openGL, GLFW, 1281错误在glShaderSource上,下一个glShaderSource工作得很好

openGL, GLFW, 1281 error on glShaderSource, the very next glShaderSource works fine somehow

本文关键字:glShaderSource 下一个 工作 很好 GLFW 1281 错误 openGL      更新时间:2023-10-16

我一直在尝试让着色器工作。他们没有。

std::cout << "a " << glGetError() <<std::endl; //making sure things work
GLuint vertexShader = 0;
GLuint fragmentShader = 1;
glCreateShader(GL_VERTEX_SHADER);
std::cout <<"b " << glGetError() <<std::endl;
glCreateShader(GL_FRAGMENT_SHADER);
std::cout <<"c " << glGetError() <<std::endl;
glShaderSource(vertexShader, GLsizei(1), (const char**)&VertexShaderSource, NULL);//ERROR
std::cout <<"d " << glGetError() <<std::endl;
glShaderSource(fragmentShader, GLsizei(1), (const char**)&FragmentShaderSource, NULL);//is somehow fine
std::cout <<"e " << glGetError() <<std::endl;
glCompileShader(vertexShader);
std::cout <<"f " << glGetError() <<std::endl;
glCompileShader(fragmentShader);
std::cout <<"g " << glGetError() <<std::endl;'

现在输出为:

a 0
b 0
c 0
d 1281
e 0
f 1281
g 0

这意味着第一个glShaderSource调用不工作,但是,由于某种原因,下一个glShaderSource工作,同时做我认为完全相同的事情。这是怎么回事?

着色器:

const std::string VertexShaderSource ="#version 330 core"
                                  ""
                                  "layout (location = 0) in vec3 position;"
                                  ""
                                  "void main()"
                                  "{"
                                  "    gl_Position = vec4(position.x, position.y, position.z, 1.0);"
                                  "}";
const std::string FragmentShaderSource ="#version 330 core"
                                    ""
                                    "out vec4 color;"
                                    ""
                                    "void main()"
                                    "{"
                                    "    color = vec4(1.0f, 1.0f, 1.0f, 1.0f);"
                                    "}";

如果以这种方式传递字符串((const char**)&VertexShaderSource)是错误的,那么为什么第二个调用工作得很好?

您必须将vertexShaderfragmentShader设置为glCreateShader的结果,而不是将其硬编码为01

GLuint vertexShader;
GLuint fragmentShader;
vertexShader = glCreateShader(GL_VERTEX_SHADER);
std::cout <<"b " << glGetError() <<std::endl;
fragmentShader = glCreateShader(GL_FRAGMENT_SHADER);
std::cout <<"c " << glGetError() <<std::endl;

函数glCreateShader如果调用失败,返回0

在你的代码中,你将vertexShader硬编码为0, opengl报告glShaderSource(vertexShader, ...错误,因为你将0作为id传递给glShaderSource

1281是一个GL_INVALID_VALUE错误,报告if shader is not a value generated by OpenGL. glShaderSource: Errors

除此之外,你(const char**)&VertexShaderSource是错误的,得到一个std::string存储数据的内存地址,你必须使用.c_str():如何传递一个std::字符串到glShaderSource?

您错误地传递了字符串,正确的方式是VertexShaderSource.c_str()