (char) temp + 0x30;
count++;
flag = 1;
}
temp = (arg0 % 100) / 10;
if ((flag!=0) || (temp != 0)) {
*arg1++ = (char) temp + 0x30;
count++;
}
temp = arg0 % 10;
*arg1 = (char) temp + 0x30;
count++;
return count;
}
}
MIPS中的byteToDec:
byteToDec: #t0= temp
#t1= flag
#v0= count
#t3= (*arg1)
bne $a0, $0, else
li $t3, '0'
sb $t3, ($a1)
li $v0, 1
jr $ra
else: div $t0, $a0, 100
beq $t0, 0, cont
bp2: addi $t3, $t0, 0x30
sb $t3, ($a1)
addi $a1, $a1, 1
addi $v0, $v0, 1
li $t1, 1
cont: rem $t3, $a0, 100
div $t0, $t3, 10
bne $t1, 0, nxtIf
beq $t0, 0, endElse
nxtIf: addi $t3, $t0, 0x30
sb $t3, ($a1)
addi $a1, $a1, 1
addi $v0, $v0, 1
endElse:rem $t0, $a0, 10
bp1: addi $t3, $t0, 0x30
sb $t3, ($a1)
addi $v0, $v0, 1
ra1: jr $ra
答案 0 :(得分:1)
您正在使用t3
作为循环计数器,然后在t3
函数中删除byteToDec
。 MIPS约定是t
寄存器是“temp”而不是像这样在函数调用中使用。您应该将循环变量放在s
寄存器(“保存”寄存器)中,如果被调用的函数需要重用相同的s
寄存器,则需要将其保存到堆栈或其他内容中。在返回被叫方之前恢复该值。
答案 1 :(得分:0)
我有一个程序,第一个循环工作,但第二个到第六个循环进入inf。我是初学者,可以使用一些帮助
#include <iostream>
#include <cmath>
#include <fstream>
using namespace std;
ofstream myfile ("Atomseries.txt");
int main()
{
float rconstant=109677.58;
int nstart, nend;
int length;
for (nstart = 1; nstart <= 6; nstart++){
for (nend= 2; nend <=nstart + 10; nend++)
length = 1/(rconstant*(1/(nstart*nstart)- 1/(nend*nend)));
myfile << length * 10000000 << endl;
}
if (nstart = 1)
myfile << "Lyman Series"<< endl;
else
;
if (nstart = 2)
myfile << "B Series" << endl;
else
;
if (nstart= 3)
myfile << " series"<< endl;
return 0;
}