Why comment glBindFragDataLocation, the GL also works correctly? - opengl

Question

const GLchar* vertexSource1 = "#version 330 core\n"
"layout (location = 0) in vec2 position;\n"
"layout (location = 1) in vec3 color;\n"
"out vec3 Color;\n"
"void main()\n"
"{\n"
"gl_Position = vec4(position, 0.0, 1.0);\n"
"Color = color;\n"
"}\0";
const GLchar* fragmentSource1 = "#version 330 core\n"
" in vec3 Color;\n"
" out vec4 outColor;\n"
" void main()\n"
" {\n"
" outColor = vec4(Color, 1.0);\n"
" }\n";
GLuint shaderProgram1 = glCreateProgram();
glAttachShader(shaderProgram1, vertexShader1);
glAttachShader(shaderProgram1, fragmentShader1);
// glBindFragDataLocation(shaderProgram1, 0, "Color");
glLinkProgram(shaderProgram1);
Whether I add glBindFragDataLocation or not, the GL works correctly, Why?

Answer 1

Because you're "lucky". The OpenGL specification provides no guarantees about how fragment shader output locations are assigned if you don't assign them. It only says that each one will have a separate location; what locations those are is up to the implementation.
However, considering the sheer volume of code that writes to a single output variable without explicitly assigning it to a location, it's highly unlikely that an OpenGL implementation would ever assign the first FS output location to anything other than 0. So while it isn't a spec guarantee, at this point, it is a de-facto requirement of implementations.
Note: That doesn't mean you shouldn't assign that location manually. It's always best to be on the safe and explicit side.
FYI: layout(location) works for fragment shader outputs too. So you should use that if you're using it on vertex attributes. Then you don't have to worry about doing it from code.