如何将 std::string 传递给 glShaderSource?

2021-12-19 00:00:00 opengl string char c++

我有以下代码:

glShaderSource(shader, 1, (const char **)data.c_str(), NULL);

但它使我的程序崩溃.如何将 std::string 转换为 const char ** ?我也试过 (const char **)& 但它说需要 l 值"，我不明白.当我使用此代码时它工作正常:

But it makes my program crash. How do I convert std::string into const char ** ? I also tried (const char **)& but it said "requires l-value" which I don't understand. It works fine when I use this code:

const char *data = "some code"; glShaderSource(shader, 1, &data, NULL);

但我不能直接从 std::string 让它工作.我可以为它分配一个新的 char 数组，但这不是很好的代码.

But I can't make it work directly from a std::string. I could allocate a new char array for it but that is not nice code.

我也尝试过 const GLchar 但显然它没有区别.

I also tried with const GLchar but obviously it makes no difference.

推荐答案

data.c_str() 返回一个 const char*，所以这样做:

data.c_str() returns a const char*, so do this:

const char *c_str = data.c_str(); glShaderSource(shader, 1, &c_str, NULL);

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