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OpenGL Creating Textures in OpenGL 4.x

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Coming from OpenGL ES 2.0, this was the typical way to create a texture:

GLuint texID = 0; // texture handle
uint8_t pixels[4] = { 0xFF, 0x00, 0xFF, 0x00 }; // 2x2 checkerboard texture

glActiveTexture(GL_TEXTURE0); // assuming you haven't already set this
glGenTextures(1, &texID); // get a free texture handle (aka, "name")
glBindTexture(GL_TEXTURE_2D, texID); // generate free texture name, and bind it
glTexImage(GL_TEXTURE_2D, 0, GL_ALPHA, 2, 2, 0, GL_ALPHA, GL_UNSIGNED_BYTE, pixels); // upload texture data to OpenGL's memory

It looks like OpenGL 4 still supports this way of doing things, but I've noticed that glTexStorage2D() is now used in the OpenGL book I've been reading. It's been stated in my book that this allows for generating mutable textures (format, dimensions, etc). It's a little simpler:

GLuint texID = 0;
uint8_t pixels[4] = { 0x00, 0xFF, 0x00, 0xFF };

glGenTextures(1, &texID);
glBindTexture(GL_TEXTURE_2D, &texID);
glTexStorage2D(GL_TEXTURE_2D, 4, GL_R8, 2, 2); // allocate storage for an 8-bit, 2x2 texture with 4 mipmap levels
glTexSubImage(GL_TEXTURE_2D, 0, 0, 0, 2, 2, GL_RED, GL_UNSIGNED_BYTE, pixels); // allocate texture data for the entire area of the texture for the first mipmap level

From what I've gathered, this is done for textures that are considered immutable in terms of changing its target, format, dimensions, etc. The data can change, however. As for generating 4 mipmaps... would I have to call glTexSubImage(), and provide 1/4-sized images for each of my mipmaps? It makes sense. I've seen somewhere that I only have to provide the initial mipmap, and I can call glGenerateMipmaps(), or for compatibility, render to FBOs to make this happen. Is this correct?

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