How to pass an std::string to glShaderSource?

Question

I have the following code:

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

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);

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.

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

Answer 1

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

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

Answer 2

The return value of std::string::c_str() is a pointer value (i.e., an address) to a static string array held inside the data-structures of the std::string object. Since the return value is just a temporary r-value (i.e., it's just a number stored in a CPU register), it is not an l-value and therefore it does not have a memory address you can actually take the address of and cast to a pointer-to-pointer. You first must save the return pointer value in a memory address. Memory-locations are l-values, and can have the address-of operator applied to them. So that is why your second method (or Dark Falcon's method) works, although keep in mind that the pointer value returned is a temporary, meaning that if you do any operations on the std::string object, it could invalidate the pointer since the std::string object internally manages the memory of its data-structures. So just because you've saved the return pointer value in a memory location doesn't mean that the pointer won't be invalidated at some later time, and at a point that you may not be capable of deterministically choosing.

Answer 3

You can get a reasonable-looking call by using a helper class. Define this class:

struct StringHelper {
  const char *p;
  StringHelper(const std::string& s) : p(s.c_str()) {}
  operator const char**() { return &p; }
};

Then, when you need to call glShaderSource, do it this way:

glShaderSource(shader, 1, StringHelper(data), NULL);

Answer 4

glShaderSource signature is, according to glShaderSource doc:

void glShaderSource(
    GLuint shader,
    GLsizei count,
    const GLchar** string,
    const GLint* length);

where string "Specifies an array of pointers to strings containing the source code to be loaded into the shader". What you're trying to pass is a pointer to a NULL terminated string (that is, a pointer to a const char*).

Unfortunately, I am not familiar with glShaderSource, but I can guess it's not expected a pointer to "some code" but something like this instead:

const char** options =
{
    "option1",
    "option2"
    // and so on
};

From opengl-redbook, you can read an example (I've shortened it in purpose):

const GLchar* shaderSrc[] = {
    "void main()",
    "{",
    "    gl_Position = gl_ModelViewProjectionMatrix * gl_Vertex;",
    "}"
};
shader = glCreateShader(GL_VERTEX_SHADER);
glShaderSource(shader, NumberOfLines(shaderSrc), shaderSrc, NULL);

Answer 5

I only want to point out that the pointer returned by c_str() is only valid as long as you don't do anything that requires reallocation of the internal buffer of std::string. That invalidates the pointer you got.

But since you really require a ** i would do this:

const char* mychararr[1] = {data.c_str()};
glShaderSource(shader, 1, mychararr, NULL);

That should work nicely as long as you don't leave the scope of mychararr.

Answer 6

Shader.cpp

#include "Shader.hpp"

Shader::Shader(GLenum type)
{
    this->_type = type;
}
Shader::~Shader() {}    

GLuint Shader::get(char* filename)
{
    GLuint shdr = glCreateShader(this->_type);
    FILE* f = 0;
    f = fopen(filename, "r+");
    char* str_tmp = 0;
    char** shdr_text = 0;
    shdr_text = (char**)malloc(sizeof(char**) * 255);
    str_tmp = (char*)malloc(sizeof(char*) * 255);
    int i = 0, ch = 0, n = 0;

    for(i = 0; i < 255; ++i){ *(shdr_text + i) = (char*)malloc(sizeof(char*) * 255); }

    i = 0;
    while((ch = fgetc(f)) != EOF)
    {
        sprintf(str_tmp, "%s%c", str_tmp, ch);
        if(ch == (int)'\n' || ch == (int)'\r')
        {
            sprintf(*(shdr_text + i), "%s", str_tmp);
            sprintf(str_tmp, "");
            ++i;
        }
    }

    free(str_tmp);
    fclose(f);

    glShaderSource(shdr, i, const_cast<const GLchar**>(shdr_text), 0);
    glCompileShader(shdr);

    free(shdr_text);

    return(shdr);
}

Shader.hpp

#ifndef SHADER_HPP
#define SHADER_HPP

#include <stdlib.h>
#include <stdio.h>
#include <GL/glew.h>
#include <GL/gl.h>

class Shader
{
    public:
        Shader(GLenum type);
        virtual ~Shader();

        GLuint get(char* filename);

    private:
        GLenum _type;
};

#endif

Answer 7

Try using the .c_str() it give you a char * that you can use as it worked for you b4

#include <string>

void ConversionSample ()
{
 std::string strTest ("This is a string");
 const char* pszConstString = strTest.c_str ();
 }