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 referenceand in function call:
some_funct(&buffer);更多推荐
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