我一直试图使一个PNG图像出现在屏幕上的SDL窗口.我正在使用Eclipse。SDL.h和SDL_image.h似乎都有正确的链接,因为函数在编译器上弹出颜色。然而,当我运行我的代码时,实际上什么都没有发生。编译器中没有错误,没有注释,没有。窗户没有出现。如果有人能在这件事上帮我,我会非常感激的。
此外,SDL以前曾在我的计算机上工作过(不使用SDL_image) --在其中,我运行了一个非常好的粒子模拟。
我的代码:
#include <iostream>
#include "SDL.h"
#include "SDL_image.h"
using namespace std;
SDL_Window *m_window; //Window upon which the game will be displayed.
SDL_Renderer *m_renderer; //Renderer used to draw the objects on the window.
SDL_Texture *playerTex;
int SCREEN_WIDTH = 600;
int SCREEN_HEIGHT = 600;
int main(int argc, char* args[]) {
if (SDL_Init(SDL_INIT_VIDEO) < 0) {
cout << "Video init failed" << endl;
return 1;
}
//Creates the actual SDL-window and stores it in the m_window variable.
m_window = SDL_CreateWindow("Marko Beocanin SDD Project",
SDL_WINDOWPOS_UNDEFINED,
SDL_WINDOWPOS_UNDEFINED, SCREEN_WIDTH, SCREEN_HEIGHT,
SDL_WINDOW_FULLSCREEN);
//Error-checking method that determines if SDL could not create a window - returns false if unsuccessful.
if (m_window == NULL) {
cout << "Window Creation failed" << endl;
SDL_Quit();
IMG_Quit();
return 2;
}
//Creates an SDL-Renderer: a tool used to actually draw objects on the Window
m_renderer = SDL_CreateRenderer(m_window, -1, 0);
//Error-checking method that determines if SDL could not create a renderer - returns false if unsuccessful.
if (m_renderer == NULL) {
cout << "Renderer creation failed." << endl;
SDL_DestroyWindow(m_window);
SDL_Quit();
IMG_Quit();
return 3;
}
SDL_Surface *tmpSurface = IMG_Load("img.png");
playerTex = SDL_CreateTextureFromSurface(m_renderer, tmpSurface);
SDL_FreeSurface(tmpSurface);
SDL_RenderClear(m_renderer);
SDL_RenderCopy(m_renderer, playerTex, NULL, NULL);
SDL_RenderPresent(m_renderer);
SDL_Delay(2000);
SDL_DestroyWindow(m_window);
SDL_Quit();
IMG_Quit();
return 0;
}发布于 2018-04-28 12:33:36
我遇到的问题是我使用了错误的SDL_image库--我使用的是x64而不是x86,这意味着它本身没有抛出错误,只是不能正常工作!
https://stackoverflow.com/questions/50020199
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