【问题标题】:Syntax error in OpenGl shader that is copied from YouTube从 YouTube 复制的 OpenGL 着色器中的语法错误
【发布时间】:2020-05-07 21:53:17
【问题描述】:

我尝试学习 OpenGL,现在我想在黑色背景上用顶点和着色器制作一个简单的红色三角形。我复制了代码from that YouTube video. 我已经更改了视图。这是我的代码:

#include "prec.h"
struct Vector2
{
    float x, y;
};

struct TrianglePos
{
    Vector2 a, b, c;
};

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);

    int result;
    glGetShaderiv(id, GL_COMPILE_STATUS, &result);
    if (!result)
    {
        int length;
        glGetShaderiv(id, GL_INFO_LOG_LENGTH, &length);
        char* message = (char*)alloca(length * sizeof(char));
        glGetShaderInfoLog(id, length, &length, message);
        std::cout << "Failed to compile shader, message: " << message << std::endl;
        glDeleteShader(id);
        return 0;
    }

    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);
    glValidateProgram(program);

    glDeleteShader(vs);
    glDeleteShader(fs);

    return program;
}

int main() {

    if (!glfwInit())
        return -1;


    GLFWwindow *window = glfwCreateWindow(640, 480, "Sand Box GL", NULL, NULL);

    if (!window) {
        std::cout << "problems with window" << std::endl;
        glfwTerminate();
        return -1;
    }

    glfwMakeContextCurrent(window);

    glewExperimental = GL_TRUE;
    if (GLEW_OK != glewInit()) {
        std::cout << "something with glew went wrong" << std::endl;
        return -1;
    }


    TrianglePos trianglePos = {
            -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, sizeof(trianglePos), &trianglePos, GL_STATIC_DRAW);

    glEnableVertexAttribArray(0);
    glVertexAttribPointer(0, 2, GL_FLOAT, GL_FALSE, sizeof(Vector2), 0);

    std::string vertexShader =
            "#version 400 core\n"
            "\n"
            "layout(location = 0) in vec4 position;\n"
            "\n"
            "void main() \n"
            "{\n"
            "   gl_Position = position;\n"
            "}\n";

    std::string fragmentShader =
            "#version 400 core\n"
            "\n"
            "layout(location = 0) out vec4 color;\n"
            "\n"
            "void main()\n"
            "{\n"
            "   color = vec4(1.0, 0.0, 0.0, 1.0);\n"
            "}\n";

    const char *versionGL;
    versionGL = (char *) (glGetString(GL_VERSION));
    std::cout << "openGl version: " << versionGL << std::endl;

    if(GL_VERSION_4_0)
        std::cout << "opengl 4.0 supported" << std::endl;

    unsigned int shader = createShader(vertexShader, fragmentShader);
    glUseProgram(shader);

    while (!glfwWindowShouldClose(window)) {
        glClear(GL_COLOR_BUFFER_BIT);

        glDrawArrays(GL_TRIANGLES, 0, 3);

        glfwSwapBuffers(window);

        glfwPollEvents();
    }
    glfwTerminate();
    return 0;
}

这是预编译的头文件(prec.h):#pragma once

#ifndef GLEWINIT_PREC_H
#define GLEWINIT_PREC_H
#endif //GLEWINIT_PREC_H

#include "GL/glew.h"
#include <GLFW/glfw3.h>

#include <iostream>

程序将其打印到控制台(“OpenGL 4.0 supported”表示GL_VERSION_4_0 == true):

openGl version: 2.1 INTEL-14.5.22 
opengl 4.0 supported

当我尝试运行它时,我从顶点和片段着色器的着色器编译器中收到以下错误消息:

ERROR: 0:1: '' :  version '400' is not supported
ERROR: 0:1: '' : syntax error: #version
ERROR: 0:3: 'layout' : syntax error: syntax error

当我将#version 400 core 更改为#version 120 时,我只收到layout 的语法错误。因此,我认为我用 glew 搞砸了一些事情。我可以尝试什么?

【问题讨论】:

  • 我不使用 GLEW,但我敢打赌GL_VERSION_4_0 是一个编译时定义,它告诉 GLEW 将尝试加载哪个版本。它没有说明硬件实际支持的 OpenGL 版本。所以可能 OpenGL 2.1 是您的硬件支持的最佳版本。
  • 由于语法错误,我该怎么办?
  • 按老方法做,没有layout 限定符。

标签: c++ opengl glsl shader glew


【解决方案1】:

您的系统不支持 OpenGL 4.0。它只支持 OpenGL 2.1。查看输出

openGl version: 2.1 INTEL-14.5.22 

GLSL 1.20 对应 OpenGL 2.1。降级着色器:

顶点着色器

#version 120

attribute vec4 position;

void main()
{
    gl_Position = position;
}

片段着色器

#version 120

void main()
{
    gl_FragColor = vec4(1.0, 0.0, 0.0, 1.0);
}

分别使用Raw string literalsC++11):

std::string vertexShader =
R"(#version 120

attribute vec4 position;

void main()
{
    gl_Position = position;
}
)";

std::string fragmentShader =
R"(#version 120

void main()
{
    gl_FragColor = vec4(1.0, 0.0, 0.0, 1.0);
}
)";

【讨论】:

  • 对我来说,原始字符串不起作用。但是使用正常的字符串它可以正常工作,非常感谢 Rabbid76!
  • @GianLaager 谢谢。别客气。自 C++11 起支持原始字符串文字
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