“int”的转换无效；到“time_t*”和另一个错误

Question

我收到以下错误：

错误：从“int”到“time_t*”的转换无效

错误：初始化“time_t time(time_t*)”的参数 1

我的代码是：

#include <iostream>
#include <cstdlib>
#include <ctime>
#include <time.h>

using namespace std;

int main()
{
char Grid[5][5];
Grid[0][0] = 'P';
Grid[0][1] = '-';
Grid[0][2] = '-';
Grid[0][3] = '-';
Grid[0][4] = '-';
Grid[1][0] = '-';
Grid[1][1] = 'T';
Grid[1][2] = '-';
Grid[1][3] = '-';
Grid[1][4] = '-';
Grid[2][0] = '-';
Grid[2][1] = '-';
Grid[2][2] = '-';
Grid[2][3] = '-';
Grid[2][4] = '-';
Grid[3][0] = '-';
Grid[3][1] = '-';
Grid[3][2] = '-';
Grid[3][3] = 'T';
Grid[3][4] = '-';
Grid[4][0] = '-';
Grid[4][1] = '-';
Grid[4][2] = '-';
Grid[4][3] = '-';
Grid[4][4] = 'X';
cout<<"P = Player (you)"<<endl<<
      "T = Trap (ouch!)"<<endl<<
      "X = Treasure (X marks the spot)"<<endl<<
      "Tell P where to go by using the commands 'right','left','up','down'."<<endl<<endl;
string command;
int x = 0;
int y = 0;
int a;
int b;
srand(time(a));
srand(time(b));
int enemy1 = rand()%4;
do
{
    cout<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl;
    cout<<Grid[0][0]<<Grid[0][1]<<Grid[0][2]<<Grid[0][3]<<Grid[0][4]<<endl<<
      Grid[1][0]<<Grid[1][1]<<Grid[1][2]<<Grid[1][3]<<Grid[1][4]<<endl<<
      Grid[2][0]<<Grid[2][1]<<Grid[2][2]<<Grid[2][3]<<Grid[2][4]<<endl<<
      Grid[3][0]<<Grid[3][1]<<Grid[3][2]<<Grid[3][3]<<Grid[3][4]<<endl<<
      Grid[4][0]<<Grid[4][1]<<Grid[4][2]<<Grid[4][3]<<Grid[4][4]<<endl;
    cout<<"Your command: ";
    cin>>command;
    cout<<endl;
    if(command=="up")
    {
        Grid[(x-1)][y] = 'P';
        x--;
        Grid[(x+1)][y] = '-';
    }
    if(command=="down")
    {
        Grid[(x+1)][y] = 'P';
        x++;
        Grid[(x-1)][y] = '-';
    }
    if(command=="right")
    {
        Grid[x][(y+1)] = 'P';
        y++;
        Grid[x][(y-1)] = '-';
    }
    if(command=="left")
    {
        Grid[x][(y-1)] = 'P';
        y--;
        Grid[x][(y+1)] = '-';
    }
    if(x==4 && y==4)
    {
        if(command=="down")
        {
            Grid[4][4] = 'P';
            cout<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl;
            cout<<Grid[0][0]<<Grid[0][1]<<Grid[0][2]<<Grid[0][3]<<Grid[0][4]<<endl<<
            Grid[1][0]<<Grid[1][1]<<Grid[1][2]<<Grid[1][3]<<Grid[1][4]<<endl<<
            Grid[2][0]<<Grid[2][1]<<Grid[2][2]<<Grid[2][3]<<Grid[2][4]<<endl<<
            Grid[3][0]<<Grid[3][1]<<Grid[3][2]<<Grid[3][3]<<Grid[3][4]<<endl<<
            Grid[4][0]<<Grid[4][1]<<Grid[4][2]<<Grid[4][3]<<Grid[4][4]<<endl;
            cout<<"You win!"<<endl;
        }
        if(command=="right")
        {
            Grid[4][4] = 'P';
            cout<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl<<endl;
            cout<<Grid[0][0]<<Grid[0][1]<<Grid[0][2]<<Grid[0][3]<<Grid[0][4]<<endl<<
            Grid[1][0]<<Grid[1][1]<<Grid[1][2]<<Grid[1][3]<<Grid[1][4]<<endl<<
            Grid[2][0]<<Grid[2][1]<<Grid[2][2]<<Grid[2][3]<<Grid[2][4]<<endl<<
            Grid[3][0]<<Grid[3][1]<<Grid[3][2]<<Grid[3][3]<<Grid[3][4]<<endl<<
            Grid[4][0]<<Grid[4][1]<<Grid[4][2]<<Grid[4][3]<<Grid[4][4]<<endl;
            cout<<"You win!"<<endl;
        }
    }
}
while(Grid[4][4] != 'P');
return 0;
}

我收到这些错误的行是第 45 行（和第 46 行，但这是相同的错误） 。 第 45 行代码如下：

srand(time(a));

Answer 1

为什么要这样使用时间？该函数声明如下：

time_t time ( time_t * timer )

向其传递一个 int 而不是一个指向 time_t 的指针肯定是一个错误，编译器会正确地抱怨。

假设您需要的只是 srand 的种子，只需使用 time(NULL) 即可。如果您确实需要在参数中设置该值，请将 a 和 b 变量设置为 time_t 类型，并使用 time(& a）。有关时间的更多信息，阅读本文。

Answer 2

我几乎确定您只想调用 srand(time(NULL)) 并只执行一次。它为伪随机数生成器设置种子，并且使用实际时间是最常见的情况（每次运行应用程序时都会不同，因此生成的数字的顺序是）。要获取实际时间，请调用 time(NULL);如果你想将时间值存储在特定的time_t变量中，你可以传递它的指针而不是NULL。

此代码：

srand(time(NULL));

和此代码：

time_t now;
time(&now);
srand(now);

甚至此代码：

time_t now;
srand(time(&now));

是等效的（出于 srand 目的）。