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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 1 和 n 之间的所有数字相加来计算此公式。

我在 HLA 中编写上述代码时遇到了挑战。我设法得到以下

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;

当我输入一个像 6 这样的数字时，输出是

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

我将如何对其进行编码以输出输入数字的总和？

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1 回答 1

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解决了

经过多次更改后，该程序起作用了。这就是我修改它的方式

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);

我曾经stdout.puti32将十六进制转换为原始的十进制数字系统

于 2019-03-02T10:04:26.760 回答