将指针传递给函数[关闭](Passing pointer on pointers into function [closed])

将指针传递给函数[关闭](Passing pointer on pointers into function [closed])

无法找到答案。 当我有指针的指针

char **buffer;

我想把它传递给一个函数

void some_func(char **buffer) {}

所以在这个函数调用缓冲区之后将会包含来自这个函数的数据，我该如何调用这个函数呢？

Couldn't find answer to this. When I have pointer on pointers

char **buffer;

and I want to pass it to a function

void some_func(char **buffer) {}

so after this function call buffer will contain data from this function how should I call this function please ?

最满意答案

在你的例子中它是正确的，但事实是，你正在传递这个指针到指针的值，所以传递它的指针：

// initialize it char** buffer; void some_func(char ***buffer) {} // by reference

并在函数调用中：

some_funct(&buffer);

in your example it is correct but the fact is that you are passing this pointer to pointer by value so pass it by pointer:

// initialize it char** buffer; void some_func(char ***buffer) {} // by reference

and in function call:

some_funct(&buffer);