它可能是我的代码,但我可以通过命令行编译和运行该程序。在Windows中,exe文件也是可运行的。但是,我无法在任何unix系统(ubuntu或osx)上运行终端外的编译代码。我对C很新,所以任何帮助都会受到赞赏。谢谢!
澄清:
在Ubuntu中我运行
gcc game.c -o game
./game
然后它完美运行。但是,如果我通过GUI浏览游戏文件并双击它,它就不会做任何事情。
我以前在Windows上会调出一个命令来运行程序,就像从cmd
运行它一样。
代码(这是一个简单的猜数游戏):
#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int guessed = 0;
int guesses = 0;
int total_guesses = 0;
int games_played = 0;
int range_top = 10;
int generate_random_number()
{
srand(time(NULL));
return (rand() % range_top);
}
void take_guess(int num)
{
int guess;
printf("what is your guess: ");
scanf(" %i",&guess);
if(guess == num)
{
guessed = 1;
}
else if(guess>num)
{
printf("Your guess was too high,\n");
}
else
{
printf("Your guess was too low.\n");
}
}
void print_stats()
{
printf("\n\n\nGames Played: %i\nTotal Guesses: %i\nAverage Guesses Per Game: %f\n\n\n\n",games_played,total_guesses,(double)total_guesses/games_played);
int i = 5;
while(i>0)
{
printf("exiting in %is\n",i);
i--;
sleep(1);
}
}
int main(void)
{
printf("This is a game in C\n");
printf("Would you like to play? (y/n): ");
char play;
scanf("%c",&play);
while(play == 'y')
{
printf("I am thinking of a number between 0 and %i\n",range_top);
int num = generate_random_number();
while(guessed ==0)
{
take_guess(num);
guesses++;
}
games_played++;
total_guesses+=guesses;
printf("It took you %i guesses to win.\n",guesses);
printf("Would you like to play again? (y/n): ");
scanf(" %c",&play);
guessed = 0;
guesses = 0;
}
print_stats();
}
答案 0 :(得分:0)
但是,如果我通过GUI浏览游戏文件并双击它,它 什么都不做
你怎么知道它没有运行?我打赌它 运行,但由于它不是在任何终端(命令行窗口)的上下文中运行,你只是看不到它的输出。这就是Linux(以及Unix)的工作方式。
Windows区分GUI和命令行应用程序,在后一种情况下会自动带来控制台窗口。不幸的是(幸运的是?),在Unix中并非如此。
如果您想实现Windows行为,您可以:
Create a launcher for your application。这将允许您指定自定义图标等。
创建一个shell脚本,该脚本将调用teminal及其中的应用程序:
game.sh
:
#!/bin/bash
gnome-terminal -e "./game"
请注意,并非每个人都安装了gnome-temrinal
,因此您可能需要调整脚本以支持更多终端模拟器(xterm
,rxvt
,也许konsole
KDE用户)。