Question

我是c的新手。我有一个问题，我无法获得指向内存中相同空间的两个指针。这是一些代码。

struct Gate* tempGate;
tempGate = (struct Gate*)malloc(sizeof(struct Gate));
struct IO* IOList;
IOList = (struct IO*)malloc(sizeof(struct IO)*20);

tempGate->inputs = (struct IO*)malloc(sizeof(struct IO*)*2);
tempGate->outputs = (struct IO*)malloc(sizeof(struct IO*));

主要是我

tempGate->inputs[j] = IOList[i];

稍后在嵌套for循环中我们有了这个

 public void send () {
    final String name,phone,fb;
    EditText one,two,three;
    one=(EditText) findViewById(R.id.editText);
    two=(EditText) findViewById(R.id.editText2);
    three=(EditText) findViewById(R.id.editText3);
    name=one.getText().toString();
    phone=two.getText().toString();
    fb=three.getText().toString();
    Thread T=new Thread(new Runnable() {
        @Override
        public void run() {
            HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://test.com/test.php");
            try {
                List <NameValuePair> nameValuePairs= new ArrayList<NameValuePair>(2);
                nameValuePairs.add(new BasicNameValuePair("name",name));
                nameValuePairs.add(new BasicNameValuePair("number",phone));
                nameValuePairs.add(new BasicNameValuePair("fbname",fb));
                httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
                HttpResponse response= httpclient.execute(httppost);

            } catch (ClientProtocolException e) {
                // TODO Auto-generated catch block
            } catch (IOException e) {
                // TODO Auto-generated catch block
            }
        }
    });
    T.start();
}

现在当我改变IOList [i]的值时，不应该tempGate-＆gt;输入[j]也会改变吗？如果不是为什么？我怎么能这样呢？帮助我Codiwan你是我唯一的希望。

Answer 1

您应该将inputs和outputs指针数组指向IO，而不是IO数组。然后，您可以使此数组的元素指向IOList的元素。

struct Gate{
    enum GateType type;
    struct Gate* next;
    int numInputs;
    struct IO** outputs;
    struct IO** inputs;
};

tempGates->inputs = malloc(sizeof(struct IO*)*2);
tempGates->outputs = malloc(sizeof(Struct IO*));

然后你的循环应该是：

tempGate->inputs[j] = &(IOList[i]);

然后，当您更改IOList[i].val时，这将更改tempGate->inputs[j]->val。

的值

我无法弄清楚我的指针错误

1 个答案: