HLA将一个和n之间的所有数字相加

Question

汇编语言

  

编写一个计算（n）（n + 1）/ 2的程序。它应该从用户那里读取值“ n”。提示：您可以通过将一个和n之间的所有数字相加来计算此公式。

用HLA编写上述代码时遇到了挑战。我设法得到了以下

program printing_n_Numbers;
    #include("stdlib.hhf");
    static
        n:int32;
        i:int32;
begin printing_n_Numbers;
    stdout.put("Enter n: ");
    stdin.get(n);
    mov(0,ecx)
    stdout.put("printing ",n," Numbers ",nl);
    for(mov(0,eax);eax<=n;add(1,eax)) do
        for(mov(0,ebx);ebx<eax;add(1,ebx)) do
            ecx = add(eax,ebx);
            stdout.put("N was = ");
            stdout.puti32(exc);
            stdout.newln();
        endfor;
    endfor;
end printing_n_Numbers;  

当我输入数字6时，输出为

Enter n: 6
printing 6 Numbers
N was = 1
N was = 2
N was = 3
N was = 4
N was = 5
N was = 2
N was = 4
N was = 6
N was = 3
N was = 6
N was = 4
N was = 8
N was = 5
N was = 6

我该如何编码以输出输入的数字总和？

Answer 1

已解决

在进行了多次更改之后，程序开始运行。这就是我修改的方式

mov(0,ecx);
    stdout.put("You Have Entered: ",n,nl);
    for(mov(0,eax);eax<=n;add(1,eax)) do
        add(eax,ecx);
    endfor;

为了打印总和，这是代码

stdout.puti32(ecx); 

我使用stdout.puti32将十六进制转换为原始十进制数字系统