我在编写c ++队列中的模板时遇到了一些麻烦。如果你们可以帮助我解决一些编译器问题,我会喜欢它!
这是我的编译器输出:
在a7_main.cpp中包含的文件中:3:0:Queue.h:在成员函数中 TYPE Queue :: pushAndPop(TYPE):
Queue.h:33:25:警告:x的名称查找已更改[启用 默认]
Queue.h:27:28:警告:在ISO标准下匹配此x 规则[默认启用]
Queue.h:30:11:警告:在旧规则下匹配此x [默认启用]
Queue.h:在全球范围内:
Queue.h:45:23:错误:无法使用模板名称队列无效 参数列表
Queue.h:复制构造函数Queue :: Queue(Queue&)[with TYPE = V2,Queue = Queue]:
a7_main.cpp:56:19:从这里实例化
Queue.h:41:5:错误:没有匹配的调用函数 队列:: pushAndPop(int)的
Queue.h:41:5:注意:候选人是:
Queue.h:27:28:注意:TYPE Queue :: pushAndPop(TYPE)[TYPE = V2]
Queue.h:27:28:注意:没有已知的转换参数1 int到V2
我已经尝试了解这些警告和错误,但它让我陷入了困境。这是我的代码:
#if !defined QUEUE_SIZE
#define QUEUE_SIZE 30
#endif
using namespace std;
template <class TYPE> class Queue
{
private:
TYPE *array;
public:
Queue(Queue& other);
Queue();
~Queue();
Queue& operator=(Queue other);
TYPE pushAndPop(TYPE x);
};
template <class TYPE> Queue<TYPE>::Queue()
{
array=new TYPE[size];
}
template <class TYPE> Queue<TYPE>::~Queue()
{
delete [] array;
}
template <class TYPE> TYPE Queue<TYPE>::pushAndPop(TYPE x)
{
TYPE item = array[0];
for(int x = 0; x<QUEUE_SIZE-1; x++){
array[x]= array[x+1];
}
array[QUEUE_SIZE-1] = x;
return item;
}
template <class TYPE> Queue<TYPE>:: Queue(Queue& other)
{
int x;
for(x=0; x<QUEUE_SIZE;x++){
array[x] = other.pushAndPop(0);
}
}
template <class TYPE> Queue& Queue<TYPE>:: operator=(Queue other)
{
int x;
for(x=0; x<QUEUE_SIZE;x++){
array[x] = other.pushAndPop(0);
}
}
编辑:这是调用代码:
#include <ncurses.h>
#include <unistd.h>
#include "Queue.h"
//----------------------------------------------------
#define START_ROW (20)
#define START_COL (20)
#define SLEEP_MICROSECONDS ((1.0 / 20.0) * 1000000.0)
//----------------------------------------------------
// Simple storage for a row, column pair.
struct V2 {
int row, col;
};
//----------------------------------------------------
// Some of this ncurses code is, unfortunately, copied from
// the man page without full understanding.
void configureNcurses()
{
initscr();
cbreak();
noecho();
nonl();
intrflush(stdscr, FALSE);
keypad(stdscr, TRUE);
nodelay(stdscr, TRUE);
curs_set(0);
start_color();
init_pair(1, COLOR_RED, COLOR_BLACK);
init_pair(2, COLOR_GREEN, COLOR_BLACK);
init_pair(3, COLOR_YELLOW, COLOR_BLACK);
init_pair(4, COLOR_BLUE, COLOR_BLACK);
init_pair(5, COLOR_MAGENTA, COLOR_BLACK);
init_pair(6, COLOR_CYAN, COLOR_BLACK);
init_pair(7, COLOR_WHITE, COLOR_BLACK);
}
//----------------------------------------------------
// Draw centipede with sequence of foreground colors, or with
// background color, depending on "erase" flag.
// (Pass centipede by value to exercise copy constructor.)
void drawCentipede(Queue<V2> centipede, bool erase)
{
int drawCharacter;
int colorNumber;
V2 currentPosition, dummyPosition;
Queue<V2> centipedeCopy;
// Make a copy of centipede to be consumed during drawing.
// (Exercises assignment operator.)
centipedeCopy = centipede;
// Prepare to draw or erase, as requested.
if (erase)
drawCharacter = ' ';
else
drawCharacter = ' ' | A_REVERSE;
// Consume centipede copy to obtain data for drawing.
for (int i = 0; i < QUEUE_SIZE; ++i) {
colorNumber = 1 + (i % 7);
currentPosition = centipede.pushAndPop(dummyPosition);
attron(COLOR_PAIR(colorNumber));
mvaddch(currentPosition.row, currentPosition.col,
drawCharacter);
attroff(COLOR_PAIR(colorNumber));
}
}
//----------------------------------------------------
// Update position based on arrow key input.
void updatePosition(V2& position, int inputChar)
{
switch (inputChar) {
case KEY_UP:
--position.row;
break;
case KEY_DOWN:
++position.row;
break;
case KEY_LEFT:
--position.col;
break;
case KEY_RIGHT:
++position.col;
break;
default:
// (Ignore all other keys.)
break;
}
}
//----------------------------------------------------
main()
{
Queue<V2> centipede;
int currentDirection;
int inputChar;
V2 currentHead;
// Configure ncurses.
configureNcurses();
// Fill queue (all centipede segments at start position).
currentHead.row = START_ROW;
currentHead.col = START_COL;
for (int i = 0; i < QUEUE_SIZE; ++i)
centipede.pushAndPop(currentHead);
// Draw instructions and initial centipede.
attron(COLOR_PAIR(2));
mvaddstr(1, 3, "USE ARROW KEYS TO MOVE, CTRL-C TO QUIT");
attroff(COLOR_PAIR(2));
drawCentipede(centipede, false);
refresh();
// Process input until killed.
currentDirection = KEY_RIGHT;
while (true) {
// Show current state, then check for input.
usleep(SLEEP_MICROSECONDS);
inputChar = getch();
if (inputChar != ERR)
currentDirection = inputChar;
// When input received, erase old centipede.
drawCentipede(centipede, true);
// Then use new input to update centipede.
updatePosition(currentHead, currentDirection);
centipede.pushAndPop(currentHead);
// Then draw new centipede, and refresh.
drawCentipede(centipede, false);
refresh();
}
// Clean up ncurses. (Re-display cursor.)
curs_set(1);
}
//----------------------------------------------------
答案 0 :(得分:1)
template <class TYPE> Queue<TYPE>::Queue()
{
array=new TYPE[size];
}
什么是size
?
template <class TYPE> TYPE Queue<TYPE>::pushAndPop(TYPE x)
^^^^^^
{
...
for(int x = 0; ...) {
^^^^^
你这里有相互矛盾的名字。这是个坏主意。以不同方式命名参数或循环变量。
template <class TYPE> Queue<TYPE>:: Queue(Queue& other)
^^^^^^
这接受了哪种Queue
?如果它只是Queue<TYPE>
,则需要准确指定。如果它可以采用其他类型,你需要另一个模板参数(但这里没有意义)。
template <class TYPE> Queue& Queue<TYPE>:: operator=(Queue other)
^^^^^^ ^^^^^
这里注意相同,加上参数看起来应该是(可能是const
)引用。
array[x] = other.pushAndPop(0);
^
(两次。)除非TYPE
是某种int
,否则这没有意义。您没有pushAndPop
重载,需要int
。
如果我正确理解您的代码,复制构造函数和复制赋值运算符都会破坏它们的参数。这非常令人惊讶。参数应由const&
进行,并且只能从。