Question

#include <stdio.h>
#include <stdlib.h>

struct test
{
    int id;
    char name[20];
};

int main()
{

struct test t1;

   t1.id=1;
   fflush(stdin);
   fgets(t1.name,20,stdin);

   print((&t1.name));          
   print1(t1.id,&(t1.name));    

}
void print(struct test *name)
{
    puts(name);
}

void print1(struct test id,struct test *name)
{

    printf("\n%d\n",id);


    puts(name);
}

当我运行此程序时，它会要求输入

测试[输入]

输出出来

测试 1 （然后程序终止）

为什么第一次投入工作以及为什么第二次投入不起作用？是的，有一个发送完整结构的选项，但我想知道这里有什么问题。

Answer 1

由于以下几个原因，您的计划无效：

您正在调用缺少前向声明的函数 - 您应该在程序编译时看到警告。不要忽视它们
您的函数正在使用不正确类型的参数 - 您应该会收到与各个字段相对应的类型，例如： void print1(int id, char *name)或者你应该通过值或指针传递整个结构，即void print1(struct test t)

一旦解决了这两个问题，并确保程序编译无警告，并启用所有编译器警告，问题就应该解决了。

Answer 2

void print(struct test *name)

应改为

void print(char name[]) // because you wish to print a null terminated array of characters.

 print((&t1.name));

应改为

print(t1.name); //name is the array you wish to print

Answer 3

你需要这个

@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container, Bundle savedInstanceState) {
    View view = inflater.inflate(R.layout.login_fragment_layout, container, false);
    final EditText password = (EditText) view.findViewById(R.id.password);

    // Workaround https://issuetracker.google.com/issues/37082815 for Android 4.4+
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.KITKAT && isRTL(getActivity())) {

        // Force a right-aligned text entry, otherwise latin character input,
        // like "abc123", will jump to the left and may even disappear!
        password.setTextDirection(View.TEXT_DIRECTION_RTL);

        // Make the "Enter password" hint display on the right hand side
        password.setInputType(InputType.TYPE_CLASS_TEXT | InputType.TYPE_TEXT_FLAG_NO_SUGGESTIONS);
    }

    password.addTextChangedListener(new TextWatcher() {

        boolean inputTypeChanged;

        @Override
        public void beforeTextChanged(CharSequence s, int start, int count, int after) {}

        @Override
        public void onTextChanged(CharSequence s, int start, int before, int count) {}

        @Override
        public void afterTextChanged(Editable s) {

            // Workaround https://code.google.com/p/android/issues/detail?id=201471 for Android 4.4+
            if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.KITKAT && isRTL(getActivity())) {
                if (s.length() > 0) {
                    if (!inputTypeChanged) {

                        // When a character is typed, dynamically change the EditText's
                        // InputType to PASSWORD, to show the dots and conceal the typed characters.
                        password.setInputType(InputType.TYPE_CLASS_TEXT |
                                InputType.TYPE_TEXT_VARIATION_PASSWORD |
                                InputType.TYPE_TEXT_FLAG_NO_SUGGESTIONS);

                        // Move the cursor to the correct place (after the typed character)
                        password.setSelection(s.length());

                        inputTypeChanged = true;
                    }
                } else {
                    // Reset EditText: Make the "Enter password" hint display on the right
                    password.setInputType(InputType.TYPE_CLASS_TEXT |
                            InputType.TYPE_TEXT_FLAG_NO_SUGGESTIONS);

                    inputTypeChanged = false;
                }
            }
        }
    });

    return view;
}

public static boolean isRTL(Context context) {
    if (Build.VERSION.SDK_INT >= Build.VERSION_CODES.JELLY_BEAN_MR1) {
        return context.getResources().getConfiguration().getLayoutDirection()
                == View.LAYOUT_DIRECTION_RTL;
        // Another way:
        // Define a boolean resource as "true" in res/values-ldrtl
        // and "false" in res/values
        // return context.getResources().getBoolean(R.bool.is_right_to_left);
    } else {
        return false;
    }
}

呼叫需要

void print(char *name)
{
    puts(name);
}

print(t1.name);需要puts（或实际为char *）。

t1.name的数据类型为const char *

同样

char *

和电话

void print1(struct test id, char *name)
{

    printf("\n%d\n",id);


    puts(name);
}

数组的名称退化为数组的第一个元素的地址。因此，当print1(t1.id,& t1.name);传递给函数时，它将成为char数组的起始地址。

为什么put不能在以下代码中工作？

3 个答案: