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c++ - SDL_BlitSurface() 返回 -1 ... 为什么?

转载 作者:太空宇宙 更新时间:2023-11-04 14:17:04 24 4
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SDL 文档说 SDL_BlitSurface() 在不成功时返回 -1,但我找不到它失败的原因。

这是我的源代码:

主要.cpp

#include <iostream>
#include <string>
#include <SDL/SDL.h>
#include <SDL/SDL_image.h>
#include "window.cpp"
#include "player.cpp"

int init();
void quit();

int main(int argc,char *args[])
{
if (init() == -1)
{
std::cerr << "Error:init()" << std::endl;
return -1;
}
window main_window("Juego","img/bg.png",800,600,32);
player player1(500,500,0,0,5,5,"img/square.png");
while (!main_window.close)
{
main_window.handle_events();
if ( main_window.draw_background() != true)
{
std::cerr << "ERROR->main_window.draw_background()" << std::endl;
main_window.close = true;
}
player1.update(main_window.screen);
SDL_Flip(main_window.screen);
SDL_Delay(60/1000);
}
quit();
return 0;
}

int init()
{
if (SDL_Init(SDL_INIT_EVERYTHING) == -1)
{
std::cerr << "Error while initialising SDL" << std::endl;
return -1;
}
return 0;
}

void quit()
{
SDL_Quit();
}

窗口.cpp

class window
{
private:
void load_image(std::string source,SDL_Surface *destination);
SDL_Surface *window_bg;
SDL_Event event_queue;
SDL_Rect window_rect;
public:
window(std::string SCREEN_TITLE,std::string SCREEN_BG,int sw,int sh,int sbpp);
void handle_events();
bool draw_background();
SDL_Surface *screen;
bool close;
};

window::window(std::string SCREEN_TITLE,std::string SCREEN_BG,int sw,int sh,int sbpp)
{
window_bg = NULL;
screen = NULL;
close = false;
window_rect.x = 0;
window_rect.y = 0;
window_rect.w = sw;
window_rect.h = sh;
screen = SDL_SetVideoMode(sw,sh,sbpp,SDL_SWSURFACE);
std::cout << "Screen created." << std::endl;
SDL_WM_SetCaption(SCREEN_TITLE.c_str(),NULL);
std::cout << "Window title: " << SCREEN_TITLE << std::endl;
load_image(SCREEN_BG,window_bg);
}

void window::handle_events()
{
while (SDL_PollEvent(&event_queue))
{
if (event_queue.type == SDL_QUIT)
{
close = true;
}
}
}

void window::load_image(std::string source,SDL_Surface *destination)
{
SDL_Surface *tmp_img = IMG_Load(source.c_str());
if (tmp_img != NULL)
{
if ((destination = SDL_DisplayFormat(tmp_img)) == NULL)
{
std::cerr << "ERROR->SDL_DisplayFormat()" << std::endl;
}
std::cout << source << " Loaded." << std::endl;
SDL_FreeSurface(tmp_img);
}
else
{
std::cerr << "Could not load: " << source << std::endl;
}
}

bool window::draw_background()
{
int error_check = SDL_BlitSurface(window_bg,&window_rect,screen,NULL);
if (error_check != 0)
{
std::cerr << "SDL_BlitSurface() == " << error_check << std::endl;
return false;
}
return true;
}

错误在 window.cpp 中(我认为):

bool window::draw_background()
{
int error_check = SDL_BlitSurface(window_bg,NULL,screen,NULL);
if (error_check != 0)
{
std::cerr << "SDL_BlitSurface() == " << error_check << std::endl;
return false;
}
return true;
}

我的程序的输出是:

Screen created.
Window title: Juego
img/bg.png Loaded.
img/square.png Loaded
SDL_BlitSurface() == -1
ERROR->main_window.draw_background()

最佳答案

load_image(SCREEN_BG,window_bg); 在窗口构造函数中加载到一个临时的 SDL_Surface 并且从不填充 destination,我假设它是一个输出参数。那么当draw_background()使用window_bg时,还是NULL。

关于c++ - SDL_BlitSurface() 返回 -1 ... 为什么?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/10289948/

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