C++ opengl2.1 中的着色器无法在 arch 中编译 linux

Question

我正在尝试使用图形 api opengl 2.1 在 C++ 中创建一个红色三角形，如下所示：

它仍然可以编译，但我的代码显示有错误，并且显示为白色，如下所示：

opengl 2.1 is supported!
error!

我的代码是：

#include <GL/gl.h>
#include <GL/glew.h>
#include <GL/glu.h>
#include <GLFW/glfw3.h>

#include <iostream>

void error_check() {
    GLenum error = glGetError();
    if (error != NULL) {
        std::cout << "error!" << std::endl;
    }
}
static unsigned int CompileShader(unsigned int type,
                                  const std::string& source) {
    unsigned int id = glCreateShader(type);
    const char* src = source.c_str();
    glShaderSource(id, 1, &src, nullptr);
    glCompileShader(id);

    return id;
}

static unsigned int CreateShader(const std::string& vertexshader,
                                 const std::string& fragmentshader) {
    unsigned int program = glCreateProgram();
    unsigned int vs = CompileShader(GL_VERTEX_SHADER, vertexshader);
    unsigned int fs = CompileShader(GL_FRAGMENT_SHADER, fragmentshader);

    glAttachShader(program, vs);
    glAttachShader(program, fs);
    glLinkProgram(program);

    return program;
}

int main() {
    glfwInit();
    GLFWwindow* engine_window =
        glfwCreateWindow(600, 600, "Senku-Engine 2", NULL, NULL);
    if (engine_window == NULL) {
        printf("window not opening!..");
        glfwTerminate();
        return -1;
    }

    glfwMakeContextCurrent(engine_window);
    glewExperimental = GL_TRUE;
    glewInit();
    if (GLEW_VERSION_2_1 == GL_TRUE) {
        std::cout << "opengl 2.1 is supported!" << std::endl;
    }

    float vertices[6] = {-0.5f, -0.5f, 0.0f, 0.5f, 0.5f, -0.5f};
    unsigned int buffer;
    glGenBuffers(1, &buffer);
    glBindBuffer(GL_ARRAY_BUFFER, buffer);
    glBufferData(GL_ARRAY_BUFFER, 6 * sizeof(float), vertices, GL_STATIC_DRAW);

    glEnableVertexAttribArray(0);
    glVertexAttribPointer(0, 2, GL_FLOAT, GL_FALSE, 2 * sizeof(float), nullptr);

    std::string vertexShader =
        "#version 120 core\n"
        "layout(location = 0) in vec4 position;\n"
        "void main()\n"
        "{\n"
        "   gl_Position = position;\n"
        "}\n";

    std::string fragmentShader =
        "#version 120 core\n"
        "layout(location = 0) out vec4 color;\n"
        "void main()\n"
        "{\n"
        "   color = vec4(1.0,0.0,0.0,1.0);\n"
        "}\n";
    unsigned int shader = CreateShader(vertexShader, fragmentShader);
    glUseProgram(shader);
    error_check();

    glViewport(0, 0, 600, 600);
    while (!glfwWindowShouldClose(engine_window)) {
        glDrawArrays(GL_TRIANGLES, 0, 3);

        glfwSwapBuffers(engine_window);
        glfwPollEvents();
    }

    glfwTerminate();
    return 0;
}

似乎我的代码在着色器源代码中仅使用一行代码时显示红色，但如果我使用多个字符串则不起作用。

Answer 1

纯 GLSL 120 实现可能不支持配置文件字符串 core，请使用

#version 120

in 和 out 是 GLSL 130，布局限定符是 GLSL 330。你需要

attribute vec4 position;

在片段着色器中，删除out变量并执行

gl_FragColor = vec4(1.0,0.0,0.0,1.0);

这些是最明显的问题。如果您还有其他错误，请使用 glGetShaderInfoLog 查询编译输出，它告诉您有关问题的信息。参见 Shader error handling。

C++ opengl2.1 中的着色器无法在 arch 中编译 linux

shader in C++ opengl2.1 doesn't compile in arch linux

c++

linux

opengl-2.0