预期 'const GLchar * const*' {aka 'const char * const*'} 但参数类型为 'char **'
expected 'const GLchar * const*' {aka 'const char * const*'} but argument is of type 'char **'
我试图将我的旧 openGl 代码(用 c++ 编写)转换为 c,当我尝试使用 glCompileShader() 时,我总是遇到关于不兼容指针类型的奇怪警告
char *buffer = 0;
long length;
FILE *f = fopen(vertexPath, "rb");
if(f) {
fseek(f, 0, SEEK_END);
length = ftell(f);
fseek(f, 0, SEEK_SET);
buffer = malloc(length);
if(buffer) {
fread(buffer, 1, length, f);
}
fclose(f);
}
if(buffer) {
unsigned int vertex;
int success;
vertex = glCreateShader(GL_VERTEX_SHADER);
glShaderSource(vertex, 1, &buffer, NULL);
glCompileShader(vertex);
glGetShaderiv(vertex, GL_COMPILE_STATUS, &success);
if(!success) {
char *infoLog;
infoLog = malloc(512);
glGetShaderInfoLog(vertex, 512, NULL, infoLog);
printf("%s%s", "Error Fragment\n", infoLog);
free(infoLog);
}
}
这是我收到的警告消息:
..\src.\shader.c: In function 'compileShader':
..\src.\shader.c:29:35: warning: passing argument 3 of 'glad_glShaderSource' from
incompatible pointer type [-Wincompatible-pointer-types]
29 | glShaderSource(vertex, 1, &buffer, NULL);
| ^~~~~~~
| |
| char **
..\src.\shader.c:29:35: note: expected 'const GLchar * const*' {aka 'const char *
const*'} but argument is of type 'char **'
我不知道这个警告是什么意思,是否重要,或者我是否可以忽略它。
谢谢 4 帮助。
I don't know what this warning means, if it is important, or if I can ignore it.
这意味着 &buffer
是一个 char **
(指向 char 指针的指针)但是该函数需要一个 char const * const *
(指向 const 指向 const char 的指针)。
基本上就是说glShaderSource
不能修改指针或它指向的缓冲区。
您可以通过强制转换来匹配签名来解决它:
(char const * const *)&buffer
我试图将我的旧 openGl 代码(用 c++ 编写)转换为 c,当我尝试使用 glCompileShader() 时,我总是遇到关于不兼容指针类型的奇怪警告
char *buffer = 0;
long length;
FILE *f = fopen(vertexPath, "rb");
if(f) {
fseek(f, 0, SEEK_END);
length = ftell(f);
fseek(f, 0, SEEK_SET);
buffer = malloc(length);
if(buffer) {
fread(buffer, 1, length, f);
}
fclose(f);
}
if(buffer) {
unsigned int vertex;
int success;
vertex = glCreateShader(GL_VERTEX_SHADER);
glShaderSource(vertex, 1, &buffer, NULL);
glCompileShader(vertex);
glGetShaderiv(vertex, GL_COMPILE_STATUS, &success);
if(!success) {
char *infoLog;
infoLog = malloc(512);
glGetShaderInfoLog(vertex, 512, NULL, infoLog);
printf("%s%s", "Error Fragment\n", infoLog);
free(infoLog);
}
}
这是我收到的警告消息:
..\src.\shader.c: In function 'compileShader':
..\src.\shader.c:29:35: warning: passing argument 3 of 'glad_glShaderSource' from
incompatible pointer type [-Wincompatible-pointer-types]
29 | glShaderSource(vertex, 1, &buffer, NULL);
| ^~~~~~~
| |
| char **
..\src.\shader.c:29:35: note: expected 'const GLchar * const*' {aka 'const char *
const*'} but argument is of type 'char **'
我不知道这个警告是什么意思,是否重要,或者我是否可以忽略它。 谢谢 4 帮助。
I don't know what this warning means, if it is important, or if I can ignore it.
这意味着 &buffer
是一个 char **
(指向 char 指针的指针)但是该函数需要一个 char const * const *
(指向 const 指向 const char 的指针)。
基本上就是说glShaderSource
不能修改指针或它指向的缓冲区。
您可以通过强制转换来匹配签名来解决它:
(char const * const *)&buffer